id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-011301 | Number Theory: Units mod m — Existence Condition | 6 | Do not skip justification steps: Find the multiplicative inverse of $367$ modulo $1573$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1573}$).
(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 $\\gcd(367,1573)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1543}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1543$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended E... | [
{
"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-011302 | Number Theory: Units mod m — Existence Condition | 6 | Explain what is being counted/optimized: Find the multiplicative inverse of $166$ modulo $267$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{267}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"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 $\\gcd(166,267)=1$, 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: $\\boxed{37}$.\nMethod 1 constructs an inverse via Bézout, producing $x=37$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Ex... | [
{
"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 $\boxed{37}$.) |
math-011303 | Computational Number Theory: Inverses and Certificates | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $1051$ modulo $1882$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1882}$).
(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": "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 $\\gcd(1051,1882)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "Generality note: Ex... | [
{
"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 $\boxed{77}$.) |
math-011304 | Computational Number Theory: Inverses and Certificates | 6 | Find the exact value: Find the multiplicative inverse of $65$ modulo $824$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{824}$).
(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 inve... | [
{
"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 $\\gcd(65,824)=1$, 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: $\\boxed{393}$.\nMethod 1 constructs an inverse via Bézout, producing $x=393$. 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... | 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 $\boxed{393}$.) |
math-011305 | Number Theory: gcd — Euclidean Algorithm | 6 | Do not skip justification steps: (a) Compute $\gcd(277,110)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 277+v\cdot 110=\gcd(277,110)$.
(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 $u=-27$ and $v=68$ with $u277+v... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-27,v=68$ satisfies $u277+v110=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: 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 $ua+vb=g$. |
math-011306 | Number Theory: Modular Inverses — Extended Euclid | 6 | Try to avoid pattern-matching; explain why: Find the multiplicative inverse of $85$ modulo $209$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{209}$).
(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": "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 $\\gcd(85,209)=1$, 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: $\\boxed{150}$.\nMethod 1 constructs an inverse via Bézout, producing $x=150$. 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. (Here the result is $\boxed{150}$.) |
math-011307 | Number Theory: Modular Inverses — Extended Euclid | 6 | Exercise: Find the multiplicative inverse of $741$ modulo $1714$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1714}$).
(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": "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 $x=1219$ and compute $741x=903279$.",
"Step 2: Reduce: $903279\\equiv 1\\pmod{1714}$ (since $903278=903278$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1219}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1219$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensit... | [
{
"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 $\boxed{1219}$.) |
math-011308 | Number Theory: gcd — Back Substitution | 6 | Answer with a short justification: (a) Compute $\gcd(1465,1030)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1465+v\cdot 1030=\gcd(1465,1030)$.
(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 $u=45$ and $v=-64$ with $u1465+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5}$.\nThe Euclidean algorithm computes $g=5$. The Bézout certificate $u=45,v=-64$ satisfies $u1465+v1030=5$, 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.",
... | 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 $\boxed{5}$.) |
math-011309 | Number Theory: Modular Inverses — Extended Euclid | 6 | Problem: Find the multiplicative inverse of $91$ modulo $101$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{101}$).
(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 $\\gcd(91,101)=1$, 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: $\\boxed{10}$.\nMethod 1 constructs an inverse via Bézout, producing $x=10$. 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... | 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 $\boxed{10}$.) |
math-011310 | Number Theory: gcd — Euclidean Algorithm | 6 | Work carefully and justify each inference: (a) Compute $\gcd(1864,535)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1864+v\cdot 535=\gcd(1864,535)$.
(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 $u=-126$ and $v=439$ with $u186... | {
"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 $u=-126,v=439$ satisfies $u1864+v535=1$, 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.",
... | 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-011311 | Number Theory: Bézout Identity — Certificates | 6 | Checkpoint: (a) Compute $\gcd(261,581)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 261+v\cdot 581=\gcd(261,581)$.
(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 $(261,581)$ to compute $g=\\gcd(261,581)$.",
"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 $u=256,v=-115$ satisfies $u261+v581=1$, 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.",
... | 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-011312 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Give a theorem-based solution: Find the multiplicative inverse of $809$ modulo $1420$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1420}$).
(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 $x=1169$ and compute $809x=945721$.",
"Step 2: Reduce: $945721\\equiv 1\\pmod{1420}$ (since $945720=945720$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1169}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1169$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended E... | [
{
"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 $\boxed{1169}$.) |
math-011313 | Computational Number Theory: Extended Euclid | 6 | Challenge: (a) Compute $\gcd(1657,1832)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1657+v\cdot 1832=\gcd(1657,1832)$.
