id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-013201 | Number Theory: Modular Inverses — Extended Euclid | 7 | Exercise: Find the multiplicative inverse of $19$ modulo $1002$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1002}$).
(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 $x=211$ and compute $19x=4009$.",
"Step 2: Reduce: $4009\\equiv 1\\pmod{1002}$ (since $4008=4008$ is divisible... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{211}$.\nMethod 1 constructs an inverse via Bézout, producing $x=211$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{211}$.) |
math-013202 | Number Theory: Units mod m — Existence Condition | 7 | Give a fully justified solution: Find the multiplicative inverse of $157$ modulo $719$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{719}$).
(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 $x=316$ and compute $157x=49612$.",
"Step 2: Reduce: $49612\\equiv 1\\pmod{719}$ (since $49611=49611$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{316}$.\nMethod 1 constructs an inverse via Bézout, producing $x=316$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{316}$.) |
math-013203 | Number Theory: Modular Inverses — Extended Euclid | 7 | Carefully track domains: Find the multiplicative inverse of $1347$ 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 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 $x=909$ and compute $1347x=1224423$.",
"Step 2: Reduce: $1224423\\equiv 1\\pmod{1858}$ (since $1224422=1224422... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{909}$.\nMethod 1 constructs an inverse via Bézout, producing $x=909$. 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. |
math-013204 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Find the exact value: Find the multiplicative inverse of $110$ modulo $1051$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1051}$).
(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": "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 $x=86$ and compute $110x=9460$.",
"Step 2: Reduce: $9460\\equiv 1\\pmod{1051}$ (since $9459=9459$ is divisible... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{86}$.\nMethod 1 constructs an inverse via Bézout, producing $x=86$. 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... | 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 $\boxed{86}$.) |
math-013205 | Number Theory: Bézout Identity — Certificates | 7 | Track quantifiers carefully: (a) Compute $\gcd(219,1797)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 219+v\cdot 1797=\gcd(219,1797)$.
(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 $u=279$ and $v=-34$ with $u219+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=279,v=-34$ satisfies $u219+v1797=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid s... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-013206 | Number Theory: Divisibility — Linear Combinations | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(1372,116)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1372+v\cdot 116=\gcd(1372,116)$.
(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 $(1372,116)$ to compute $g=\\gcd(1372,116)$.",
"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 $u=-6,v=71$ satisfies $u1372+v116=4$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid scal... | [
{
"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-013207 | Number Theory: Bézout Identity — Certificates | 7 | Track quantifiers carefully: (a) Compute $\gcd(1986,206)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1986+v\cdot 206=\gcd(1986,206)$.
(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 $(1986,206)$ to compute $g=\\gcd(1986,206)$.",
"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 $u=-39,v=376$ satisfies $u1986+v206=2$, 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.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013208 | Number Theory: Divisibility — Linear Combinations | 7 | Proceed methodically: (a) Compute $\gcd(1668,1822)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1668+v\cdot 1822=\gcd(1668,1822)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-71$ and $v=65$ with $u1668+... | {
"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 $u=-71,v=65$ satisfies $u1668+v1822=2$, 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.",
... | 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-013209 | Computational Number Theory: Extended Euclid | 7 | Work this out carefully: (a) Compute $\gcd(1346,1313)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1346+v\cdot 1313=\gcd(1346,1313)$.
(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 $(1346,1313)$ to compute $g=\\gcd(1346,1313)$.",
"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 $u=-557,v=571$ satisfies $u1346+v1313=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013210 | Number Theory: Modular Inverses — Extended Euclid | 7 | Problem: Find the multiplicative inverse of $73$ modulo $858$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{858}$).
(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 $\\gcd(73,858)=1$, 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: $\\boxed{811}$.\nMethod 1 constructs an inverse via Bézout, producing $x=811$. 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 $\boxed{811}$.) |
math-013211 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve and sanity-check: Find the multiplicative inverse of $1094$ modulo $1823$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1823}$).
(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": "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 $\\gcd(1094,1823)=1$, 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: $\\boxed{5}$.\nMethod 1 constructs an inverse via Bézout, producing $x=5$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity ... | [
{
"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 $\boxed{5}$.) |
math-013212 | Number Theory: gcd — Back Substitution | 7 | Do not skip justification steps: (a) Compute $\gcd(1700,704)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1700+v\cdot 704=\gcd(1700,704)$.
