id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-013101 | Number Theory: Divisibility — Linear Combinations | 7 | Determine the requested value: (a) Compute $\gcd(1615,984)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1615+v\cdot 984=\gcd(1615,984)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1615,984)$ to compute $g=\\gcd(1615,984)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=223,v=-366$ satisfies $u1615+v984=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustn... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013102 | Number Theory: Units mod m — Existence Condition | 7 | Track quantifiers carefully: Find the multiplicative inverse of $599$ modulo $989$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{989}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(599,989)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{530}$.\nMethod 1 constructs an inverse via Bézout, producing $x=530$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{530}$.) |
math-013103 | Number Theory: Divisibility — Linear Combinations | 7 | Derive the result step-by-step: (a) Compute $\gcd(1312,1875)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1312+v\cdot 1875=\gcd(1312,1875)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution cha... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1312,1875)$ to compute $g=\\gcd(1312,1875)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=373,v=-261$ satisfies $u1312+v1875=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013104 | Number Theory: Units mod m — Existence Condition | 7 | Give reasoning, not just computation: Find the multiplicative inverse of $121$ modulo $420$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{420}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient cond... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=361$ and compute $121x=43681$.",
"Step 2: Reduce: $43681\\equiv 1\\pmod{420}$ (since $43680=43680$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{361}$.\nMethod 1 constructs an inverse via Bézout, producing $x=361$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{361}$.) |
math-013105 | Number Theory: Modular Inverses — Extended Euclid | 7 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $110$ modulo $273$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{273}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=206$ and compute $110x=22660$.",
"Step 2: Reduce: $22660\\equiv 1\\pmod{273}$ (since $22659=22659$ is divis... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{206}$.\nMethod 1 constructs an inverse via Bézout, producing $x=206$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013106 | Number Theory: gcd — Euclidean Algorithm | 7 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(790,988)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 790+v\cdot 988=\gcd(790,988)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backwa... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(790,988)$ to compute $g=\\gcd(790,988)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-5,v=4$ satisfies $u790+v988=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013107 | Number Theory: gcd — Euclidean Algorithm | 7 | Work this out carefully: (a) Compute $\gcd(757,1788)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 757+v\cdot 1788=\gcd(757,1788)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=385$ and $v=-163$ with $u757... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=385,v=-163$ satisfies $u757+v1788=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013108 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Solve (and briefly cross-validate): Find the multiplicative inverse of $1006$ modulo $1103$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1103}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient con... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=705$ and compute $1006x=709230$.",
"Step 2: Reduce: $709230\\equiv 1\\pmod{1103}$ (since $709229=709229$ is... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{705}$.\nMethod 1 constructs an inverse via Bézout, producing $x=705$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{705}$.) |
math-013109 | Number Theory: gcd — Back Substitution | 7 | Find the exact value: (a) Compute $\gcd(1073,1444)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1073+v\cdot 1444=\gcd(1073,1444)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=397$ and $v=-295$ with $u107... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=397,v=-295$ satisfies $u1073+v1444=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013110 | Number Theory: Bézout Identity — Certificates | 7 | Work this out carefully: (a) Compute $\gcd(236,1193)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 236+v\cdot 1193=\gcd(236,1193)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-551$ and $v=109$ with $u236... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-551,v=109$ satisfies $u236+v1193=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013111 | Number Theory: Modular Inverses — Extended Euclid | 7 | Work carefully and justify each inference: Find the multiplicative inverse of $54$ modulo $137$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{137}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=33$ and compute $54x=1782$.",
"Step 2: Reduce: $1782\\equiv 1\\pmod{137}$ (since $1781=1781$ is divisible b... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{33}$.\nMethod 1 constructs an inverse via Bézout, producing $x=33$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Exten... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{33}$.) |