(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": "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 $u=-335$ and $v=303$ with $u165... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-335,v=303$ satisfies $u1657+v1832=1$, 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.",
... | 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-011314 | Number Theory: Divisibility — Linear Combinations | 6 | Determine the requested value: (a) Compute $\gcd(289,1144)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 289+v\cdot 1144=\gcd(289,1144)$.
(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 $(289,1144)$ to compute $g=\\gcd(289,1144)$.",
"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 $u=-95,v=24$ satisfies $u289+v1144=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustnes... | [
{
"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-011315 | Number Theory: gcd — Back Substitution | 6 | Work this out carefully: (a) Compute $\gcd(1783,1512)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1783+v\cdot 1512=\gcd(1783,1512)$.
(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 $u=703$ and $v=-829$ with $u178... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=703,v=-829$ satisfies $u1783+v1512=1$, 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.",
... | 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-011316 | Number Theory: gcd — Back Substitution | 6 | Challenge: (a) Compute $\gcd(1359,1055)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1359+v\cdot 1055=\gcd(1359,1055)$.
(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 $(1359,1055)$ to compute $g=\\gcd(1359,1055)$.",
"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 $u=59,v=-76$ satisfies $u1359+v1055=1$, 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.",
... | 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-011317 | Computational Number Theory: Inverses and Certificates | 6 | Warm-up: Find the multiplicative inverse of $567$ modulo $806$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{806}$).
(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 $x=317$ and compute $567x=179739$.",
"Step 2: Reduce: $179739\\equiv 1\\pmod{806}$ (since $179738=179738$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{317}$.\nMethod 1 constructs an inverse via Bézout, producing $x=317$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness 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 $\boxed{317}$.) |
math-011318 | Number Theory: gcd — Back Substitution | 6 | Give an answer and a quick verification: (a) Compute $\gcd(1344,1237)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1344+v\cdot 1237=\gcd(1344,1237)$.
(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 $(1344,1237)$ to compute $g=\\gcd(1344,1237)$.",
"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 $u=474,v=-515$ satisfies $u1344+v1237=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness 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.",
... | 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-011319 | Number Theory: Units mod m — Existence Condition | 6 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $295$ modulo $663$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{663}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficie... | [
{
"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 $x=445$ and compute $295x=131275$.",
"Step 2: Reduce: $131275\\equiv 1\\pmod{663}$ (since $131274=131274$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{445}$.\nMethod 1 constructs an inverse via Bézout, producing $x=445$. 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 $\boxed{445}$.) |
math-011320 | Number Theory: Divisibility — Linear Combinations | 6 | Give a theorem-based solution: (a) Compute $\gcd(1849,925)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1849+v\cdot 925=\gcd(1849,925)$.
(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 $u=-1$ and $v=2$ with $u1849+v9... | {
"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 $u=-1,v=2$ satisfies $u1849+v925=1$, 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.",
... | 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-011321 | Number Theory: gcd — Euclidean Algorithm | 6 | Use two approaches if possible: (a) Compute $\gcd(718,121)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 718+v\cdot 121=\gcd(718,121)$.
(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 $(718,121)$ to compute $g=\\gcd(718,121)$.",
"Step 2: Record the remainder e... | {
"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 $u=15,v=-89$ satisfies $u718+v121=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "I... | [
{
"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-011322 | Number Theory: Bézout Identity — Certificates | 6 | Solve and sanity-check: (a) Compute $\gcd(1788,873)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1788+v\cdot 873=\gcd(1788,873)$.
(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": "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 $u=104$ and $v=-213$ with $u178... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=104,v=-213$ satisfies $u1788+v873=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: 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 $ua+vb=g$. |
math-011323 | Number Theory: gcd — Euclidean Algorithm | 6 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(220,1033)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 220+v\cdot 1033=\gcd(220,1033)$.
(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 $(220,1033)$ to compute $g=\\gcd(220,1033)$.",
"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 $u=108,v=-23$ satisfies $u220+v1033=1$, 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.",
... | 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-011324 | Number Theory: gcd — Back Substitution | 6 | Track units/moduli carefully: (a) Compute $\gcd(522,738)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 522+v\cdot 738=\gcd(522,738)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inc... | [
{
"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 $(522,738)$ to compute $g=\\gcd(522,738)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{18}$.\nThe Euclidean algorithm computes $g=18$. The Bézout certificate $u=17,v=-12$ satisfies $u522+v738=18$, 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 $\boxed{18}$.) |
math-011325 | Computational Number Theory: Inverses and Certificates | 6 | Challenge: Find the multiplicative inverse of $670$ modulo $889$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{889}$).
