id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001101 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Problem: Compute the remainder when $75^77$ 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... | [
{
"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: $\\boxed{15}$.\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. Both yield remainder 15."... | [
{
"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-001102 | Number Theory: Residues — Last Digits / Two Digits | 1 | Start by stating any domain restrictions: Compute the remainder when $4^239$ 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": "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: $\\boxed{4}$.\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. Both yield remainder 4.",
"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. (Here the result is $\boxed{4}$.) |
math-001103 | Number Theory: Congruences — Efficient Exponentiation | 1 | Compute the requested quantity: Compute the remainder when $64^187$ 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: $\\boxed{4}$.\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. 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 ... | 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-001104 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Do not skip justification steps: Compute the remainder when $52^27$ 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": "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 ... | 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-001105 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Warm-up: Compute the remainder when $5^91$ 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-001106 | Number Theory: Congruences — Efficient Exponentiation | 1 | Be explicit about assumptions: Compute the remainder when $96^36$ 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": "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-001107 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Track quantifiers carefully: Compute the remainder when $38^47$ 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/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 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{92}$.\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 92.... | [
{
"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 $\boxed{92}$.) |
math-001108 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Explain what is being counted/optimized: Compute the remainder when $36^204$ 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., parit... | [
{
"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: $\\boxed{16}$.\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 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 $\boxed{16}$.) |
math-001109 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Checkpoint: Compute the remainder when $79^26$ 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).
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 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\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 ... | 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-001110 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide a rigorous solution: Compute the remainder when $55^248$ 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/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": "Both approaches agree after simplification. 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 yield remainder 5.",
"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-001111 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Explain why your operations are valid: Compute the remainder when $44^167$ 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 o... | [
{
"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-001112 | Number Theory: Congruences — Efficient Exponentiation | 1 | Where appropriate, name the theorem you use: Compute the remainder when $40^66$ 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., par... | [
{
"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-001113 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Complete the analysis: Compute the remainder when $25^124$ 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": "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 ... | 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 $\boxed{5}$.) |
math-001114 | Number Theory: Residues — Last Digits / Two Digits | 1 | Use two approaches if possible: Compute the remainder when $37^166$ 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": "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 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. |
math-001115 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Where appropriate, name the theorem you use: Compute the remainder when $22^119$ 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": "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 ... | 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-001116 | Number Theory: Congruences — Efficient Exponentiation | 1 | Try to avoid pattern-matching; explain why: Compute the remainder when $53^72$ 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., 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 10, a^2\\bmod 10... | {
"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 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 $\boxed{1}$.) |
math-001117 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Show all reasoning: Compute the remainder when $43^97$ 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 reason... | [
{
"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: $\\boxed{3}$.\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 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. (Here the result is $\boxed{3}$.) |
math-001118 | Number Theory: Residues — Last Digits / Two Digits | 1 | Use two approaches if possible: Compute the remainder when $57^60$ 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": "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 ... | 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-001119 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Answer with a short justification: Compute the remainder when $55^39$ 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: $\\boxed{15}$.\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. Both yield remainder 15."... | [
{
"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 $\boxed{15}$.) |
math-001120 | Number Theory: Congruences — Efficient Exponentiation | 1 | Explain what is being counted/optimized: Compute the remainder when $89^242$ 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., parit... | [
{
"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: $\\boxed{21}$.\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... | [
{
"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-001121 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Do not skip justification steps: Compute the remainder when $90^54$ 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": "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 ... | 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-001122 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give reasoning, not just computation: Compute the remainder when $45^46$ 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": "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-001123 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Provide a rigorous solution: Compute the remainder when $54^245$ 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": "Both approaches agree after simplification. 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. Both yield remainder 4.",
"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-001124 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give an answer and a quick verification: Compute the remainder when $23^167$ 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., 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 100, a^2\\bmod 1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{47}$.\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 47.",
"... | [
{
"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 $\boxed{47}$.) |
math-001125 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve with verification: Compute the remainder when $62^128$ 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-001126 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Write the solution set clearly: Compute the remainder when $79^228$ 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: $\\boxed{1}$.\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 ... | 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-001127 | Number Theory: Residues — Last Digits / Two Digits | 1 | Prompt: Compute the remainder when $43^86$ 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": "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 ... | 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-001128 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Use two approaches if possible: Compute the remainder when $46^96$ 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": "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 ... | 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-001129 | Number Theory: Residues — Last Digits / Two Digits | 1 | Challenge: Compute the remainder when $18^230$ 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).
