id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-001201 | Number Theory: Congruences — Efficient Exponentiation | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $19^100$ 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 ch... | [
{
"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: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both 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-001202 | Number Theory: Residues — Last Digits / Two Digits | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $51^38$ 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 c... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 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: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yie... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001203 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Carefully track domains: Compute the remainder when $3^54$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 re... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001204 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Warm-up: Compute the remainder when $12^79$ 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: $\\boxed{8}$.\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. ... | [
{
"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-001205 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Start by stating any domain restrictions: Compute the remainder when $8^171$ 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... | [
{
"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: $\\boxed{12}$.\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.",
"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 ... | 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 $\boxed{12}$.) |
math-001206 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Give a theorem-based solution: Compute the remainder when $88^165$ 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: $\\boxed{68}$.\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 ... | 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 $\boxed{68}$.) |
math-001207 | Number Theory: Residues — Last Digits / Two Digits | 1 | Where appropriate, name the theorem you use: Compute the remainder when $98^155$ 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: $\\boxed{32}$.\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... | [
{
"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-001208 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Problem: Compute the remainder when $95^145$ 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).
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 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 ... | 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-001209 | Number Theory: Residues — Last Digits / Two Digits | 1 | Solve with verification: Compute the remainder when $14^160$ 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": "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 ... | 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-001210 | Number Theory: Congruences — Efficient Exponentiation | 1 | Problem: Compute the remainder when $29^182$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
You... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001211 | Number Theory: Residues — Last Digits / Two Digits | 1 | Task: Compute the remainder when $43^188$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your w... | [
{
"method_name": "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-001212 | Number Theory: Residues — Last Digits / Two Digits | 1 | Answer with a short justification: Compute the remainder when $15^90$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{5}$.) |
math-001213 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Keep the final answer in boxed form: Compute the remainder when $28^203$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{52}$.\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 52.... | [
{
"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-001214 | Number Theory: Congruences — Efficient Exponentiation | 1 | Find the exact value: Compute the remainder when $42^147$ 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: $\\boxed{8}$.\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. (Here the result is $\boxed{8}$.) |
math-001215 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Give a fully justified solution: Compute the remainder when $54^152$ 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 ... | [
{
"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 ... | 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-001216 | Number Theory: Residues — Last Digits / Two Digits | 1 | Show all reasoning: Compute the remainder when $40^95$ 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: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001217 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Keep the final answer in boxed form: Compute the remainder when $83^98$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001218 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Track quantifiers carefully: Compute the remainder when $42^58$ 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... | [
{
"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 ... | 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-001219 | Number Theory: Residues — Last Digits / Two Digits | 1 | Answer using clear logical steps: Compute the remainder when $2^142$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\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 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-001220 | Number Theory: Congruences — Efficient Exponentiation | 1 | Exercise: Compute the remainder when $53^208$ 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/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: $\\boxed{1}$.\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-001221 | Number Theory: Residues — Last Digits / Two Digits | 1 | Give reasoning, not just computation: Compute the remainder when $35^89$ 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: $\\boxed{75}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001222 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Provide a rigorous solution: Compute the remainder when $3^38$ 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": "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 ... | 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-001223 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve and justify each step: Compute the remainder when $90^50$ 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... | [
{
"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: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
"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-001224 | Number Theory: Congruences — Efficient Exponentiation | 1 | Explain why your operations are valid: Compute the remainder when $20^119$ 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., parity o... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001225 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Start by stating any domain restrictions: Compute the remainder when $6^194$ 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: $\\boxed{96}$.\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 96.",
"... | [
{
"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-001226 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Track units/moduli carefully: Compute the remainder when $18^176$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{76}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"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-001227 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Exercise: Compute the remainder when $84^201$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Yo... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001228 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Work carefully and justify each inference: Compute the remainder when $36^94$ 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: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001229 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Problem: Compute the remainder when $82^103$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 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 100, a^2\\bmod 1... | {
"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-001230 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve and justify each step: Compute the remainder when $83^93$ 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... | [
{
"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: $\\boxed{3}$.\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 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-001231 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Find the exact value: Compute the remainder when $38^43$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reas... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. 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.",