math-013112 | Number Theory: gcd — Back Substitution | 7 | Give an answer and a quick verification: (a) Compute $\gcd(762,238)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 762+v\cdot 238=\gcd(762,238)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(762,238)$ to compute $g=\\gcd(762,238)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=5,v=-16$ satisfies $u762+v238=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Ge... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013113 | Computational Number Theory: Inverses and Certificates | 7 | Answer with a short justification: Find the multiplicative inverse of $16$ modulo $51$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{51}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(16,51)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{16}$.\nMethod 1 constructs an inverse via Bézout, producing $x=16$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{16}$.) |
math-013114 | Number Theory: gcd — Back Substitution | 7 | Indicate where a theorem is used: (a) Compute $\gcd(1396,1695)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1396+v\cdot 1695=\gcd(1396,1695)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution c... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1396,1695)$ to compute $g=\\gcd(1396,1695)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-839,v=691$ satisfies $u1396+v1695=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scale... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013115 | Computational Number Theory: Inverses and Certificates | 7 | Give an answer and a quick verification: Find the multiplicative inverse of $60$ modulo $161$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{161}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=51$ and compute $60x=3060$.",
"Step 2: Reduce: $3060\\equiv 1\\pmod{161}$ (since $3059=3059$ is divisible b... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{51}$.\nMethod 1 constructs an inverse via Bézout, producing $x=51$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysi... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013116 | Computational Number Theory: Extended Euclid | 7 | State any required conditions first: (a) Compute $\gcd(1630,646)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1630+v\cdot 646=\gcd(1630,646)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution c... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1630,646)$ to compute $g=\\gcd(1630,646)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-86,v=217$ satisfies $u1630+v646=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013117 | Number Theory: Divisibility — Linear Combinations | 7 | Warm-up: (a) Compute $\gcd(969,861)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 969+v\cdot 861=\gcd(969,861)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief verifica... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(969,861)$ to compute $g=\\gcd(969,861)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=8,v=-9$ satisfies $u969+v861=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Rob... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013118 | Number Theory: Bézout Identity — Certificates | 7 | Do not skip justification steps: (a) Compute $\gcd(702,1676)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 702+v\cdot 1676=\gcd(702,1676)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(702,1676)$ to compute $g=\\gcd(702,1676)$.",
"Step 2: Record the remainder... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=191,v=-80$ satisfies $u702+v1676=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid s... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013119 | Number Theory: Units mod m — Existence Condition | 7 | Show all reasoning: Find the multiplicative inverse of $259$ modulo $1195$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1195}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=729$ and compute $259x=188811$.",
"Step 2: Reduce: $188811\\equiv 1\\pmod{1195}$ (since $188810=188810$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{729}$.\nMethod 1 constructs an inverse via Bézout, producing $x=729$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013120 | Number Theory: gcd — Back Substitution | 7 | Give reasoning, not just computation: (a) Compute $\gcd(978,1404)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 978+v\cdot 1404=\gcd(978,1404)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-89$ and $v=62$ with $u978+v... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6}$.\nThe Euclidean algorithm computes $g=6$. The Bézout certificate $u=-89,v=62$ satisfies $u978+v1404=6$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{6}$.) |
math-013121 | Number Theory: Divisibility — Linear Combinations | 7 | Give a fully justified solution: (a) Compute $\gcd(279,612)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 279+v\cdot 612=\gcd(279,612)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(279,612)$ to compute $g=\\gcd(279,612)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nThe Euclidean algorithm computes $g=9$. The Bézout certificate $u=11,v=-5$ satisfies $u279+v612=9$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Ge... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{9}$.) |
math-013122 | Number Theory: Modular Inverses — Extended Euclid | 7 | Track units/moduli carefully: Find the multiplicative inverse of $172$ modulo $761$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{761}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(172,761)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{615}$.\nMethod 1 constructs an inverse via Bézout, producing $x=615$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{615}$.) |
math-013123 | Number Theory: Divisibility — Linear Combinations | 7 | Complete the analysis: (a) Compute $\gcd(1940,1786)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1940+v\cdot 1786=\gcd(1940,1786)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Incl... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=58$ and $v=-63$ with $u1940+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=58,v=-63$ satisfies $u1940+v1786=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid sc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013124 | Number Theory: gcd — Euclidean Algorithm | 7 | Derive the result step-by-step: (a) Compute $\gcd(211,102)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 211+v\cdot 102=\gcd(211,102)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
I... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(211,102)$ to compute $g=\\gcd(211,102)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-29,v=60$ satisfies $u211+v102=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013125 | Number Theory: Divisibility — Linear Combinations | 7 | Checkpoint: (a) Compute $\gcd(1835,1074)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1835+v\cdot 1074=\gcd(1835,1074)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1835,1074)$ to compute $g=\\gcd(1835,1074)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-175,v=299$ satisfies $u1835+v1074=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013126 | Number Theory: Units mod m — Existence Condition | 7 | Track units/moduli carefully: Find the multiplicative inverse of $183$ modulo $853$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{853}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=289$ and compute $183x=52887$.",
"Step 2: Reduce: $52887\\equiv 1\\pmod{853}$ (since $52886=52886$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{289}$.\nMethod 1 constructs an inverse via Bézout, producing $x=289$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013127 | Computational Number Theory: Inverses and Certificates | 7 | Answer with a short justification: Find the multiplicative inverse of $1352$ modulo $1539$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1539}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient cond... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(1352,1539)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{716}$.\nMethod 1 constructs an inverse via Bézout, producing $x=716$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{716}$.) |
math-013128 | Number Theory: Divisibility — Linear Combinations | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(986,164)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 986+v\cdot 164=\gcd(986,164)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substituti... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=1$ and $v=-6$ with $u986+v16... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=1,v=-6$ satisfies $u986+v164=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013129 | Number Theory: Modular Inverses — Extended Euclid | 7 | Use two approaches if possible: Find the multiplicative inverse of $105$ modulo $172$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{172}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(105,172)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{77}$.\nMethod 1 constructs an inverse via Bézout, producing $x=77$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is f... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{77}$.) |
math-013130 | Computational Number Theory: Inverses and Certificates | 7 | Proceed methodically: Find the multiplicative inverse of $135$ modulo $1487$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1487}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an i... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(135,1487)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{738}$.\nMethod 1 constructs an inverse via Bézout, producing $x=738$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013131 | Number Theory: Bézout Identity — Certificates | 7 | Solve and include a self-check: (a) Compute $\gcd(1107,1217)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1107+v\cdot 1217=\gcd(1107,1217)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution cha... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1107,1217)$ to compute $g=\\gcd(1107,1217)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-520,v=473$ satisfies $u1107+v1217=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013132 | Number Theory: Modular Inverses — Extended Euclid | 7 | Keep the final answer in boxed form: Find the multiplicative inverse of $4$ modulo $229$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{229}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditi... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(4,229)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{172}$.\nMethod 1 constructs an inverse via Bézout, producing $x=172$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013133 | Number Theory: Bézout Identity — Certificates | 7 | Give a theorem-based solution: (a) Compute $\gcd(1557,1132)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1557+v\cdot 1132=\gcd(1557,1132)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chai... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1557,1132)$ to compute $g=\\gcd(1557,1132)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=285,v=-392$ satisfies $u1557+v1132=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013134 | Number Theory: Units mod m — Existence Condition | 7 | Challenge: Find the multiplicative inverse of $93$ modulo $1864$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1864}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ex... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(93,1864)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{461}$.\nMethod 1 constructs an inverse via Bézout, producing $x=461$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013135 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Solve and sanity-check: Find the multiplicative inverse of $1350$ modulo $1537$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1537}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for a... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=600$ and compute $1350x=810000$.",
"Step 2: Reduce: $810000\\equiv 1\\pmod{1537}$ (since $809999=809999$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{600}$.\nMethod 1 constructs an inverse via Bézout, producing $x=600$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013136 | Computational Number Theory: Inverses and Certificates | 7 | Question: Find the multiplicative inverse of $42$ modulo $53$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{53}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist.
... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(42,53)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24}$.\nMethod 1 constructs an inverse via Bézout, producing $x=24$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended Eu... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{24}$.) |
math-013137 | Number Theory: Modular Inverses — Extended Euclid | 7 | Be explicit about assumptions: Find the multiplicative inverse of $239$ modulo $640$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{640}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition f... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(239,640)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{399}$.\nMethod 1 constructs an inverse via Bézout, producing $x=399$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{399}$.) |
math-013138 | Number Theory: Bézout Identity — Certificates | 7 | State any required conditions first: (a) Compute $\gcd(943,387)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 943+v\cdot 387=\gcd(943,387)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chai... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(943,387)$ to compute $g=\\gcd(943,387)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-158,v=385$ satisfies $u943+v387=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013139 | Number Theory: gcd — Back Substitution | 7 | Do not skip justification steps: (a) Compute $\gcd(1025,1157)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1025+v\cdot 1157=\gcd(1025,1157)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=149$ and $v=-132$ with $u102... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=149,v=-132$ satisfies $u1025+v1157=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013140 | Number Theory: Units mod m — Existence Condition | 7 | Track quantifiers carefully: Find the multiplicative inverse of $1085$ modulo $1289$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1289}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=714$ and compute $1085x=774690$.",
"Step 2: Reduce: $774690\\equiv 1\\pmod{1289}$ (since $774689=774689$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{714}$.\nMethod 1 constructs an inverse via Bézout, producing $x=714$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{714}$.) |
math-013141 | Number Theory: Divisibility — Linear Combinations | 7 | Solve with verification: (a) Compute $\gcd(605,1398)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 605+v\cdot 1398=\gcd(605,1398)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-409$ and $v=177$ with $u605... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-409,v=177$ satisfies $u605+v1398=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013142 | Computational Number Theory: Extended Euclid | 7 | Question: (a) Compute $\gcd(1400,1283)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1400+v\cdot 1283=\gcd(1400,1283)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief v... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1400,1283)$ to compute $g=\\gcd(1400,1283)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-318,v=347$ satisfies $u1400+v1283=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensit... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013143 | Number Theory: gcd — Euclidean Algorithm | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(204,215)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 204+v\cdot 215=\gcd(204,215)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substituti... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=39$ and $v=-37$ with $u204+v... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=39,v=-37$ satisfies $u204+v215=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the pro... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013144 | Number Theory: Modular Inverses — Extended Euclid | 7 | Question: Find the multiplicative inverse of $745$ modulo $933$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{933}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=799$ and compute $745x=595255$.",
"Step 2: Reduce: $595255\\equiv 1\\pmod{933}$ (since $595254=595254$ is d... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{799}$.\nMethod 1 constructs an inverse via Bézout, producing $x=799$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013145 | Computational Number Theory: Inverses and Certificates | 7 | Give a fully justified solution: Find the multiplicative inverse of $21$ modulo $671$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{671}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(21,671)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{32}$.\nMethod 1 constructs an inverse via Bézout, producing $x=32$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{32}$.) |
math-013146 | Computational Number Theory: Extended Euclid | 7 | Solve and sanity-check: (a) Compute $\gcd(139,1142)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 139+v\cdot 1142=\gcd(139,1142)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Includ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(139,1142)$ to compute $g=\\gcd(139,1142)$.",
"Step 2: Record the remainder... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-419,v=51$ satisfies $u139+v1142=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertu... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013147 | Computational Number Theory: Extended Euclid | 7 | Determine the requested value: (a) Compute $\gcd(807,1343)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 807+v\cdot 1343=\gcd(807,1343)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-223$ and $v=134$ with $u807... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-223,v=134$ satisfies $u807+v1343=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013148 | Computational Number Theory: Extended Euclid | 7 | Answer with a short justification: (a) Compute $\gcd(957,455)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 957+v\cdot 455=\gcd(957,455)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(957,455)$ to compute $g=\\gcd(957,455)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=213,v=-448$ satisfies $u957+v455=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013149 | Computational Number Theory: Extended Euclid | 7 | Question: (a) Compute $\gcd(1839,442)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1839+v\cdot 442=\gcd(1839,442)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief veri... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1839,442)$ to compute $g=\\gcd(1839,442)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=193,v=-803$ satisfies $u1839+v442=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "General... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013150 | Number Theory: gcd — Back Substitution | 7 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(216,885)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 216+v\cdot 885=\gcd(216,885)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear back... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-127$ and $v=31$ with $u216+... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=-127,v=31$ satisfies $u216+v885=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generalit... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-013151 | Computational Number Theory: Inverses and Certificates | 7 | Give a fully justified solution: Find the multiplicative inverse of $479$ modulo $715$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{715}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(479,715)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{409}$.\nMethod 1 constructs an inverse via Bézout, producing $x=409$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{409}$.) |