... | [
{
"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: $\\boxed{24}$.\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 24.",
"... | [
{
"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 $\boxed{24}$.) |
math-001130 | Number Theory: Residues — Last Digits / Two Digits | 1 | Exercise: Compute the remainder when $13^166$ 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": "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 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 ... | 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-001131 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve with verification: Compute the remainder when $61^19$ 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": "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 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-001132 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Proceed methodically: Compute the remainder when $89^96$ 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 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 10, a^2\\bmod 10... | {
"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. (Here the result is $\boxed{1}$.) |
math-001133 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Problem: Compute the remainder when $7^164$ 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": "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 ... | 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 $\boxed{1}$.) |
math-001134 | Number Theory: Residues — Last Digits / Two Digits | 1 | Keep the final answer in boxed form: Compute the remainder when $57^243$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{13}$.\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 13.",
"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 $\boxed{13}$.) |
math-001135 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve with verification: Compute the remainder when $43^136$ 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": "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: $\\boxed{1}$.\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 ... | 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 $\boxed{1}$.) |
math-001136 | Number Theory: Congruences — Efficient Exponentiation | 1 | Keep the final answer in boxed form: Compute the remainder when $45^78$ 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 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 ... | 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 $\boxed{5}$.) |
math-001137 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide a rigorous solution: Compute the remainder when $36^205$ 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 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 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. |
math-001138 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Track quantifiers carefully: Compute the remainder when $20^76$ 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": "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-001139 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve with verification: Compute the remainder when $19^112$ 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 ... | [
{
"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: $\\boxed{1}$.\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 ... | 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-001140 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Show all reasoning: Compute the remainder when $15^115$ 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 reaso... | [
{
"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: $\\boxed{15}$.\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-001141 | Number Theory: Residues — Last Digits / Two Digits | 1 | Carefully track domains: Compute the remainder when $9^154$ 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": "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: $\\boxed{1}$.\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 ... | 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 $\boxed{1}$.) |
math-001142 | Number Theory: Congruences — Efficient Exponentiation | 1 | Work carefully and justify each inference: Compute the remainder when $48^175$ 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": "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 ... | 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-001143 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Indicate where a theorem is used: Compute the remainder when $36^196$ 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": "Cross-check: both derivations land on the same invariant quantity. 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 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 $\boxed{36}$.) |
math-001144 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a fully justified solution: Compute the remainder when $77^154$ 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": "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 ... | 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-001145 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Proceed methodically: Compute the remainder when $2^69$ 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 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 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-001146 | Number Theory: Residues — Last Digits / Two Digits | 1 | Be explicit about assumptions: Compute the remainder when $88^66$ 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": "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: $\\boxed{84}$.\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 $\boxed{84}$.) |
math-001147 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Answer with a short justification: Compute the remainder when $80^123$ 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 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": "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 ... | 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-001148 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Use two approaches if possible: Compute the remainder when $11^46$ 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: $\\boxed{1}$.\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 ... | 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-001149 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Work carefully and justify each inference: Compute the remainder when $98^65$ 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": "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: $\\boxed{68}$.\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 68.",
"... | [
{
"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 $\boxed{68}$.) |
math-001150 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Solve (and briefly cross-validate): Compute the remainder when $98^131$ 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 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": "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 ... | 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 $\boxed{2}$.) |
