id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001301 | Number Theory: Congruences — Efficient Exponentiation | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $68^193$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency c... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 8.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001302 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve with verification: Compute the remainder when $83^25$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{43}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001303 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Proceed methodically: Compute the remainder when $56^195$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 re... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{76}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 76.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{76}$.) |
math-001304 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Warm-up: Compute the remainder when $51^131$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{11}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{11}$.) |
math-001305 | Number Theory: Residues — Last Digits / Two Digits | 1 | Derive the result step-by-step: Compute the remainder when $73^158$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001306 | Number Theory: Congruences — Efficient Exponentiation | 1 | State any required conditions first: Compute the remainder when $98^215$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001307 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Explain why your operations are valid: Compute the remainder when $3^82$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001308 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Where appropriate, name the theorem you use: Compute the remainder when $73^100$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pa... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001309 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Prompt: Compute the remainder when $49^157$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001310 | Number Theory: Congruences — Efficient Exponentiation | 1 | Give reasoning, not just computation: Compute the remainder when $48^139$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{12}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 12."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001311 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Where appropriate, name the theorem you use: Compute the remainder when $56^158$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pa... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001312 | Number Theory: Congruences — Efficient Exponentiation | 1 | Compute the requested quantity: Compute the remainder when $25^250$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{25}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{25}$.) |
math-001313 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Track quantifiers carefully: Compute the remainder when $52^127$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mo... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{8}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 8.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001314 | Number Theory: Residues — Last Digits / Two Digits | 1 | Keep the final answer in boxed form: Compute the remainder when $38^98$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{4}$.) |
math-001315 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give reasoning, not just computation: Compute the remainder when $9^8$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001316 | Number Theory: Congruences — Efficient Exponentiation | 1 | Task: Compute the remainder when $66^65$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your wo... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001317 | Number Theory: Residues — Last Digits / Two Digits | 1 | Where appropriate, name the theorem you use: Compute the remainder when $46^188$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., p... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{36}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 36.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001318 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Explain why your operations are valid: Compute the remainder when $69^172$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{61}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001319 | Number Theory: Residues — Last Digits / Two Digits | 1 | Solve and sanity-check: Compute the remainder when $60^249$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 r... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001320 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Write the solution set clearly: Compute the remainder when $98^68$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001321 | Number Theory: Congruences — Efficient Exponentiation | 1 | State any required conditions first: Compute the remainder when $34^127$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{44}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 44.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{44}$.) |
math-001322 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Task: Compute the remainder when $29^98$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your wo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001323 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Explain why your operations are valid: Compute the remainder when $31^37$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{11}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{11}$.) |
math-001324 | Number Theory: Congruences — Efficient Exponentiation | 1 | Be explicit about assumptions: Compute the remainder when $8^163$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{12}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{12}$.) |
math-001325 | Number Theory: Residues — Last Digits / Two Digits | 1 | Problem: Compute the remainder when $2^47$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{8}$.) |
math-001326 | Number Theory: Congruences — Efficient Exponentiation | 1 | Complete the analysis: Compute the remainder when $50^145$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 re... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001327 | Number Theory: Congruences — Efficient Exponentiation | 1 | Exercise: Compute the remainder when $44^77$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001328 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Show all reasoning: Compute the remainder when $47^131$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reas... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 3.",... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001329 | Number Theory: Residues — Last Digits / Two Digits | 1 | Compute the requested quantity: Compute the remainder when $26^232$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001330 | Number Theory: Residues — Last Digits / Two Digits | 1 | Use two approaches if possible: Compute the remainder when $15^143$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{75}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001331 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Checkpoint: Compute the remainder when $18^43$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Y... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 2.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{2}$.) |
math-001332 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Give reasoning, not just computation: Compute the remainder when $96^144$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity o... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{56}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001333 | Number Theory: Residues — Last Digits / Two Digits | 1 | Complete the analysis: Compute the remainder when $18^24$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.",
"r... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001334 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve (and briefly cross-validate): Compute the remainder when $47^246$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or m... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001335 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Do not skip justification steps: Compute the remainder when $92^125$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{12}$.) |
math-001336 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a theorem-based solution: Compute the remainder when $45^65$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 5.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001337 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Challenge: Compute the remainder when $75^25$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Yo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{15}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{15}$.) |
math-001338 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Exercise: Compute the remainder when $3^225$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 3.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{3}$.) |
math-001339 | Number Theory: Residues — Last Digits / Two Digits | 1 | Proceed methodically: Compute the remainder when $25^122$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001340 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $44^148$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001341 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $32^118$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency che... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{4}$.) |
math-001342 | Number Theory: Residues — Last Digits / Two Digits | 1 | Track units/moduli carefully: Compute the remainder when $44^131$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 4.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001343 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Work carefully and justify each inference: Compute the remainder when $97^250$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., par... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{49}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001344 | Number Theory: Residues — Last Digits / Two Digits | 1 | Explain each transformation: Compute the remainder when $90^186$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001345 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Carefully track domains: Compute the remainder when $74^110$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001346 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Track quantifiers carefully: Compute the remainder when $78^103$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{12}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{12}$.) |
math-001347 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a fully justified solution: Compute the remainder when $13^205$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 3.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001348 | Number Theory: Congruences — Efficient Exponentiation | 1 | Warm-up: Compute the remainder when $83^119$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 7.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001349 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Work carefully and justify each inference: Compute the remainder when $69^88$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pari... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{41}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 41.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{41}$.) |