(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 $u=41$ and $v=-99$ with $u1700+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4}$.\nThe Euclidean algorithm computes $g=4$. The Bézout certificate $u=41,v=-99$ satisfies $u1700+v704=4$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid sca... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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 $\boxed{4}$.) |
math-013213 | Number Theory: Bézout Identity — Certificates | 7 | Track units/moduli carefully: (a) Compute $\gcd(128,1529)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 128+v\cdot 1529=\gcd(128,1529)$.
(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 $u=657$ and $v=-55$ with $u128+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=657,v=-55$ satisfies $u128+v1529=1$, 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.",
... | 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-013214 | Number Theory: Units mod m — Existence Condition | 7 | Give a fully justified solution: Find the multiplicative inverse of $673$ modulo $1006$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1006}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=719$ and compute $673x=483887$.",
"Step 2: Reduce: $483887\\equiv 1\\pmod{1006}$ (since $483886=483886$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{719}$.\nMethod 1 constructs an inverse via Bézout, producing $x=719$. 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 $\boxed{719}$.) |
math-013215 | Computational Number Theory: Inverses and Certificates | 7 | Solve and include a self-check: Find the multiplicative inverse of $44$ modulo $785$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{785}$).
(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 $\\gcd(44,785)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{339}$.\nMethod 1 constructs an inverse via Bézout, producing $x=339$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{339}$.) |
math-013216 | Number Theory: Units mod m — Existence Condition | 7 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $400$ modulo $1407$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1407}$).
(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 $\\gcd(400,1407)=1$, 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: $\\boxed{904}$.\nMethod 1 constructs an inverse via Bézout, producing $x=904$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{904}$.) |
math-013217 | Number Theory: gcd — Euclidean Algorithm | 7 | Do not skip justification steps: (a) Compute $\gcd(1627,590)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1627+v\cdot 590=\gcd(1627,590)$.
(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 $u=33$ and $v=-91$ with $u1627+... | {
"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 $u=33,v=-91$ satisfies $u1627+v590=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013218 | Computational Number Theory: Inverses and Certificates | 7 | Give an answer and a quick verification: Find the multiplicative inverse of $161$ modulo $1167$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1167}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=29$ and compute $161x=4669$.",
"Step 2: Reduce: $4669\\equiv 1\\pmod{1167}$ (since $4668=4668$ is divisible... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{29}$.\nMethod 1 constructs an inverse via Bézout, producing $x=29$. 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... | 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 $\boxed{29}$.) |
math-013219 | Number Theory: Modular Inverses — Extended Euclid | 7 | Problem: Find the multiplicative inverse of $93$ modulo $98$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{98}$).
(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 $x=39$ and compute $93x=3627$.",
"Step 2: Reduce: $3627\\equiv 1\\pmod{98}$ (since $3626=3626$ is divisible by... | {
"consistency_check": "The two methods are consistent and must coincide. 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": "Robustness note: Extended Euclid is ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013220 | Number Theory: Modular Inverses — Extended Euclid | 7 | State any required conditions first: Find the multiplicative inverse of $1826$ modulo $1979$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1979}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1384$ and compute $1826x=2527184$.",
"Step 2: Reduce: $2527184\\equiv 1\\pmod{1979}$ (since $2527183=252718... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1384}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1384$. 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. |
math-013221 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Track units/moduli carefully: Find the multiplicative inverse of $11$ 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 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 $x=170$ and compute $11x=1870$.",
"Step 2: Reduce: $1870\\equiv 1\\pmod{267}$ (since $1869=1869$ is divisible ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{170}$.\nMethod 1 constructs an inverse via Bézout, producing $x=170$. 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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013222 | Number Theory: Bézout Identity — Certificates | 7 | Answer using clear logical steps: (a) Compute $\gcd(428,1462)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 428+v\cdot 1462=\gcd(428,1462)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chai... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(428,1462)$ to compute $g=\\gcd(428,1462)$.",
"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 $u=-345,v=101$ satisfies $u428+v1462=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"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-013223 | Computational Number Theory: Extended Euclid | 7 | Complete the analysis: (a) Compute $\gcd(1540,743)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1540+v\cdot 743=\gcd(1540,743)$.