(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 $x=479$ and compute $670x=320930$.",
"Step 2: Reduce: $320930\\equiv 1\\pmod{889}$ (since $320929=320929$ is d... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{479}$.\nMethod 1 constructs an inverse via Bézout, producing $x=479$. 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 $\boxed{479}$.) |
math-011326 | Number Theory: Modular Inverses — Extended Euclid | 6 | Work this out carefully: Find the multiplicative inverse of $35$ modulo $554$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{554}$).
(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 $\\gcd(35,554)=1$, 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: $\\boxed{95}$.\nMethod 1 constructs an inverse via Bézout, producing $x=95$. 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... | 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 $\boxed{95}$.) |
math-011327 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $246$ modulo $493$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{493}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessar... | [
{
"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 $\\gcd(246,493)=1$, 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: $\\boxed{491}$.\nMethod 1 constructs an inverse via Bézout, producing $x=491$. 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... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011328 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve and include a self-check: Find the multiplicative inverse of $282$ modulo $509$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{509}$).
(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 $\\gcd(282,509)=1$, 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: $\\boxed{435}$.\nMethod 1 constructs an inverse via Bézout, producing $x=435$. 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 $\boxed{435}$.) |
math-011329 | Number Theory: Bézout Identity — Certificates | 6 | Solve (and briefly cross-validate): (a) Compute $\gcd(498,129)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 498+v\cdot 129=\gcd(498,129)$.
(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 $u=7$ and $v=-27$ with $u498+v1... | {
"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 $u=7,v=-27$ satisfies $u498+v129=3$, 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 $ua+vb=g$. |
math-011330 | Number Theory: gcd — Back Substitution | 6 | State any required conditions first: (a) Compute $\gcd(946,1707)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 946+v\cdot 1707=\gcd(946,1707)$.
(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 $(946,1707)$ to compute $g=\\gcd(946,1707)$.",
"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 $u=406,v=-225$ satisfies $u946+v1707=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: 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 $ua+vb=g$. |
math-011331 | Computational Number Theory: Inverses and Certificates | 6 | Carefully track domains: Find the multiplicative inverse of $1159$ modulo $1610$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1610}$).
(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": "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 $x=639$ and compute $1159x=740601$.",
"Step 2: Reduce: $740601\\equiv 1\\pmod{1610}$ (since $740600=740600$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{639}$.\nMethod 1 constructs an inverse via Bézout, producing $x=639$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"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 $\boxed{639}$.) |
math-011332 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $1383$ modulo $1396$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1396}$).
(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": "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 $\\gcd(1383,1396)=1$, 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: $\\boxed{859}$.\nMethod 1 constructs an inverse via Bézout, producing $x=859$. 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... | 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 $\boxed{859}$.) |
math-011333 | Number Theory: Bézout Identity — Certificates | 6 | Start by stating any domain restrictions: (a) Compute $\gcd(473,1702)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 473+v\cdot 1702=\gcd(473,1702)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitut... | [
{
"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 $(473,1702)$ to compute $g=\\gcd(473,1702)$.",
"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 $u=-421,v=117$ satisfies $u473+v1702=1$, 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-011334 | Number Theory: Bézout Identity — Certificates | 6 | Compute the requested quantity: (a) Compute $\gcd(347,1978)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 347+v\cdot 1978=\gcd(347,1978)$.
(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 $(347,1978)$ to compute $g=\\gcd(347,1978)$.",
"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 $u=-57,v=10$ satisfies $u347+v1978=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: 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.",
... | 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-011335 | Number Theory: Bézout Identity — Certificates | 6 | Work this out carefully: (a) Compute $\gcd(638,900)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 638+v\cdot 900=\gcd(638,900)$.
(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": "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 $u=79$ and $v=-56$ with $u638+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 $u=79,v=-56$ satisfies $u638+v900=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "R... | [
{
"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-011336 | Number Theory: gcd — Euclidean Algorithm | 6 | Determine the requested value: (a) Compute $\gcd(554,391)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 554+v\cdot 391=\gcd(554,391)$.
(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 $(554,391)$ to compute $g=\\gcd(554,391)$.",
"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 $u=12,v=-17$ satisfies $u554+v391=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivit... | [
{
"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-011337 | Number Theory: gcd — Back Substitution | 6 | Explain each transformation: (a) Compute $\gcd(1181,641)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1181+v\cdot 641=\gcd(1181,641)$.
(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 $(1181,641)$ to compute $g=\\gcd(1181,641)$.",
"Step 2: Record the remainder... | {
"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 $u=165,v=-304$ satisfies $u1181+v641=1$, 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-011338 | Number Theory: gcd — Euclidean Algorithm | 6 | Explain why your operations are valid: (a) Compute $\gcd(213,1898)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 213+v\cdot 1898=\gcd(213,1898)$.