math-001350 | Number Theory: Congruences — Efficient Exponentiation | 1 | Track quantifiers carefully: Compute the remainder when $40^200$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001351 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | State any required conditions first: Compute the remainder when $69^90$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001352 | Number Theory: Residues — Last Digits / Two Digits | 1 | Where appropriate, name the theorem you use: Compute the remainder when $37^88$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pa... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{21}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 21.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{21}$.) |
math-001353 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give an answer and a quick verification: Compute the remainder when $43^82$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{49}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 49.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{49}$.) |
math-001354 | Number Theory: Residues — Last Digits / Two Digits | 1 | Explain what is being counted/optimized: Compute the remainder when $80^102$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001355 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give reasoning, not just computation: Compute the remainder when $64^186$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001356 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give an answer and a quick verification: Compute the remainder when $2^217$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{72}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001357 | Number Theory: Residues — Last Digits / Two Digits | 1 | Work carefully and justify each inference: Compute the remainder when $54^212$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pari... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001358 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Indicate where a theorem is used: Compute the remainder when $25^63$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 25.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{25}$.) |
math-001359 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a fully justified solution: Compute the remainder when $93^42$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{49}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 49.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{49}$.) |
math-001360 | Number Theory: Congruences — Efficient Exponentiation | 1 | Question: Compute the remainder when $97^220$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Y... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001361 | Number Theory: Congruences — Efficient Exponentiation | 1 | Exercise: Compute the remainder when $99^85$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001362 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Carefully track domains: Compute the remainder when $31^5$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 re... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{11}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 11.",
"r... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{11}$.) |
math-001363 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Use two approaches if possible: Compute the remainder when $11^129$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{11}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001364 | Number Theory: Residues — Last Digits / Two Digits | 1 | Indicate where a theorem is used: Compute the remainder when $30^142$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001365 | Number Theory: Congruences — Efficient Exponentiation | 1 | Work this out carefully: Compute the remainder when $86^223$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001366 | Number Theory: Congruences — Efficient Exponentiation | 1 | State any required conditions first: Compute the remainder when $95^117$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{75}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001367 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Determine the requested value: Compute the remainder when $36^114$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001368 | Number Theory: Residues — Last Digits / Two Digits | 1 | Give reasoning, not just computation: Compute the remainder when $25^49$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001369 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Work carefully and justify each inference: Compute the remainder when $96^207$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pari... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001370 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | State any required conditions first: Compute the remainder when $56^40$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or m... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001371 | Number Theory: Residues — Last Digits / Two Digits | 1 | Find the exact value: Compute the remainder when $75^98$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reas... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{5}$.) |
math-001372 | Number Theory: Residues — Last Digits / Two Digits | 1 | Show all reasoning: Compute the remainder when $73^73$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reaso... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{33}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{33}$.) |
math-001373 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Proceed methodically: Compute the remainder when $88^152$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001374 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Track units/moduli carefully: Compute the remainder when $30^63$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001375 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give a fully justified solution: Compute the remainder when $77^217$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{17}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 17."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001376 | Number Theory: Residues — Last Digits / Two Digits | 1 | Solve and then verify: Compute the remainder when $60^43$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001377 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide a rigorous solution: Compute the remainder when $81^9$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001378 | Number Theory: Congruences — Efficient Exponentiation | 1 | Give a fully justified solution: Compute the remainder when $30^24$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001379 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Warm-up: Compute the remainder when $88^224$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.",
"r... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001380 | Number Theory: Congruences — Efficient Exponentiation | 1 | Start by stating any domain restrictions: Compute the remainder when $58^67$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{2}$.) |
math-001381 | Number Theory: Congruences — Efficient Exponentiation | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $94^148$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency che... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001382 | Number Theory: Residues — Last Digits / Two Digits | 1 | Find the exact value: Compute the remainder when $66^108$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001383 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $10^189$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency c... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001384 | Number Theory: Congruences — Efficient Exponentiation | 1 | Work this out carefully: Compute the remainder when $90^37$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 r... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001385 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Task: Compute the remainder when $76^224$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your w... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001386 | Number Theory: Congruences — Efficient Exponentiation | 1 | Problem: Compute the remainder when $90^113$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001387 | Number Theory: Congruences — Efficient Exponentiation | 1 | Prompt: Compute the remainder when $95^178$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{5}$.) |
math-001388 | Number Theory: Residues — Last Digits / Two Digits | 1 | Solve and include a self-check: Compute the remainder when $14^103$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{44}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 44.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{44}$.) |
math-001389 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Exercise: Compute the remainder when $90^224$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Yo... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001390 | Number Theory: Residues — Last Digits / Two Digits | 1 | Challenge: Compute the remainder when $92^158$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Y... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001391 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Start by stating any domain restrictions: Compute the remainder when $76^199$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parit... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001392 | Number Theory: Congruences — Efficient Exponentiation | 1 | Answer using clear logical steps: Compute the remainder when $10^106$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001393 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Question: Compute the remainder when $43^22$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001394 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Task: Compute the remainder when $81^77$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your wo... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001395 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Keep the final answer in boxed form: Compute the remainder when $72^244$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or ... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001396 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve and sanity-check: Compute the remainder when $91^29$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 re... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001397 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Start by stating any domain restrictions: Compute the remainder when $30^248$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parit... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001398 | Number Theory: Congruences — Efficient Exponentiation | 1 | Compute the requested quantity: Compute the remainder when $6^148$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001399 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Answer with a short justification: Compute the remainder when $74^246$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001400 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Carefully track domains: Compute the remainder when $34^248$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16."... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
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