(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 $u=344$ and $v=-713$ with $u154... | {
"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 $u=344,v=-713$ satisfies $u1540+v743=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"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-013224 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Work this out carefully: Find the multiplicative inverse of $107$ modulo $157$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{157}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(107,157)=1$, 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: $\\boxed{135}$.\nMethod 1 constructs an inverse via Bézout, producing $x=135$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{135}$.) |
math-013225 | Number Theory: Divisibility — Linear Combinations | 7 | Exercise: (a) Compute $\gcd(1720,1290)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1720+v\cdot 1290=\gcd(1720,1290)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief v... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1720,1290)$ to compute $g=\\gcd(1720,1290)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{430}$.\nThe Euclidean algorithm computes $g=430$. The Bézout certificate $u=1,v=-1$ satisfies $u1720+v1290=430$, and divisibility shows no larger common divisor can exist.",
"robustness_analysi... | [
{
"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-013226 | Number Theory: Divisibility — Linear Combinations | 7 | Explain what is being counted/optimized: (a) Compute $\gcd(1418,1436)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1418+v\cdot 1436=\gcd(1418,1436)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substit... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=319$ and $v=-315$ with $u141... | {
"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 $u=319,v=-315$ satisfies $u1418+v1436=2$, 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 $ua+vb=g$. |
math-013227 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Solve and justify each step: Find the multiplicative inverse of $289$ modulo $545$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{545}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(289,545)=1$, 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: $\\boxed{479}$.\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": "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... | 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 $\boxed{479}$.) |
math-013228 | Number Theory: gcd — Back Substitution | 7 | Where appropriate, name the theorem you use: (a) Compute $\gcd(902,1092)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 902+v\cdot 1092=\gcd(902,1092)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substi... | [
{
"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 $(902,1092)$ to compute $g=\\gcd(902,1092)$.",
"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 $u=-23,v=19$ satisfies $u902+v1092=2$, 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 $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013229 | Number Theory: gcd — Euclidean Algorithm | 7 | Solve and justify each step: (a) Compute $\gcd(112,750)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 112+v\cdot 750=\gcd(112,750)$.
(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 $(112,750)$ to compute $g=\\gcd(112,750)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-154,v=23$ satisfies $u112+v750=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertur... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013230 | Number Theory: gcd — Euclidean Algorithm | 7 | Derive the result step-by-step: (a) Compute $\gcd(1880,1982)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1880+v\cdot 1982=\gcd(1880,1982)$.
(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 $u=136$ and $v=-129$ with $u188... | {
"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 $u=136,v=-129$ satisfies $u1880+v1982=2$, 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 $ua+vb=g$. |
math-013231 | Computational Number Theory: Inverses and Certificates | 7 | Solve with verification: Find the multiplicative inverse of $141$ modulo $1043$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1043}$).
(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 $x=540$ and compute $141x=76140$.",
"Step 2: Reduce: $76140\\equiv 1\\pmod{1043}$ (since $76139=76139$ is divi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{540}$.\nMethod 1 constructs an inverse via Bézout, producing $x=540$. 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... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013232 | Computational Number Theory: Extended Euclid | 7 | Give a fully justified solution: (a) Compute $\gcd(1398,1184)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1398+v\cdot 1184=\gcd(1398,1184)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=83$ and $v=-98$ with $u1398+... | {
"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 $u=83,v=-98$ satisfies $u1398+v1184=2$, 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.",
... | 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-013233 | Number Theory: gcd — Euclidean Algorithm | 7 | Compute the requested quantity: (a) Compute $\gcd(1968,1883)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1968+v\cdot 1883=\gcd(1968,1883)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution cha... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1968,1883)$ to compute $g=\\gcd(1968,1883)$.",
"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 $u=288,v=-301$ satisfies $u1968+v1883=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013234 | Number Theory: gcd — Euclidean Algorithm | 7 | Explain each transformation: (a) Compute $\gcd(1040,642)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1040+v\cdot 642=\gcd(1040,642)$.
(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 $u=-50$ and $v=81$ with $u1040+... | {
"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 $u=-50,v=81$ satisfies $u1040+v642=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivi... | [
{
"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-013235 | Number Theory: Divisibility — Linear Combinations | 7 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(417,1779)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 417+v\cdot 1779=\gcd(417,1779)$.
(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 $(417,1779)$ to compute $g=\\gcd(417,1779)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=64,v=-15$ satisfies $u417+v1779=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales 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.",
... | 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 $\boxed{3}$.) |
math-013236 | Computational Number Theory: Inverses and Certificates | 7 | Do not skip justification steps: Find the multiplicative inverse of $21$ modulo $200$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{200}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(21,200)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{181}$.\nMethod 1 constructs an inverse via Bézout, producing $x=181$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013237 | Number Theory: Bézout Identity — Certificates | 7 | Warm-up: (a) Compute $\gcd(995,1196)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 995+v\cdot 1196=\gcd(995,1196)$.
(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 $(995,1196)$ to compute $g=\\gcd(995,1196)$.",
"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 $u=119,v=-99$ satisfies $u995+v1196=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generali... | [
{
"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-013238 | Number Theory: Units mod m — Existence Condition | 7 | Carefully track domains: Find the multiplicative inverse of $556$ modulo $1255$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1255}$).