(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 $u=-499$ and $v=56$ with $u213+... | {
"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 $u=-499,v=56$ satisfies $u213+v1898=1$, 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-011339 | Number Theory: Modular Inverses — Extended Euclid | 6 | Keep the final answer in boxed form: Find the multiplicative inverse of $31$ modulo $81$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{81}$).
(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": "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 $x=34$ and compute $31x=1054$.",
"Step 2: Reduce: $1054\\equiv 1\\pmod{81}$ (since $1053=1053$ is divisible by... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{34}$.\nMethod 1 constructs an inverse via Bézout, producing $x=34$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness 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... | 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 $\boxed{34}$.) |
math-011340 | Number Theory: gcd — Back Substitution | 6 | Complete the analysis: (a) Compute $\gcd(1146,1722)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1146+v\cdot 1722=\gcd(1146,1722)$.
(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": "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 $(1146,1722)$ to compute $g=\\gcd(1146,1722)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6}$.\nThe Euclidean algorithm computes $g=6$. The Bézout certificate $u=-3,v=2$ satisfies $u1146+v1722=6$, 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 $\boxed{6}$.) |
math-011341 | Computational Number Theory: Extended Euclid | 6 | Answer using clear logical steps: (a) Compute $\gcd(1255,1991)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1255+v\cdot 1991=\gcd(1255,1991)$.
(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 $(1255,1991)$ to compute $g=\\gcd(1255,1991)$.",
"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 $u=-890,v=561$ satisfies $u1255+v1991=1$, 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 $ua+vb=g$. |
math-011342 | Computational Number Theory: Inverses and Certificates | 6 | Challenge: Find the multiplicative inverse of $223$ modulo $903$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{903}$).
(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": "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 $\\gcd(223,903)=1$, 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: $\\boxed{328}$.\nMethod 1 constructs an inverse via Bézout, producing $x=328$. 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 $\boxed{328}$.) |
math-011343 | Number Theory: Units mod m — Existence Condition | 6 | Explain what is being counted/optimized: Find the multiplicative inverse of $289$ modulo $483$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{483}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"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 $289x=104329$.",
"Step 2: Reduce: $104329\\equiv 1\\pmod{483}$ (since $104328=104328$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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: ... | [
{
"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-011344 | Number Theory: Divisibility — Linear Combinations | 6 | Solve with verification: (a) Compute $\gcd(554,689)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 554+v\cdot 689=\gcd(554,689)$.
(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": "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 $v=-119$ with $u554... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=148,v=-119$ satisfies $u554+v689=1$, 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.",
... | 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-011345 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $364$ modulo $1559$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1559}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necess... | [
{
"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 $\\gcd(364,1559)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{227}$.\nMethod 1 constructs an inverse via Bézout, producing $x=227$. 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... | 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 $\boxed{227}$.) |
math-011346 | Number Theory: Units mod m — Existence Condition | 6 | Give a fully justified solution: Find the multiplicative inverse of $73$ modulo $280$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{280}$).
(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 $\\gcd(73,280)=1$, 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: $\\boxed{257}$.\nMethod 1 constructs an inverse via Bézout, producing $x=257$. 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-011347 | Number Theory: gcd — Euclidean Algorithm | 6 | Solve and then verify: (a) Compute $\gcd(207,229)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 207+v\cdot 229=\gcd(207,229)$.
(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 ... | [
{
"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 $(207,229)$ to compute $g=\\gcd(207,229)$.",
"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 $u=52,v=-47$ satisfies $u207+v229=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: 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 $ua+vb=g$. |
math-011348 | Number Theory: Modular Inverses — Extended Euclid | 6 | Problem: Find the multiplicative inverse of $1221$ modulo $1717$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1717}$).
(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 $\\gcd(1221,1717)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{45}$.\nMethod 1 constructs an inverse via Bézout, producing $x=45$. 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... | 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 $\boxed{45}$.) |
math-011349 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Checkpoint: Find the multiplicative inverse of $647$ modulo $1974$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1974}$).
(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 ... | [
{
"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 $x=299$ and compute $647x=193453$.",
"Step 2: Reduce: $193453\\equiv 1\\pmod{1974}$ (since $193452=193452$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{299}$.\nMethod 1 constructs an inverse via Bézout, producing $x=299$. 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... | 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 $\boxed{299}$.) |
math-011350 | Number Theory: Units mod m — Existence Condition | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $256$ modulo $361$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{361}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessar... | [
{
"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 $\\gcd(256,361)=1$, 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: $\\boxed{55}$.\nMethod 1 constructs an inverse via Bézout, producing $x=55$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011351 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $947$ modulo $1014$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1014}$).
(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 $\\gcd(947,1014)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{227}$.\nMethod 1 constructs an inverse via Bézout, producing $x=227$. 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. |
math-011352 | Number Theory: Units mod m — Existence Condition | 6 | Solve with verification: Find the multiplicative inverse of $209$ modulo $1085$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1085}$).