"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 ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{12}$.) |
math-001232 | Number Theory: Congruences — Efficient Exponentiation | 1 | Carefully track domains: Compute the remainder when $54^221$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "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 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 ... | 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-001233 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Solve and justify each step: Compute the remainder when $9^82$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity 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 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: $\\boxed{81}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{81}$.) |
math-001234 | Number Theory: Residues — Last Digits / Two Digits | 1 | Work carefully and justify each inference: Compute the remainder when $40^233$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., pari... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",
"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-001235 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Explain each transformation: Compute the remainder when $31^101$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{31}$.\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 $\boxed{31}$.) |
math-001236 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Show all reasoning: Compute the remainder when $39^120$ 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001237 | Number Theory: Residues — Last Digits / Two Digits | 1 | Try to avoid pattern-matching; explain why: Compute the remainder when $27^179$ 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": "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: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 3.",
... | [
{
"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 $\boxed{3}$.) |
math-001238 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Keep the final answer in boxed form: Compute the remainder when $4^89$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mo... | [
{
"method_name": "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: $\\boxed{4}$.\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.",
"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-001239 | Number Theory: Residues — Last Digits / Two Digits | 1 | Track units/moduli carefully: Compute the remainder when $86^43$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001240 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Give a theorem-based solution: Compute the remainder when $45^123$ 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 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: $\\boxed{25}$.\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... | [
{
"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-001241 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Keep the final answer in boxed form: Compute the remainder when $38^203$ 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: $\\boxed{12}$.\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.",
"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 $\boxed{12}$.) |
math-001242 | Number Theory: Residues — Last Digits / Two Digits | 1 | Keep the final answer in boxed form: Compute the remainder when $71^61$ 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": "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: $\\boxed{11}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 11."... | [
{
"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-001243 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Determine the requested value: Compute the remainder when $96^53$ 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/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: $\\boxed{16}$.\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 ... | Remember: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001244 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Work carefully and justify each inference: Compute the remainder when $89^17$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parit... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "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 ... | 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-001245 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Checkpoint: Compute the remainder when $28^63$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Y... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "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 ... | 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-001246 | Number Theory: Residues — Last Digits / Two Digits | 1 | Derive the result step-by-step: Compute the remainder when $61^250$ 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: $\\boxed{1}$.\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 ... | 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-001247 | Number Theory: Congruences — Efficient Exponentiation | 1 | Explain why your operations are valid: Compute the remainder when $76^57$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.",
"r... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001248 | Number Theory: Residues — Last Digits / Two Digits | 1 | Explain each transformation: Compute the remainder when $53^70$ 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": "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: $\\boxed{9}$.\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.",
... | [
{
"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-001249 | Number Theory: Congruences — Efficient Exponentiation | 1 | Provide both a computational and a conceptual explanation: Compute the remainder when $67^241$ 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 c... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 20 throughout; replace $a$ by $a\\bmod 20$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\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-001250 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give an answer and a quick verification: Compute the remainder when $33^92$ 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": "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-001251 | Number Theory: Residues — Last Digits / Two Digits | 1 | Start by stating any domain restrictions: Compute the remainder when $35^56$ 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": "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 ... | 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-001252 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Answer using clear logical steps: Compute the remainder when $54^223$ 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001253 | Number Theory: Congruences — Efficient Exponentiation | 1 | Write the solution set clearly: Compute the remainder when $85^107$ 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5}$.\nBoth methods compute the same residue class of $a^b$ modulo 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 $\boxed{5}$.) |
math-001254 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve and include a self-check: Compute the remainder when $50^148$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001255 | Number Theory: Residues — Last Digits / Two Digits | 1 | Solve and then verify: Compute the remainder when $25^241$ 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: $\\boxed{25}$.\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-001256 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Track quantifiers carefully: Compute the remainder when $24^73$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/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": "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 ... | 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-001257 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $4^149$ 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 che... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{44}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 44.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{44}$.) |
math-001258 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Problem: Compute the remainder when $81^21$ 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: $\\boxed{1}$.\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. ... | [
{
"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-001259 | Number Theory: Residues — Last Digits / Two Digits | 1 | Challenge: Compute the remainder when $2^45$ 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).