(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": "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 $\\gcd(556,1255)=1$, 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: $\\boxed{1176}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1176$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid... | [
{
"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 $\boxed{1176}$.) |
math-013239 | Number Theory: Bézout Identity — Certificates | 7 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(716,1573)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 716+v\cdot 1573=\gcd(716,1573)$.
(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 $(716,1573)$ to compute $g=\\gcd(716,1573)$.",
"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 $u=-714,v=325$ satisfies $u716+v1573=1$, 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.",
... | 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-013240 | Computational Number Theory: Inverses and Certificates | 7 | Explain why your operations are valid: Find the multiplicative inverse of $92$ modulo $121$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{121}$).
(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 $\\gcd(92,121)=1$, 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: $\\boxed{25}$.\nMethod 1 constructs an inverse via Bézout, producing $x=25$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Eucli... | [
{
"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 $\boxed{25}$.) |
math-013241 | Number Theory: Divisibility — Linear Combinations | 7 | Provide a rigorous solution: (a) Compute $\gcd(1117,510)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1117+v\cdot 510=\gcd(1117,510)$.
(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 $u=163$ and $v=-357$ with $u111... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=163,v=-357$ satisfies $u1117+v510=1$, 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.",
... | 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-013242 | Number Theory: Modular Inverses — Extended Euclid | 7 | Challenge: Find the multiplicative inverse of $1275$ modulo $1382$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1382}$).
(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": "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 $\\gcd(1275,1382)=1$, 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: $\\boxed{1227}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1227$. 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 $\boxed{1227}$.) |
math-013243 | Number Theory: Modular Inverses — Extended Euclid | 7 | Give a theorem-based solution: Find the multiplicative inverse of $44$ modulo $127$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{127}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=26$ and compute $44x=1144$.",
"Step 2: Reduce: $1144\\equiv 1\\pmod{127}$ (since $1143=1143$ is divisible b... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{26}$.\nMethod 1 constructs an inverse via Bézout, producing $x=26$. 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... | 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 $\boxed{26}$.) |
math-013244 | Number Theory: Modular Inverses — Extended Euclid | 7 | Complete the analysis: Find the multiplicative inverse of $204$ modulo $545$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{545}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an in... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(204,545)=1$, 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: $\\boxed{179}$.\nMethod 1 constructs an inverse via Bézout, producing $x=179$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{179}$.) |
math-013245 | Computational Number Theory: Inverses and Certificates | 7 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $118$ modulo $281$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{281}$).
(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": "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 $\\gcd(118,281)=1$, 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: $\\boxed{231}$.\nMethod 1 constructs an inverse via Bézout, producing $x=231$. 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... | 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 $\boxed{231}$.) |
math-013246 | Computational Number Theory: Inverses and Certificates | 7 | Solve and include a self-check: Find the multiplicative inverse of $891$ modulo $1964$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1964}$).
(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 $x=723$ and compute $891x=644193$.",
"Step 2: Reduce: $644193\\equiv 1\\pmod{1964}$ (since $644192=644192$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{723}$.\nMethod 1 constructs an inverse via Bézout, producing $x=723$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{723}$.) |
math-013247 | Computational Number Theory: Inverses and Certificates | 7 | Track units/moduli carefully: Find the multiplicative inverse of $838$ modulo $1089$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1089}$).
(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 $x=820$ and compute $838x=687160$.",
"Step 2: Reduce: $687160\\equiv 1\\pmod{1089}$ (since $687159=687159$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{820}$.\nMethod 1 constructs an inverse via Bézout, producing $x=820$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{820}$.) |
math-013248 | Computational Number Theory: Extended Euclid | 7 | Proceed methodically: (a) Compute $\gcd(1204,1071)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1204+v\cdot 1071=\gcd(1204,1071)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-8$ and $v=9$ with $u1204+v1... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{7}$.\nThe Euclidean algorithm computes $g=7$. The Bézout certificate $u=-8,v=9$ satisfies $u1204+v1071=7$, 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.",
... | 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 $\boxed{7}$.) |
math-013249 | Computational Number Theory: Inverses and Certificates | 7 | Solve and include a self-check: Find the multiplicative inverse of $8$ modulo $669$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{669}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=251$ and compute $8x=2008$.",
"Step 2: Reduce: $2008\\equiv 1\\pmod{669}$ (since $2007=2007$ is divisible b... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{251}$.\nMethod 1 constructs an inverse via Bézout, producing $x=251$. 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 $\boxed{251}$.) |
math-013250 | Number Theory: Modular Inverses — Extended Euclid | 7 | Task: Find the multiplicative inverse of $485$ modulo $799$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{799}$).