math-001151 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a fully justified solution: Compute the remainder when $30^116$ 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": "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: $\\boxed{0}$.\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-001152 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Where appropriate, name the theorem you use: Compute the remainder when $18^230$ 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": "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: $\\boxed{4}$.\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-001153 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give an answer and a quick verification: Compute the remainder when $10^226$ 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., parit... | [
{
"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: $\\boxed{0}$.\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.",
"ro... | [
{
"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-001154 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Checkpoint: Compute the remainder when $74^3$ 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": "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: $\\boxed{4}$.\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. 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 $\boxed{4}$.) |
math-001155 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Exercise: Compute the remainder when $31^191$ 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": "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: $\\boxed{1}$.\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. ... | [
{
"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-001156 | Number Theory: Residues — Last Digits / Two Digits | 1 | Give reasoning, not just computation: Compute the remainder when $92^244$ 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": "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: $\\boxed{16}$.\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 ... | 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-001157 | Number Theory: Residues — Last Digits / Two Digits | 1 | Proceed methodically: Compute the remainder when $57^57$ 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 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 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: $\\boxed{7}$.\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.",
... | [
{
"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 $\boxed{7}$.) |
math-001158 | Number Theory: Congruences — Efficient Exponentiation | 1 | Explain what is being counted/optimized: Compute the remainder when $24^33$ 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": "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 ... | 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 $\boxed{4}$.) |
math-001159 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Solve and justify each step: Compute the remainder when $80^57$ 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/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": "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 ... | 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-001160 | Number Theory: Residues — Last Digits / Two Digits | 1 | Work this out carefully: Compute the remainder when $99^41$ 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": "Both approaches agree after simplification. 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 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 ... | 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-001161 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Checkpoint: Compute the remainder when $10^152$ 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 reasoning).
... | [
{
"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-001162 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Proceed methodically: Compute the remainder when $63^130$ 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": "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 ... | 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-001163 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve and sanity-check: Compute the remainder when $68^39$ 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 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 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{32}$.\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 32.",
"... | [
{
"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 $\boxed{32}$.) |
math-001164 | Number Theory: Residues — Last Digits / Two Digits | 1 | Explain what is being counted/optimized: Compute the remainder when $82^134$ 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., parit... | [
{
"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: $\\boxed{24}$.\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 $\boxed{24}$.) |
math-001165 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Provide a rigorous solution: Compute the remainder when $74^166$ 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/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 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 ... | 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-001166 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Task: Compute the remainder when $82^13$ 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": "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 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 $\boxed{12}$.) |
math-001167 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Answer using clear logical steps: Compute the remainder when $48^206$ 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": "Both approaches agree after simplification. 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. Both yield remainder 4.",
"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 $\boxed{4}$.) |
math-001168 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Where appropriate, name the theorem you use: Compute the remainder when $48^125$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{68}$.\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. |
math-001169 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Challenge: Compute the remainder when $90^167$ 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: $\\boxed{0}$.\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 ... | 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-001170 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve (and briefly cross-validate): Compute the remainder when $62^171$ 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": "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 ... | 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-001171 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Derive the result step-by-step: Compute the remainder when $94^137$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{64}$.\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. |
math-001172 | Number Theory: Congruences — Efficient Exponentiation | 1 | Prompt: Compute the remainder when $77^162$ 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).