math-013152 | Number Theory: Bézout Identity — Certificates | 7 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(133,1414)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 133+v\cdot 1414=\gcd(133,1414)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear bac... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(133,1414)$ to compute $g=\\gcd(133,1414)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nThe Euclidean algorithm computes $g=7$. The Bézout certificate $u=-85,v=8$ satisfies $u133+v1414=7$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013153 | Number Theory: gcd — Euclidean Algorithm | 7 | Do not skip justification steps: (a) Compute $\gcd(1172,1862)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1172+v\cdot 1862=\gcd(1172,1862)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=367$ and $v=-231$ with $u117... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=367,v=-231$ satisfies $u1172+v1862=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013154 | Number Theory: gcd — Euclidean Algorithm | 7 | Checkpoint: (a) Compute $\gcd(1664,1599)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1664+v\cdot 1599=\gcd(1664,1599)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1664,1599)$ to compute $g=\\gcd(1664,1599)$.",
"Step 2: Record the remaind... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13}$.\nThe Euclidean algorithm computes $g=13$. The Bézout certificate $u=-49,v=51$ satisfies $u1664+v1599=13$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{13}$.) |
math-013155 | Number Theory: Divisibility — Linear Combinations | 7 | Task: (a) Compute $\gcd(1539,1621)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1539+v\cdot 1621=\gcd(1539,1621)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief verif... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1539,1621)$ to compute $g=\\gcd(1539,1621)$.",
"Step 2: Record the remaind... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-257,v=244$ satisfies $u1539+v1621=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: E... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013156 | Computational Number Theory: Inverses and Certificates | 7 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $904$ modulo $1621$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1621}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necess... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1569$ and compute $904x=1418376$.",
"Step 2: Reduce: $1418376\\equiv 1\\pmod{1621}$ (since $1418375=1418375... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1569}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1569$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1569}$.) |
math-013157 | Number Theory: Units mod m — Existence Condition | 7 | Problem: Find the multiplicative inverse of $263$ modulo $298$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{298}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(263,298)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{17}$.\nMethod 1 constructs an inverse via Bézout, producing $x=17$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast a... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013158 | Number Theory: Divisibility — Linear Combinations | 7 | Start by stating any domain restrictions: (a) Compute $\gcd(545,583)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 545+v\cdot 583=\gcd(545,583)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=46$ and $v=-43$ with $u545+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=46,v=-43$ satisfies $u545+v583=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013159 | Computational Number Theory: Inverses and Certificates | 7 | Checkpoint: Find the multiplicative inverse of $445$ modulo $946$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{946}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ex... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(445,946)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{625}$.\nMethod 1 constructs an inverse via Bézout, producing $x=625$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013160 | Number Theory: gcd — Back Substitution | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(1969,213)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1969+v\cdot 213=\gcd(1969,213)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substit... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1969,213)$ to compute $g=\\gcd(1969,213)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-86,v=795$ satisfies $u1969+v213=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013161 | Number Theory: Bézout Identity — Certificates | 7 | Work carefully and justify each inference: (a) Compute $\gcd(1366,850)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1366+v\cdot 850=\gcd(1366,850)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-28$ and $v=45$ with $u1366+... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-28,v=45$ satisfies $u1366+v850=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013162 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve and include a self-check: Find the multiplicative inverse of $1305$ modulo $1576$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1576}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=977$ and compute $1305x=1274985$.",
"Step 2: Reduce: $1274985\\equiv 1\\pmod{1576}$ (since $1274984=1274984... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{977}$.\nMethod 1 constructs an inverse via Bézout, producing $x=977$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{977}$.) |
math-013163 | Number Theory: gcd — Euclidean Algorithm | 7 | Give a fully justified solution: (a) Compute $\gcd(463,1370)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 463+v\cdot 1370=\gcd(463,1370)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(463,1370)$ to compute $g=\\gcd(463,1370)$.",
"Step 2: Record the remainder... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=577,v=-195$ satisfies $u463+v1370=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013164 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Challenge: Find the multiplicative inverse of $268$ modulo $337$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{337}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=210$ and compute $268x=56280$.",
"Step 2: Reduce: $56280\\equiv 1\\pmod{337}$ (since $56279=56279$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{210}$.\nMethod 1 constructs an inverse via Bézout, producing $x=210$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{210}$.) |