(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 $x=257$ and compute $485x=124645$.",
"Step 2: Reduce: $124645\\equiv 1\\pmod{799}$ (since $124644=124644$ is d... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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 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-013251 | Number Theory: Divisibility — Linear Combinations | 7 | Find the exact value: (a) Compute $\gcd(167,1272)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 167+v\cdot 1272=\gcd(167,1272)$.
(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 $u=-457$ and $v=60$ with $u167+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-457,v=60$ satisfies $u167+v1272=1$, 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.",
... | 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-013252 | Computational Number Theory: Extended Euclid | 7 | Write the solution set clearly: (a) Compute $\gcd(84,354)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 84+v\cdot 354=\gcd(84,354)$.
(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 $(84,354)$ to compute $g=\\gcd(84,354)$.",
"Step 2: Record the remainder equ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6}$.\nThe Euclidean algorithm computes $g=6$. The Bézout certificate $u=-21,v=5$ satisfies $u84+v354=6$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sen... | [
{
"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 $\boxed{6}$.) |
math-013253 | Number Theory: Divisibility — Linear Combinations | 7 | Warm-up: (a) Compute $\gcd(621,1578)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 621+v\cdot 1578=\gcd(621,1578)$.
(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": "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 $u=-155$ and $v=61$ with $u621+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=-155,v=61$ satisfies $u621+v1578=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-013254 | Number Theory: Modular Inverses — Extended Euclid | 7 | Give an answer and a quick verification: Find the multiplicative inverse of $99$ modulo $850$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{850}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "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 $\\gcd(99,850)=1$, 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: $\\boxed{249}$.\nMethod 1 constructs an inverse via Bézout, producing $x=249$. 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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013255 | Number Theory: gcd — Back Substitution | 7 | Give reasoning, not just computation: (a) Compute $\gcd(1679,176)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1679+v\cdot 176=\gcd(1679,176)$.
(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 $(1679,176)$ to compute $g=\\gcd(1679,176)$.",
"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 $u=63,v=-601$ satisfies $u1679+v176=1$, 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.",
... | 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-013256 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $1195$ modulo $1484$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1484}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the neces... | [
{
"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 $\\gcd(1195,1484)=1$, 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: $\\boxed{647}$.\nMethod 1 constructs an inverse via Bézout, producing $x=647$. 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. |
math-013257 | Computational Number Theory: Inverses and Certificates | 7 | Explain what is being counted/optimized: Find the multiplicative inverse of $1047$ modulo $1904$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1904}$).
(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 $\\gcd(1047,1904)=1$, 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: $\\boxed{471}$.\nMethod 1 constructs an inverse via Bézout, producing $x=471$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013258 | Number Theory: gcd — Euclidean Algorithm | 7 | Solve and include a self-check: (a) Compute $\gcd(1999,1598)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1999+v\cdot 1598=\gcd(1999,1598)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution cha... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1999,1598)$ to compute $g=\\gcd(1999,1598)$.",
"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 $u=267,v=-334$ satisfies $u1999+v1598=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013259 | Number Theory: Divisibility — Linear Combinations | 7 | Give a fully justified solution: (a) Compute $\gcd(623,1927)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 623+v\cdot 1927=\gcd(623,1927)$.
(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 $(623,1927)$ to compute $g=\\gcd(623,1927)$.",
"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 $u=897,v=-290$ satisfies $u623+v1927=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013260 | Number Theory: Divisibility — Linear Combinations | 7 | Work this out carefully: (a) Compute $\gcd(1425,175)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1425+v\cdot 175=\gcd(1425,175)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=1$ and $v=-8$ with $u1425+v1... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25}$.\nThe Euclidean algorithm computes $g=25$. The Bézout certificate $u=1,v=-8$ satisfies $u1425+v175=25$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013261 | Number Theory: Bézout Identity — Certificates | 7 | Use two approaches if possible: (a) Compute $\gcd(273,499)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 273+v\cdot 499=\gcd(273,499)$.
(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 $(273,499)$ to compute $g=\\gcd(273,499)$.",
"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 $u=223,v=-122$ satisfies $u273+v499=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013262 | Number Theory: Divisibility — Linear Combinations | 7 | Question: (a) Compute $\gcd(929,561)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 929+v\cdot 561=\gcd(929,561)$.