(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 $x=244$ and compute $209x=50996$.",
"Step 2: Reduce: $50996\\equiv 1\\pmod{1085}$ (since $50995=50995$ is divi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{244}$.\nMethod 1 constructs an inverse via Bézout, producing $x=244$. 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... | 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 $\boxed{244}$.) |
math-011353 | Computational Number Theory: Extended Euclid | 6 | Write the solution set clearly: (a) Compute $\gcd(1254,386)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1254+v\cdot 386=\gcd(1254,386)$.
(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 $u=-4$ and $v=13$ with $u1254+v... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-4,v=13$ satisfies $u1254+v386=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturb... | [
{
"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-011354 | Computational Number Theory: Inverses and Certificates | 6 | Give an answer and a quick verification: Find the multiplicative inverse of $747$ modulo $1682$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1682}$).
(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": "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 $\\gcd(747,1682)=1$, 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: $\\boxed{671}$.\nMethod 1 constructs an inverse via Bézout, producing $x=671$. 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... | 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 $\boxed{671}$.) |
math-011355 | Number Theory: Units mod m — Existence Condition | 6 | Use two approaches if possible: Find the multiplicative inverse of $64$ modulo $195$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{195}$).
(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 $\\gcd(64,195)=1$, 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: $\\boxed{64}$.\nMethod 1 constructs an inverse via Bézout, producing $x=64$. 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... | 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 $\boxed{64}$.) |
math-011356 | Number Theory: gcd — Euclidean Algorithm | 6 | Derive the result step-by-step: (a) Compute $\gcd(1521,1243)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1521+v\cdot 1243=\gcd(1521,1243)$.
(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": "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 $u=389$ and $v=-476$ with $u152... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=389,v=-476$ satisfies $u1521+v1243=1$, 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.",
... | 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-011357 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $72$ modulo $1433$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1433}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessa... | [
{
"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 $x=617$ and compute $72x=44424$.",
"Step 2: Reduce: $44424\\equiv 1\\pmod{1433}$ (since $44423=44423$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{617}$.\nMethod 1 constructs an inverse via Bézout, producing $x=617$. 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 $\boxed{617}$.) |
math-011358 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Keep the final answer in boxed form: Find the multiplicative inverse of $757$ modulo $1569$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1569}$).
(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 $x=1084$ and compute $757x=820588$.",
"Step 2: Reduce: $820588\\equiv 1\\pmod{1569}$ (since $820587=820587$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1084}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1084$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: E... | [
{
"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 $\boxed{1084}$.) |
math-011359 | Computational Number Theory: Inverses and Certificates | 6 | Checkpoint: Find the multiplicative inverse of $2$ modulo $431$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{431}$).
(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": "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 $\\gcd(2,431)=1$, 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: $\\boxed{216}$.\nMethod 1 constructs an inverse via Bézout, producing $x=216$. 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... | 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 $\boxed{216}$.) |
math-011360 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Explain what is being counted/optimized: Find the multiplicative inverse of $163$ modulo $544$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{544}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"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 $\\gcd(163,544)=1$, 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: $\\boxed{267}$.\nMethod 1 constructs an inverse via Bézout, producing $x=267$. 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 $\boxed{267}$.) |
math-011361 | Number Theory: Bézout Identity — Certificates | 6 | Problem: (a) Compute $\gcd(1762,1812)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1762+v\cdot 1812=\gcd(1762,1812)$.
(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 ve... | [
{
"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 $u=-145$ and $v=141$ with $u176... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-145,v=141$ satisfies $u1762+v1812=2$, 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.",
... | 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-011362 | Number Theory: gcd — Back Substitution | 6 | Start by stating any domain restrictions: (a) Compute $\gcd(447,1651)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 447+v\cdot 1651=\gcd(447,1651)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitut... | [
{
"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 $(447,1651)$ to compute $g=\\gcd(447,1651)$.",
"Step 2: Record the remainder... | {
"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 $u=229,v=-62$ satisfies $u447+v1651=1$, 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-011363 | Number Theory: Bézout Identity — Certificates | 6 | Explain why your operations are valid: (a) Compute $\gcd(776,1711)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 776+v\cdot 1711=\gcd(776,1711)$.
(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 $(776,1711)$ to compute $g=\\gcd(776,1711)$.",
"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 $u=-721,v=327$ satisfies $u776+v1711=1$, 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.",
... | 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-011364 | Number Theory: Modular Inverses — Extended Euclid | 6 | Give a theorem-based solution: Find the multiplicative inverse of $541$ modulo $637$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{637}$).