math-013165 | Number Theory: Bézout Identity — Certificates | 7 | State any required conditions first: (a) Compute $\gcd(194,1106)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 194+v\cdot 1106=\gcd(194,1106)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution c... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-57$ and $v=10$ with $u194+v... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-57,v=10$ satisfies $u194+v1106=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013166 | Number Theory: Units mod m — Existence Condition | 7 | Warm-up: Find the multiplicative inverse of $605$ modulo $791$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{791}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(605,791)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{404}$.\nMethod 1 constructs an inverse via Bézout, producing $x=404$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{404}$.) |
math-013167 | Number Theory: gcd — Euclidean Algorithm | 7 | Be explicit about assumptions: (a) Compute $\gcd(1427,1552)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1427+v\cdot 1552=\gcd(1427,1552)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chai... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1427,1552)$ to compute $g=\\gcd(1427,1552)$.",
"Step 2: Record the remaind... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-149,v=137$ satisfies $u1427+v1552=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: E... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013168 | Number Theory: Modular Inverses — Extended Euclid | 7 | Try to avoid pattern-matching; explain why: Find the multiplicative inverse of $547$ modulo $935$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{935}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficien... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(547,935)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{788}$.\nMethod 1 constructs an inverse via Bézout, producing $x=788$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013169 | Number Theory: gcd — Back Substitution | 7 | Give an answer and a quick verification: (a) Compute $\gcd(1067,396)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1067+v\cdot 396=\gcd(1067,396)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substituti... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=13$ and $v=-35$ with $u1067+... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{11}$.\nThe Euclidean algorithm computes $g=11$. The Bézout certificate $u=13,v=-35$ satisfies $u1067+v396=11$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013170 | Number Theory: gcd — Back Substitution | 7 | Challenge: (a) Compute $\gcd(939,1197)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 939+v\cdot 1197=\gcd(939,1197)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief ver... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(939,1197)$ to compute $g=\\gcd(939,1197)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=-116,v=91$ satisfies $u939+v1197=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013171 | Number Theory: Units mod m — Existence Condition | 7 | Task: Find the multiplicative inverse of $413$ modulo $1549$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1549}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist.... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1534$ and compute $413x=633542$.",
"Step 2: Reduce: $633542\\equiv 1\\pmod{1549}$ (since $633541=633541$ is... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1534}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1534$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fa... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1534}$.) |
math-013172 | Number Theory: Modular Inverses — Extended Euclid | 7 | Explain each transformation: Find the multiplicative inverse of $355$ modulo $421$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{421}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(355,421)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{236}$.\nMethod 1 constructs an inverse via Bézout, producing $x=236$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{236}$.) |
math-013173 | Computational Number Theory: Inverses and Certificates | 7 | Solve and then verify: Find the multiplicative inverse of $832$ modulo $1731$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1731}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(832,1731)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{310}$.\nMethod 1 constructs an inverse via Bézout, producing $x=310$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{310}$.) |
math-013174 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve (and briefly cross-validate): Find the multiplicative inverse of $241$ modulo $331$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{331}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condit... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=114$ and compute $241x=27474$.",
"Step 2: Reduce: $27474\\equiv 1\\pmod{331}$ (since $27473=27473$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{114}$.\nMethod 1 constructs an inverse via Bézout, producing $x=114$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{114}$.) |
math-013175 | Computational Number Theory: Inverses and Certificates | 7 | Solve and include a self-check: Find the multiplicative inverse of $889$ modulo $1369$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1369}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditio... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(889,1369)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{482}$.\nMethod 1 constructs an inverse via Bézout, producing $x=482$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{482}$.) |
math-013176 | Number Theory: gcd — Back Substitution | 7 | Solve with verification: (a) Compute $\gcd(1563,1654)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1563+v\cdot 1654=\gcd(1563,1654)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
In... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1563,1654)$ to compute $g=\\gcd(1563,1654)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-309,v=292$ satisfies $u1563+v1654=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013177 | Computational Number Theory: Inverses and Certificates | 7 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $1564$ modulo $1703$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1703}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=49$ and compute $1564x=76636$.",
"Step 2: Reduce: $76636\\equiv 1\\pmod{1703}$ (since $76635=76635$ is divi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{49}$.\nMethod 1 constructs an inverse via Bézout, producing $x=49$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013178 | Computational Number Theory: Inverses and Certificates | 7 | Task: Find the multiplicative inverse of $763$ modulo $823$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{823}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist.
... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=96$ and compute $763x=73248$.",
"Step 2: Reduce: $73248\\equiv 1\\pmod{823}$ (since $73247=73247$ is divisi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{96}$.\nMethod 1 constructs an inverse via Bézout, producing $x=96$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{96}$.) |
math-013179 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Start by stating any domain restrictions: Find the multiplicative inverse of $185$ modulo $1477$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1477}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficien... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=495$ and compute $185x=91575$.",
"Step 2: Reduce: $91575\\equiv 1\\pmod{1477}$ (since $91574=91574$ is divi... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{495}$.\nMethod 1 constructs an inverse via Bézout, producing $x=495$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013180 | Number Theory: Bézout Identity — Certificates | 7 | Track quantifiers carefully: (a) Compute $\gcd(1432,104)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1432+v\cdot 104=\gcd(1432,104)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
I... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=4$ and $v=-55$ with $u1432+v... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nThe Euclidean algorithm computes $g=8$. The Bézout certificate $u=4,v=-55$ satisfies $u1432+v104=8$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{8}$.) |
math-013181 | Number Theory: Units mod m — Existence Condition | 7 | Keep the final answer in boxed form: Find the multiplicative inverse of $580$ modulo $921$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{921}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condi... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(580,921)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{316}$.\nMethod 1 constructs an inverse via Bézout, producing $x=316$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{316}$.) |
math-013182 | Computational Number Theory: Inverses and Certificates | 7 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $20$ modulo $1877$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1877}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(20,1877)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{657}$.\nMethod 1 constructs an inverse via Bézout, producing $x=657$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{657}$.) |
math-013183 | Number Theory: Modular Inverses — Extended Euclid | 7 | Complete the analysis: Find the multiplicative inverse of $584$ modulo $1251$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1251}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=422$ and compute $584x=246448$.",
"Step 2: Reduce: $246448\\equiv 1\\pmod{1251}$ (since $246447=246447$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{422}$.\nMethod 1 constructs an inverse via Bézout, producing $x=422$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{422}$.) |
math-013184 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Find the exact value: Find the multiplicative inverse of $29$ modulo $1952$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1952}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an in... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(29,1952)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1077}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1077$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Genera... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1077}$.) |
math-013185 | Number Theory: Divisibility — Linear Combinations | 7 | Solve (and briefly cross-validate): (a) Compute $\gcd(907,471)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 907+v\cdot 471=\gcd(907,471)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=148$ and $v=-285$ with $u907... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=148,v=-285$ satisfies $u907+v471=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013186 | Computational Number Theory: Extended Euclid | 7 | Do not skip justification steps: (a) Compute $\gcd(1620,863)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1620+v\cdot 863=\gcd(1620,863)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-57$ and $v=107$ with $u1620... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-57,v=107$ satisfies $u1620+v863=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013187 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve with verification: Find the multiplicative inverse of $482$ modulo $525$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{525}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=293$ and compute $482x=141226$.",
"Step 2: Reduce: $141226\\equiv 1\\pmod{525}$ (since $141225=141225$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{293}$.\nMethod 1 constructs an inverse via Bézout, producing $x=293$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{293}$.) |
math-013188 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $158$ modulo $247$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{247}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(158,247)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{111}$.\nMethod 1 constructs an inverse via Bézout, producing $x=111$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{111}$.) |
math-013189 | Computational Number Theory: Inverses and Certificates | 7 | Solve and justify each step: Find the multiplicative inverse of $854$ modulo $1523$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1523}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition f... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=708$ and compute $854x=604632$.",