(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 verific... | [
{
"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 $u=218$ and $v=-361$ with $u929... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=218,v=-361$ satisfies $u929+v561=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013263 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Write the solution set clearly: Find the multiplicative inverse of $1174$ modulo $1419$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1419}$).
(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 $\\gcd(1174,1419)=1$, 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: $\\boxed{139}$.\nMethod 1 constructs an inverse via Bézout, producing $x=139$. 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-013264 | Number Theory: gcd — Euclidean Algorithm | 7 | Track units/moduli carefully: (a) Compute $\gcd(476,209)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 476+v\cdot 209=\gcd(476,209)$.
(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": "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 $v=41$ with $u476+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-18,v=41$ satisfies $u476+v209=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: Eu... | [
{
"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-013265 | Number Theory: gcd — Euclidean Algorithm | 7 | Give an answer and a quick verification: (a) Compute $\gcd(1472,250)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1472+v\cdot 250=\gcd(1472,250)$.
(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 $(1472,250)$ to compute $g=\\gcd(1472,250)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-9,v=53$ satisfies $u1472+v250=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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 $\boxed{2}$.) |
math-013266 | Computational Number Theory: Extended Euclid | 7 | Give a theorem-based solution: (a) Compute $\gcd(1171,557)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1171+v\cdot 557=\gcd(1171,557)$.
(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 $u=215$ and $v=-452$ with $u117... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=215,v=-452$ satisfies $u1171+v557=1$, 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.",
... | 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-013267 | Computational Number Theory: Extended Euclid | 7 | Explain why your operations are valid: (a) Compute $\gcd(393,138)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 393+v\cdot 138=\gcd(393,138)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=13$ and $v=-37$ with $u393+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=13,v=-37$ satisfies $u393+v138=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013268 | Computational Number Theory: Extended Euclid | 7 | Work carefully and justify each inference: (a) Compute $\gcd(1290,197)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1290+v\cdot 197=\gcd(1290,197)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitu... | [
{
"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 $(1290,197)$ to compute $g=\\gcd(1290,197)$.",
"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 $u=-31,v=203$ satisfies $u1290+v197=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness 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.",
... | 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-013269 | Computational Number Theory: Inverses and Certificates | 7 | Be explicit about assumptions: Find the multiplicative inverse of $244$ modulo $829$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{829}$).
(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 $\\gcd(244,829)=1$, 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: $\\boxed{282}$.\nMethod 1 constructs an inverse via Bézout, producing $x=282$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013270 | Number Theory: gcd — Euclidean Algorithm | 7 | Explain why your operations are valid: (a) Compute $\gcd(395,223)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 395+v\cdot 223=\gcd(395,223)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "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 $(395,223)$ to compute $g=\\gcd(395,223)$.",
"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 $u=-35,v=62$ satisfies $u395+v223=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013271 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Use two approaches if possible: Find the multiplicative inverse of $18$ modulo $131$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{131}$).
(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 $x=51$ and compute $18x=918$.",
"Step 2: Reduce: $918\\equiv 1\\pmod{131}$ (since $917=917$ is divisible by 13... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{51}$.\nMethod 1 constructs an inverse via Bézout, producing $x=51$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{51}$.) |
math-013272 | Number Theory: gcd — Euclidean Algorithm | 7 | Give a theorem-based solution: (a) Compute $\gcd(1514,227)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1514+v\cdot 227=\gcd(1514,227)$.
(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 $(1514,227)$ to compute $g=\\gcd(1514,227)$.",
"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 $u=-112,v=747$ satisfies $u1514+v227=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-013273 | Number Theory: Bézout Identity — Certificates | 7 | Solve (and briefly cross-validate): (a) Compute $\gcd(220,1418)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 220+v\cdot 1418=\gcd(220,1418)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-58$ and $v=9$ with $u220+v1... | {
"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 $u=-58,v=9$ satisfies $u220+v1418=2$, 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.",
... | 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-013274 | Number Theory: Units mod m — Existence Condition | 7 | Compute the requested quantity: Find the multiplicative inverse of $401$ modulo $1932$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1932}$).
(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 $x=53$ and compute $401x=21253$.",
"Step 2: Reduce: $21253\\equiv 1\\pmod{1932}$ (since $21252=21252$ is divis... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{53}$.\nMethod 1 constructs an inverse via Bézout, producing $x=53$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysi... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{53}$.) |
math-013275 | Computational Number Theory: Inverses and Certificates | 7 | Prompt: Find the multiplicative inverse of $67$ modulo $406$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{406}$).