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 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 ... | 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 $\boxed{32}$.) |
math-001260 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Task: Compute the remainder when $54^71$ 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{4}$.) |
math-001261 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Answer using clear logical steps: Compute the remainder when $42^102$ 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": "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 ... | 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 $\boxed{64}$.) |
math-001262 | Number Theory: Congruences — Efficient Exponentiation | 1 | Work carefully and justify each inference: Compute the remainder when $48^226$ 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., par... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. 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 ... | 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 $\boxed{64}$.) |
math-001263 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Derive the result step-by-step: Compute the remainder when $78^4$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{6}$.) |
math-001264 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Answer with a short justification: Compute the remainder when $13^105$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{93}$.\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 93.",
"... | [
{
"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-001265 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Track quantifiers carefully: Compute the remainder when $99^5$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{9}$.) |
math-001266 | Number Theory: Congruences — Efficient Exponentiation | 1 | Determine the requested value: Compute the remainder when $50^103$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 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 ... | 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-001267 | Number Theory: Residues — Last Digits / Two Digits | 1 | Prompt: Compute the remainder when $26^174$ 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": "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 ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001268 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Track quantifiers carefully: Compute the remainder when $20^90$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or 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: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 0.",... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001269 | Number Theory: Congruences — Efficient Exponentiation | 1 | Give reasoning, not just computation: Compute the remainder when $45^190$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5}$.\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.",
... | [
{
"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-001270 | Number Theory: Congruences — Efficient Exponentiation | 1 | Give a theorem-based solution: Compute the remainder when $9^51$ 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": "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 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. |
math-001271 | Number Theory: Residues — Last Digits / Two Digits | 1 | Answer using clear logical steps: Compute the remainder when $49^181$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001272 | Number Theory: Congruences — Efficient Exponentiation | 1 | Track quantifiers carefully: Compute the remainder when $72^49$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 2.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001273 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Try to avoid pattern-matching; explain why: Compute the remainder when $98^94$ 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": "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-001274 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Start by stating any domain restrictions: Compute the remainder when $83^165$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parit... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 3.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{3}$.) |
math-001275 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $95^61$ 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 chec... | [
{
"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: $\\boxed{15}$.\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.",
"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 $\boxed{15}$.) |
math-001276 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Where appropriate, name the theorem you use: Compute the remainder when $35^242$ 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., 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": "Both approaches agree after simplification. Final answer: $\\boxed{25}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 25.",
"... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001277 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Write the solution set clearly: Compute the remainder when $78^107$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 2.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001278 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Warm-up: Compute the remainder when $57^133$ 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).
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 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: $\\boxed{57}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{57}$.) |
math-001279 | Number Theory: Congruences — Efficient Exponentiation | 1 | Solve and sanity-check: Compute the remainder when $21^238$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{61}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yi... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{61}$.) |
math-001280 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Solve and justify each step: Compute the remainder when $60^60$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Key idea: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{0}$.) |
math-001281 | Number Theory: Congruences — Efficient Exponentiation | 1 | Checkpoint: Compute the remainder when $14^186$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{6}$.) |
math-001282 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Use two approaches if possible: Compute the remainder when $11^226$ 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": "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 ... | 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-001283 | Number Theory: Congruences — Efficient Exponentiation | 1 | Compute the requested quantity: Compute the remainder when $72^232$ 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: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 16.... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{16}$.) |
math-001284 | Number Theory: Residues — Last Digits / Two Digits | 1 | Do not skip justification steps: Compute the remainder when $4^14$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. ... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{6}$.) |
math-001285 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Solve and sanity-check: Compute the remainder when $9^5$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reas... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001286 | Number Theory: Congruences — Efficient Exponentiation | 1 | Task: Compute the remainder when $93^144$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 reasoning).