(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 $x=564$ and compute $541x=305124$.",
"Step 2: Reduce: $305124\\equiv 1\\pmod{637}$ (since $305123=305123$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{564}$.\nMethod 1 constructs an inverse via Bézout, producing $x=564$. 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. |
math-011365 | Number Theory: Divisibility — Linear Combinations | 6 | Solve and justify each step: (a) Compute $\gcd(1986,878)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1986+v\cdot 878=\gcd(1986,878)$.
(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 $(1986,878)$ to compute $g=\\gcd(1986,878)$.",
"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 $u=42,v=-95$ satisfies $u1986+v878=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: 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.",
... | 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-011366 | Computational Number Theory: Inverses and Certificates | 6 | Show all reasoning: Find the multiplicative inverse of $525$ modulo $1343$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1343}$).
(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 $x=110$ and compute $525x=57750$.",
"Step 2: Reduce: $57750\\equiv 1\\pmod{1343}$ (since $57749=57749$ is divi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{110}$.\nMethod 1 constructs an inverse via Bézout, producing $x=110$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"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-011367 | Computational Number Theory: Inverses and Certificates | 6 | Compute the requested quantity: Find the multiplicative inverse of $813$ modulo $1640$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1640}$).
(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": "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 $x=117$ and compute $813x=95121$.",
"Step 2: Reduce: $95121\\equiv 1\\pmod{1640}$ (since $95120=95120$ is divi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{117}$.\nMethod 1 constructs an inverse via Bézout, producing $x=117$. 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... | 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 $\boxed{117}$.) |
math-011368 | Number Theory: Units mod m — Existence Condition | 6 | Give an answer and a quick verification: Find the multiplicative inverse of $1062$ modulo $1373$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1373}$).
(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 $\\gcd(1062,1373)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{777}$.\nMethod 1 constructs an inverse via Bézout, producing $x=777$. 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. |
math-011369 | Number Theory: Bézout Identity — Certificates | 6 | Carefully track domains: (a) Compute $\gcd(1424,1147)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1424+v\cdot 1147=\gcd(1424,1147)$.
(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 $(1424,1147)$ to compute $g=\\gcd(1424,1147)$.",
"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 $u=294,v=-365$ satisfies $u1424+v1147=1$, 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.",
... | 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-011370 | Computational Number Theory: Inverses and Certificates | 6 | Problem: Find the multiplicative inverse of $159$ modulo $511$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{511}$).
(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 $\\gcd(159,511)=1$, 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: $\\boxed{45}$.\nMethod 1 constructs an inverse via Bézout, producing $x=45$. 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... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011371 | Number Theory: gcd — Euclidean Algorithm | 6 | Provide a rigorous solution: (a) Compute $\gcd(1311,912)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1311+v\cdot 912=\gcd(1311,912)$.
(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 $u=7$ and $v=-10$ with $u1311+v... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{57}$.\nThe Euclidean algorithm computes $g=57$. The Bézout certificate $u=7,v=-10$ satisfies $u1311+v912=57$, 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.",
... | 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-011372 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Question: Find the multiplicative inverse of $807$ modulo $1504$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1504}$).
(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": "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 $x=1463$ and compute $807x=1180641$.",
"Step 2: Reduce: $1180641\\equiv 1\\pmod{1504}$ (since $1180640=1180640... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1463}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1463$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extende... | [
{
"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 $\boxed{1463}$.) |
math-011373 | Computational Number Theory: Inverses and Certificates | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $498$ modulo $1471$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1471}$).
(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 $x=192$ and compute $498x=95616$.",
"Step 2: Reduce: $95616\\equiv 1\\pmod{1471}$ (since $95615=95615$ is divi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{192}$.\nMethod 1 constructs an inverse via Bézout, producing $x=192$. 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... | 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 $\boxed{192}$.) |
math-011374 | Computational Number Theory: Extended Euclid | 6 | Give an answer and a quick verification: (a) Compute $\gcd(527,1976)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 527+v\cdot 1976=\gcd(527,1976)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substituti... | [
{
"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 $(527,1976)$ to compute $g=\\gcd(527,1976)$.",
"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 $u=15,v=-4$ satisfies $u527+v1976=1$, 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.",
... | 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-011375 | Number Theory: Units mod m — Existence Condition | 6 | Compute the requested quantity: Find the multiplicative inverse of $229$ modulo $659$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{659}$).
(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 $x=259$ and compute $229x=59311$.",
"Step 2: Reduce: $59311\\equiv 1\\pmod{659}$ (since $59310=59310$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{259}$.\nMethod 1 constructs an inverse via Bézout, producing $x=259$. 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... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011376 | Number Theory: Divisibility — Linear Combinations | 6 | Exercise: (a) Compute $\gcd(970,1312)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 970+v\cdot 1312=\gcd(970,1312)$.