(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 $\\gcd(67,406)=1$, 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: $\\boxed{303}$.\nMethod 1 constructs an inverse via Bézout, producing $x=303$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013276 | Number Theory: Modular Inverses — Extended Euclid | 7 | Give a fully justified solution: Find the multiplicative inverse of $275$ modulo $441$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{441}$).
(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 $x=263$ and compute $275x=72325$.",
"Step 2: Reduce: $72325\\equiv 1\\pmod{441}$ (since $72324=72324$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{263}$.\nMethod 1 constructs an inverse via Bézout, producing $x=263$. 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-013277 | Computational Number Theory: Extended Euclid | 7 | Use two approaches if possible: (a) Compute $\gcd(708,1138)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 708+v\cdot 1138=\gcd(708,1138)$.
(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 $(708,1138)$ to compute $g=\\gcd(708,1138)$.",
"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 $u=262,v=-163$ satisfies $u708+v1138=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Eu... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-013278 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Explain why your operations are valid: Find the multiplicative inverse of $454$ modulo $573$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{573}$).
(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 $\\gcd(454,573)=1$, 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: $\\boxed{130}$.\nMethod 1 constructs an inverse via Bézout, producing $x=130$. 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... | 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 $\boxed{130}$.) |
math-013279 | Number Theory: gcd — Back Substitution | 7 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(1009,1066)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1009+v\cdot 1066=\gcd(1009,1066)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-subs... | [
{
"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 $u=187$ and $v=-177$ with $u100... | {
"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 $u=187,v=-177$ satisfies $u1009+v1066=1$, 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.",
... | 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-013280 | Number Theory: Units mod m — Existence Condition | 7 | Use two approaches if possible: Find the multiplicative inverse of $685$ modulo $722$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{722}$).
(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 $\\gcd(685,722)=1$, 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: $\\boxed{39}$.\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": "Robustness note: Extended Euclid is ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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-013281 | Number Theory: Congruences — Solving $ax\equiv 1$ | 7 | Provide a rigorous solution: Find the multiplicative inverse of $685$ modulo $1604$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1604}$).
(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 $x=377$ and compute $685x=258245$.",
"Step 2: Reduce: $258245\\equiv 1\\pmod{1604}$ (since $258244=258244$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{377}$.\nMethod 1 constructs an inverse via Bézout, producing $x=377$. 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 $\boxed{377}$.) |
math-013282 | Number Theory: Bézout Identity — Certificates | 7 | Give an answer and a quick verification: (a) Compute $\gcd(396,365)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 396+v\cdot 365=\gcd(396,365)$.
(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 $u=106$ and $v=-115$ with $u396... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=106,v=-115$ satisfies $u396+v365=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid sc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-013283 | Number Theory: gcd — Euclidean Algorithm | 7 | Solve with verification: (a) Compute $\gcd(1483,1413)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1483+v\cdot 1413=\gcd(1483,1413)$.
(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 $(1483,1413)$ to compute $g=\\gcd(1483,1413)$.",
"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 $u=-545,v=572$ satisfies $u1483+v1413=1$, 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.",
... | 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-013284 | Number Theory: Modular Inverses — Extended Euclid | 7 | Be explicit about assumptions: Find the multiplicative inverse of $178$ modulo $567$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{567}$).
(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 $\\gcd(178,567)=1$, 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: $\\boxed{481}$.\nMethod 1 constructs an inverse via Bézout, producing $x=481$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{481}$.) |
math-013285 | Computational Number Theory: Inverses and Certificates | 7 | Find the exact value: Find the multiplicative inverse of $347$ modulo $1847$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1847}$).
(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 $\\gcd(347,1847)=1$, 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: $\\boxed{841}$.\nMethod 1 constructs an inverse via Bézout, producing $x=841$. 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-013286 | Computational Number Theory: Extended Euclid | 7 | Work carefully and justify each inference: (a) Compute $\gcd(897,563)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 897+v\cdot 563=\gcd(897,563)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitutio... | [
{
"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 $u=59$ and $v=-94$ with $u897+v... | {
"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 $u=59,v=-94$ satisfies $u897+v563=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "G... | [
{
"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-013287 | Number Theory: Units mod m — Existence Condition | 7 | Show all reasoning: Find the multiplicative inverse of $1075$ modulo $1589$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1589}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an in... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(1075,1589)=1$, 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: $\\boxed{541}$.\nMethod 1 constructs an inverse via Bézout, producing $x=541$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013288 | Computational Number Theory: Inverses and Certificates | 7 | Give an answer and a quick verification: Find the multiplicative inverse of $1078$ modulo $1171$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1171}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficien... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=277$ and compute $1078x=298606$.",
"Step 2: Reduce: $298606\\equiv 1\\pmod{1171}$ (since $298605=298605$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{277}$.\nMethod 1 constructs an inverse via Bézout, producing $x=277$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013289 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve and then verify: Find the multiplicative inverse of $5$ modulo $426$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{426}$).