Your ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 100, a^2\\bmod 1... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001287 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Proceed methodically: Compute the remainder when $16^135$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 rea... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "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 ... | 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-001288 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Solve and justify each step: Compute the remainder when $56^141$ is divided by $100$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/m... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 100 throughout; replace $a$ by $a\\bmod 100$ to start.",
"Step 2: Compute successive square... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{56}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001289 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Do not skip justification steps: Compute the remainder when $69^51$ 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{69}$.\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 69.... | [
{
"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 $\boxed{69}$.) |
math-001290 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Solve with verification: Compute the remainder when $84^67$ 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 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": "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. (Here the result is $\boxed{4}$.) |
math-001291 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Explain why your operations are valid: Compute the remainder when $58^86$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 10, a^2\\bmod 10... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001292 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Work carefully and justify each inference: Compute the remainder when $77^3$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity... | [
{
"method_name": "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: $\\boxed{3}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both 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-001293 | Number Theory: Residues — Last Digits / Two Digits | 1 | Warm-up: Compute the remainder when $67^146$ 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).
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 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: $\\boxed{69}$.\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. |
math-001294 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Work this out carefully: Compute the remainder when $89^5$ 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": "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 ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{9}$.) |
math-001295 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Task: Compute the remainder when $37^149$ 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 ... | 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 $\boxed{77}$.) |
math-001296 | Number Theory: Modular Arithmetic — Repeated Squaring | 1 | Make each step logically reversible (or explain if not): Compute the remainder when $41^92$ 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 chec... | [
{
"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: $\\boxed{1}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 1.",
... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Core principle: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. (Here the result is $\boxed{1}$.) |
math-001297 | Discrete Math: Modular Computation — Avoid Huge Integers | 1 | Work carefully and justify each inference: Compute the remainder when $60^228$ 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., 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{0}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yiel... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | Takeaway: In modular arithmetic, reduce early and often: compute large powers via repeated squaring, and use periodicity/cycles as a conceptual shortcut when the modulus is small. |
math-001298 | Number Theory: Modular Arithmetic — Cycles and Periods | 1 | Give a theorem-based solution: Compute the remainder when $26^132$ is divided by $10$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/... | [
{
"method_name": "Repeated Squaring",
"approach": "Use exponentiation by squaring to compute $a^b\\bmod m$ efficiently, reducing modulo $m$ after each multiplication.",
"steps": [
"Step 1: Work modulo 10 throughout; replace $a$ by $a\\bmod 10$ to start.",
"Step 2: Compute successive squares ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nBoth methods compute the same residue class of $a^b$ modulo 10: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 6.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{6}$.) |
math-001299 | Number Theory: Residues — Last Digits / Two Digits | 1 | Carefully track domains: Compute the remainder when $77^170$ is divided by $20$.
(a) Solve using modular reduction and repeated squaring.
(b) Solve using a cycle/period argument for residues modulo $m$ (e.g., last-digit cycles).
(c) Verify your final remainder by a quick consistency check (e.g., parity or mod 5/mod 4 ... | [
{
"method_name": "Residue Cycle / Euler–Totient Heuristic",
"approach": "Use that residues modulo $m$ often repeat with a period; for $m\\in\\{10,20,100\\}$, last-digit/two-digit cycles are short and can be tracked explicitly.",
"steps": [
"Step 1: Consider the sequence $a^1\\bmod 20, a^2\\bmod 20... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{9}$.\nBoth methods compute the same residue class of $a^b$ modulo 20: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity. Both yield remainder 9.",
"rob... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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-001300 | Number Theory: Congruences — Efficient Exponentiation | 1 | Answer with a short justification: Compute the remainder when $84^16$ 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": "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: $\\boxed{16}$.\nBoth methods compute the same residue class of $a^b$ modulo 100: repeated squaring is an exact algorithm, and the cycle argument is a conceptual description of the same modular periodicity... | [
{
"error_description": "Reduced $a^b$ by reducing $b$ modulo $m$ (e.g., replaced $b$ with $b\\bmod m$).",
"why_plausible": "It resembles the valid rule $a\\equiv a'\\pmod m$.",
"why_wrong": "Exponents do not reduce modulo $m$ in general; they reduce modulo a period (like $\\varphi(m)$) only under extra ... | 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 $\boxed{16}$.) |
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