(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": "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 $u=-211$ and $v=156$ with $u970... | {
"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 $u=-211,v=156$ satisfies $u970+v1312=2$, 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.",
... | 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-011377 | Number Theory: gcd — Euclidean Algorithm | 6 | Be explicit about assumptions: (a) Compute $\gcd(516,986)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 516+v\cdot 986=\gcd(516,986)$.
(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 $(516,986)$ to compute $g=\\gcd(516,986)$.",
"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 $u=193,v=-101$ satisfies $u516+v986=2$, 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.",
... | 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-011378 | Number Theory: Units mod m — Existence Condition | 6 | Task: Find the multiplicative inverse of $1725$ modulo $1819$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1819}$).
(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 $x=1103$ and compute $1725x=1902675$.",
"Step 2: Reduce: $1902675\\equiv 1\\pmod{1819}$ (since $1902674=190267... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1103}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1103$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensit... | [
{
"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-011379 | Computational Number Theory: Extended Euclid | 6 | Where appropriate, name the theorem you use: (a) Compute $\gcd(1309,1227)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1309+v\cdot 1227=\gcd(1309,1227)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-sub... | [
{
"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 $(1309,1227)$ to compute $g=\\gcd(1309,1227)$.",
"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 $u=-404,v=431$ satisfies $u1309+v1227=1$, 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.",
... | 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-011380 | Number Theory: gcd — Back Substitution | 6 | Solve and sanity-check: (a) Compute $\gcd(1989,1843)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1989+v\cdot 1843=\gcd(1989,1843)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inc... | [
{
"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 $(1989,1843)$ to compute $g=\\gcd(1989,1843)$.",
"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 $u=-871,v=940$ satisfies $u1989+v1843=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011381 | Number Theory: gcd — Back Substitution | 6 | Solve and include a self-check: (a) Compute $\gcd(1758,1424)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1758+v\cdot 1424=\gcd(1758,1424)$.
(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": "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 $u=-81$ and $v=100$ with $u1758... | {
"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 $u=-81,v=100$ satisfies $u1758+v1424=2$, 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.",
... | 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-011382 | Number Theory: gcd — Euclidean Algorithm | 6 | Show all reasoning: (a) Compute $\gcd(1341,853)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1341+v\cdot 853=\gcd(1341,853)$.
(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 ... | [
{
"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 $u=215$ and $v=-338$ with $u134... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=215,v=-338$ satisfies $u1341+v853=1$, 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.",
... | 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-011383 | Computational Number Theory: Inverses and Certificates | 6 | State any required conditions first: Find the multiplicative inverse of $655$ modulo $838$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{838}$).
(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": "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 $x=87$ and compute $655x=56985$.",
"Step 2: Reduce: $56985\\equiv 1\\pmod{838}$ (since $56984=56984$ is divisi... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{87}$.\nMethod 1 constructs an inverse via Bézout, producing $x=87$. 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... | 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 $\boxed{87}$.) |
math-011384 | Computational Number Theory: Inverses and Certificates | 6 | Answer with a short justification: Find the multiplicative inverse of $559$ modulo $1858$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1858}$).
(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": "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 $x=349$ and compute $559x=195091$.",
"Step 2: Reduce: $195091\\equiv 1\\pmod{1858}$ (since $195090=195090$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{349}$.\nMethod 1 constructs an inverse via Bézout, producing $x=349$. 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... | 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 $\boxed{349}$.) |
math-011385 | Computational Number Theory: Extended Euclid | 6 | Give reasoning, not just computation: (a) Compute $\gcd(614,1670)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 614+v\cdot 1670=\gcd(614,1670)$.
(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 $u=68$ and $v=-25$ with $u614+v... | {
"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 $u=68,v=-25$ satisfies $u614+v1670=2$, 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.",
... | 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-011386 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Try to avoid pattern-matching; explain why: Find the multiplicative inverse of $811$ modulo $1632$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1632}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffici... | [
{
"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 $x=163$ and compute $811x=132193$.",
"Step 2: Reduce: $132193\\equiv 1\\pmod{1632}$ (since $132192=132192$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{163}$.\nMethod 1 constructs an inverse via Bézout, producing $x=163$. 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-011387 | Number Theory: gcd — Euclidean Algorithm | 6 | Question: (a) Compute $\gcd(1703,161)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1703+v\cdot 161=\gcd(1703,161)$.
(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": "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 $u=-45$ and $v=476$ with $u1703... | {
"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 $u=-45,v=476$ satisfies $u1703+v161=1$, 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.",
... | 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-011388 | Computational Number Theory: Inverses and Certificates | 6 | Use two approaches if possible: Find the multiplicative inverse of $248$ modulo $381$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{381}$).