"Step 2: Reduce: $604632\\equiv 1\\pmod{1523}$ (since $604631=604631$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{708}$.\nMethod 1 constructs an inverse via Bézout, producing $x=708$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013190 | Number Theory: Divisibility — Linear Combinations | 7 | Determine the requested value: (a) Compute $\gcd(1841,226)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1841+v\cdot 226=\gcd(1841,226)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1841,226)$ to compute $g=\\gcd(1841,226)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-89,v=725$ satisfies $u1841+v226=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the p... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013191 | Number Theory: Modular Inverses — Extended Euclid | 7 | Exercise: Find the multiplicative inverse of $138$ modulo $1673$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1673}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ex... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(138,1673)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{885}$.\nMethod 1 constructs an inverse via Bézout, producing $x=885$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{885}$.) |
math-013192 | Number Theory: Bézout Identity — Certificates | 7 | Where appropriate, name the theorem you use: (a) Compute $\gcd(1896,124)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1896+v\cdot 124=\gcd(1896,124)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substi... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=7$ and $v=-107$ with $u1896+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nThe Euclidean algorithm computes $g=4$. The Bézout certificate $u=7,v=-107$ satisfies $u1896+v124=4$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid sca... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{4}$.) |
math-013193 | Computational Number Theory: Extended Euclid | 7 | Find the exact value: (a) Compute $\gcd(1150,200)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1150+v\cdot 200=\gcd(1150,200)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1150,200)$ to compute $g=\\gcd(1150,200)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{50}$.\nThe Euclidean algorithm computes $g=50$. The Bézout certificate $u=-1,v=6$ satisfies $u1150+v200=50$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{50}$.) |
math-013194 | Number Theory: gcd — Back Substitution | 7 | Solve and then verify: (a) Compute $\gcd(531,1281)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 531+v\cdot 1281=\gcd(531,1281)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(531,1281)$ to compute $g=\\gcd(531,1281)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=193,v=-80$ satisfies $u531+v1281=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013195 | Number Theory: gcd — Euclidean Algorithm | 7 | Work this out carefully: (a) Compute $\gcd(1077,1255)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1077+v\cdot 1255=\gcd(1077,1255)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
In... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-557$ and $v=478$ with $u107... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-557,v=478$ satisfies $u1077+v1255=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013196 | Number Theory: gcd — Euclidean Algorithm | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(1414,526)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1414+v\cdot 526=\gcd(1414,526)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substit... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=93$ and $v=-250$ with $u1414... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=93,v=-250$ satisfies $u1414+v526=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid s... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013197 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $879$ modulo $1240$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1240}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffic... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(879,1240)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{79}$.\nMethod 1 constructs an inverse via Bézout, producing $x=79$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013198 | Computational Number Theory: Extended Euclid | 7 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(1668,410)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1668+v\cdot 410=\gcd(1668,410)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear b... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1668,410)$ to compute $g=\\gcd(1668,410)$.",
"Step 2: Record the remainder... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=44,v=-179$ satisfies $u1668+v410=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013199 | Number Theory: Modular Inverses — Extended Euclid | 7 | Track units/moduli carefully: Find the multiplicative inverse of $245$ modulo $268$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{268}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=233$ and compute $245x=57085$.",
"Step 2: Reduce: $57085\\equiv 1\\pmod{268}$ (since $57084=57084$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{233}$.\nMethod 1 constructs an inverse via Bézout, producing $x=233$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013200 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Be explicit about assumptions: Find the multiplicative inverse of $7$ modulo $1482$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1482}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition f... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(7,1482)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{847}$.\nMethod 1 constructs an inverse via Bézout, producing $x=847$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{847}$.) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.