(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": "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 $x=341$ and compute $5x=1705$.",
"Step 2: Reduce: $1705\\equiv 1\\pmod{426}$ (since $1704=1704$ is divisible b... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{341}$.\nMethod 1 constructs an inverse via Bézout, producing $x=341$. 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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-013290 | Computational Number Theory: Extended Euclid | 7 | Solve and then verify: (a) Compute $\gcd(1022,633)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1022+v\cdot 633=\gcd(1022,633)$.
(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 $u=179$ and $v=-289$ with $u102... | {
"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 $u=179,v=-289$ satisfies $u1022+v633=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"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-013291 | Number Theory: Bézout Identity — Certificates | 7 | Work carefully and justify each inference: (a) Compute $\gcd(1021,135)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1021+v\cdot 135=\gcd(1021,135)$.
(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 $u=16$ and $v=-121$ with $u1021... | {
"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 $u=16,v=-121$ satisfies $u1021+v135=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generali... | [
{
"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-013292 | Number Theory: Bézout Identity — Certificates | 7 | Track quantifiers carefully: (a) Compute $\gcd(1674,1707)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1674+v\cdot 1707=\gcd(1674,1707)$.
(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 $u=-207$ and $v=203$ with $u167... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=-207,v=203$ satisfies $u1674+v1707=3$, 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 $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-013293 | Number Theory: Modular Inverses — Extended Euclid | 7 | Solve and sanity-check: Find the multiplicative inverse of $1549$ modulo $1655$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1655}$).
(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": "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 $\\gcd(1549,1655)=1$, 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: $\\boxed{484}$.\nMethod 1 constructs an inverse via Bézout, producing $x=484$. 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... | 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 $\boxed{484}$.) |
math-013294 | Computational Number Theory: Inverses and Certificates | 7 | Indicate where a theorem is used: Find the multiplicative inverse of $979$ modulo $1234$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1234}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condit... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=963$ and compute $979x=942777$.",
"Step 2: Reduce: $942777\\equiv 1\\pmod{1234}$ (since $942776=942776$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{963}$.\nMethod 1 constructs an inverse via Bézout, producing $x=963$. 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. |
math-013295 | Number Theory: Modular Inverses — Extended Euclid | 7 | Find the exact value: Find the multiplicative inverse of $1396$ modulo $1891$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1891}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1551$ and compute $1396x=2165196$.",
"Step 2: Reduce: $2165196\\equiv 1\\pmod{1891}$ (since $2165195=216519... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1551}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1551$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fa... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1551}$.) |
math-013296 | Number Theory: Bézout Identity — Certificates | 7 | Use two approaches if possible: (a) Compute $\gcd(1951,613)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1951+v\cdot 613=\gcd(1951,613)$.
(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 $(1951,613)$ to compute $g=\\gcd(1951,613)$.",
"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 $u=104,v=-331$ satisfies $u1951+v613=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"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-013297 | Number Theory: gcd — Euclidean Algorithm | 7 | Explain what is being counted/optimized: (a) Compute $\gcd(1807,1746)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1807+v\cdot 1746=\gcd(1807,1746)$.
(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 $(1807,1746)$ to compute $g=\\gcd(1807,1746)$.",
"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 $u=229,v=-237$ satisfies $u1807+v1746=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis"... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-013298 | Computational Number Theory: Inverses and Certificates | 7 | Answer with a short justification: Find the multiplicative inverse of $883$ modulo $1436$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1436}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condi... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(883,1436)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1275}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1275$. 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... | 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 $\boxed{1275}$.) |
math-013299 | Number Theory: gcd — Euclidean Algorithm | 7 | Determine the requested value: (a) Compute $\gcd(925,253)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 925+v\cdot 253=\gcd(925,253)$.
(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 $(925,253)$ to compute $g=\\gcd(925,253)$.",
"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 $u=-32,v=117$ satisfies $u925+v253=1$, 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-013300 | Number Theory: gcd — Back Substitution | 7 | Give an answer and a quick verification: (a) Compute $\gcd(501,219)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 501+v\cdot 219=\gcd(501,219)$.
(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 $u=7$ and $v=-16$ with $u501+v2... | {
"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 $u=7,v=-16$ satisfies $u501+v219=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity... | [
{
"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 $\boxed{3}$.) |
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