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 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{29}$.\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 29.",
"... | [
{
"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 $\boxed{29}$.) |
math-001173 | Number Theory: Residues — Last Digits / Two Digits | 1 | Keep the final answer in boxed form: Compute the remainder when $90^61$ 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": "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-001174 | Number Theory: Residues — Last Digits / Two Digits | 1 | Derive the result step-by-step: Compute the remainder when $74^193$ 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 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": "The two methods are consistent and must coincide. 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. 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 ... | 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-001175 | Number Theory: Residues — Last Digits / Two Digits | 1 | Indicate where a theorem is used: Compute the remainder when $92^193$ 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": "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-001176 | Number Theory: Residues — Last Digits / Two Digits | 1 | Carefully track domains: Compute the remainder when $44^112$ 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": "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: $\\boxed{36}$.\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 ... | 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 $\boxed{36}$.) |
math-001177 | Number Theory: Congruences — Efficient Exponentiation | 1 | Start by stating any domain restrictions: Compute the remainder when $72^167$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{88}$.\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. |
math-001178 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Work this out carefully: Compute the remainder when $4^58$ 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": "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-001179 | Number Theory: Congruences — Efficient Exponentiation | 1 | Task: Compute the remainder when $56^83$ 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": "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-001180 | Number Theory: Residues — Last Digits / Two Digits | 1 | Track units/moduli carefully: Compute the remainder when $84^84$ 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": "Both approaches agree after simplification. 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 ... | 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-001181 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give a fully justified solution: Compute the remainder when $40^215$ 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": "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-001182 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Use two approaches if possible: Compute the remainder when $38^14$ 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": "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: $\\boxed{64}$.\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 64.",
"... | [
{
"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 $\boxed{64}$.) |
math-001183 | Number Theory: Residues — Last Digits / Two Digits | 1 | Give a theorem-based solution: Compute the remainder when $50^85$ 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": "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 ... | 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-001184 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve and justify each step: Compute the remainder when $14^241$ 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": "Both approaches agree after simplification. 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. Both yield remainder 4.",
"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-001185 | Number Theory: Congruences — Efficient Exponentiation | 1 | Work carefully and justify each inference: Compute the remainder when $42^91$ 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": "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: $\\boxed{8}$.\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. |
math-001186 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Derive the result step-by-step: Compute the remainder when $42^212$ 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": "Consistency verification shows both paths yield the identical boxed result. 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... | [
{
"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 $\boxed{36}$.) |
math-001187 | Number Theory: Congruences — Efficient Exponentiation | 1 | Compute the requested quantity: Compute the remainder when $45^226$ 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: $\\boxed{25}$.\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 ... | 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-001188 | Number Theory: Residues — Last Digits / Two Digits | 1 | Checkpoint: Compute the remainder when $75^70$ 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).
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 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. 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 yield remainder 5.",
"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-001189 | Number Theory: Congruences — Efficient Exponentiation | 1 | Be explicit about assumptions: Compute the remainder when $19^208$ 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": "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 ... | 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-001190 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Where appropriate, name the theorem you use: Compute the remainder when $44^49$ 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., 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 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 ... | 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 $\boxed{4}$.) |
math-001191 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve and sanity-check: Compute the remainder when $68^67$ 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": "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 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 ... | 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-001192 | Number Theory: Residues — Last Digits / Two Digits | 1 | Prompt: Compute the remainder when $44^26$ 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": "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. |
math-001193 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Use two approaches if possible: Compute the remainder when $59^73$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{19}$.\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. |
math-001194 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $33^15$ 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 ch... | [
{
"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: $\\boxed{7}$.\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. ... | [
{
"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 $\boxed{7}$.) |
math-001195 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Carefully track domains: Compute the remainder when $74^183$ 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": "Both approaches agree after simplification. 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. Both yield remainder 4.",
"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. (Here the result is $\boxed{4}$.) |
math-001196 | Number Theory: Congruences — Efficient Exponentiation | 1 | Give a theorem-based solution: Compute the remainder when $15^249$ 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": "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: $\\boxed{75}$.\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. (Here the result is $\boxed{75}$.) |
math-001197 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Do not skip justification steps: Compute the remainder when $5^41$ 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": "The two methods are consistent and must coincide. 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 ... | 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-001198 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Work this out carefully: Compute the remainder when $19^29$ 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": "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: $\\boxed{79}$.\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 79.",
"... | [
{
"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 $\boxed{79}$.) |
math-001199 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Solve (and briefly cross-validate): Compute the remainder when $28^72$ 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 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": "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. |
math-001200 | Number Theory: Residues — Last Digits / Two Digits | 1 | Prompt: Compute the remainder when $77^61$ 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).
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 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{77}$.\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 77.",
"... | [
{
"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. |
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