(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 $\\gcd(248,381)=1$, 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: $\\boxed{275}$.\nMethod 1 constructs an inverse via Bézout, producing $x=275$. 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... | 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 $\boxed{275}$.) |
math-011389 | Number Theory: Divisibility — Linear Combinations | 6 | State any required conditions first: (a) Compute $\gcd(488,1426)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 488+v\cdot 1426=\gcd(488,1426)$.
(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 $(488,1426)$ to compute $g=\\gcd(488,1426)$.",
"Step 2: Record the remainder... | {
"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 $u=-225,v=77$ satisfies $u488+v1426=2$, 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.",
... | 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-011390 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Task: Find the multiplicative inverse of $1621$ modulo $1775$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1775}$).
(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 $x=1556$ and compute $1621x=2522276$.",
"Step 2: Reduce: $2522276\\equiv 1\\pmod{1775}$ (since $2522275=252227... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1556}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1556$. 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. |
math-011391 | Number Theory: Bézout Identity — Certificates | 6 | Solve and justify each step: (a) Compute $\gcd(605,1223)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 605+v\cdot 1223=\gcd(605,1223)$.
(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 $(605,1223)$ to compute $g=\\gcd(605,1223)$.",
"Step 2: Record the remainder... | {
"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 $u=188,v=-93$ satisfies $u605+v1223=1$, 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.",
... | 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-011392 | Number Theory: gcd — Back Substitution | 6 | Answer with a short justification: (a) Compute $\gcd(614,1703)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 614+v\cdot 1703=\gcd(614,1703)$.
(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": "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 $u=-147$ and $v=53$ with $u614+... | {
"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 $u=-147,v=53$ satisfies $u614+v1703=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitiv... | [
{
"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-011393 | Number Theory: Divisibility — Linear Combinations | 6 | Prompt: (a) Compute $\gcd(430,1628)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 430+v\cdot 1628=\gcd(430,1628)$.
(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 verifi... | [
{
"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 $(430,1628)$ to compute $g=\\gcd(430,1628)$.",
"Step 2: Record the remainder... | {
"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 $u=-53,v=14$ satisfies $u430+v1628=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustnes... | [
{
"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-011394 | Number Theory: Bézout Identity — Certificates | 6 | Keep the final answer in boxed form: (a) Compute $\gcd(256,1428)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 256+v\cdot 1428=\gcd(256,1428)$.
(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 $(256,1428)$ to compute $g=\\gcd(256,1428)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nThe Euclidean algorithm computes $g=4$. The Bézout certificate $u=106,v=-19$ satisfies $u256+v1428=4$, 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.",
... | 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-011395 | Number Theory: Modular Inverses — Extended Euclid | 6 | Challenge: Find the multiplicative inverse of $100$ modulo $417$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{417}$).
(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": "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 $\\gcd(100,417)=1$, 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: $\\boxed{196}$.\nMethod 1 constructs an inverse via Bézout, producing $x=196$. 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... | 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 $\boxed{196}$.) |
math-011396 | Computational Number Theory: Extended Euclid | 6 | Solve and justify each step: (a) Compute $\gcd(1073,1830)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1073+v\cdot 1830=\gcd(1073,1830)$.
(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 $(1073,1830)$ to compute $g=\\gcd(1073,1830)$.",
"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 $u=527,v=-309$ satisfies $u1073+v1830=1$, 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-011397 | Computational Number Theory: Extended Euclid | 6 | Track units/moduli carefully: (a) Compute $\gcd(90,443)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 90+v\cdot 443=\gcd(90,443)$.
(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": "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 $u=64$ and $v=-13$ with $u90+v4... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=64,v=-13$ satisfies $u90+v443=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: 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 $ua+vb=g$. |
math-011398 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $77$ modulo $79$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{79}$).
(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 $\\gcd(77,79)=1$, 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: $\\boxed{39}$.\nMethod 1 constructs an inverse via Bézout, producing $x=39$. 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... | 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 $\boxed{39}$.) |
math-011399 | Number Theory: Divisibility — Linear Combinations | 6 | Keep the final answer in boxed form: (a) Compute $\gcd(344,364)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 344+v\cdot 364=\gcd(344,364)$.
(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": "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 $u=18$ and $v=-17$ with $u344+v... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\nThe Euclidean algorithm computes $g=4$. The Bézout certificate $u=18,v=-17$ satisfies $u344+v364=4$, 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.",
... | 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 $\boxed{4}$.) |
math-011400 | Number Theory: gcd — Back Substitution | 6 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(1249,640)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1249+v\cdot 640=\gcd(1249,640)$.
(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 $(1249,640)$ to compute $g=\\gcd(1249,640)$.",
"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 $u=289,v=-564$ satisfies $u1249+v640=1$, 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.",
... | 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 $\boxed{1}$.) |
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