id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-015101 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Solve and sanity-check: Find the multiplicative inverse of $595$ modulo $634$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{634}$).
(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 $x=65$ and compute $595x=38675$.",
"Step 2: Reduce: $38675\\equiv 1\\pmod{634}$ (since $38674=38674$ is divisi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{65}$.\nMethod 1 constructs an inverse via Bézout, producing $x=65$. 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 $\boxed{65}$.) |
math-015102 | Number Theory: gcd — Back Substitution | 8 | Task: (a) Compute $\gcd(1999,1196)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1999+v\cdot 1196=\gcd(1999,1196)$.
(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": "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 $u=563$ and $v=-941$ with $u199... | {
"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 $u=563,v=-941$ satisfies $u1999+v1196=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Genera... | [
{
"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-015103 | Number Theory: gcd — Euclidean Algorithm | 8 | Proceed methodically: (a) Compute $\gcd(697,87)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 697+v\cdot 87=\gcd(697,87)$.
(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 brie... | [
{
"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 $(697,87)$ to compute $g=\\gcd(697,87)$.",
"Step 2: Record the remainder equ... | {
"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 $u=1,v=-8$ satisfies $u697+v87=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sens... | [
{
"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-015104 | Number Theory: Modular Inverses — Extended Euclid | 8 | Do not skip justification steps: Find the multiplicative inverse of $123$ modulo $787$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{787}$).
(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 $\\gcd(123,787)=1$, 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: $\\boxed{32}$.\nMethod 1 constructs an inverse via Bézout, producing $x=32$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015105 | Computational Number Theory: Inverses and Certificates | 8 | Solve with verification: Find the multiplicative inverse of $1058$ modulo $1367$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1367}$).
(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 $\\gcd(1058,1367)=1$, 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: $\\boxed{407}$.\nMethod 1 constructs an inverse via Bézout, producing $x=407$. 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. (Here the result is $\boxed{407}$.) |
math-015106 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Complete the analysis: Find the multiplicative inverse of $574$ modulo $807$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{807}$).
(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 $\\gcd(574,807)=1$, 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: $\\boxed{142}$.\nMethod 1 constructs an inverse via Bézout, producing $x=142$. 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... | 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 $\boxed{142}$.) |
math-015107 | Number Theory: gcd — Euclidean Algorithm | 8 | Write the solution set clearly: (a) Compute $\gcd(1437,671)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1437+v\cdot 671=\gcd(1437,671)$.
(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 $(1437,671)$ to compute $g=\\gcd(1437,671)$.",
"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 $u=-113,v=242$ satisfies $u1437+v671=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pert... | [
{
"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-015108 | Computational Number Theory: Inverses and Certificates | 8 | Compute the requested quantity: Find the multiplicative inverse of $539$ modulo $613$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{613}$).
(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 $x=555$ and compute $539x=299145$.",
"Step 2: Reduce: $299145\\equiv 1\\pmod{613}$ (since $299144=299144$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{555}$.\nMethod 1 constructs an inverse via Bézout, producing $x=555$. 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... | 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 $\boxed{555}$.) |
math-015109 | Number Theory: Units mod m — Existence Condition | 8 | Start by stating any domain restrictions: Find the multiplicative inverse of $845$ modulo $991$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{991}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(845,991)=1$, 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: $\\boxed{767}$.\nMethod 1 constructs an inverse via Bézout, producing $x=767$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015110 | Number Theory: Units mod m — Existence Condition | 8 | Do not skip justification steps: Find the multiplicative inverse of $331$ modulo $1158$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1158}$).
(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 $\\gcd(331,1158)=1$, 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: $\\boxed{7}$.\nMethod 1 constructs an inverse via Bézout, producing $x=7$. 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 $\boxed{7}$.) |
math-015111 | Computational Number Theory: Extended Euclid | 8 | Answer using clear logical steps: (a) Compute $\gcd(1763,1269)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1763+v\cdot 1269=\gcd(1763,1269)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution c... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1763,1269)$ to compute $g=\\gcd(1763,1269)$.",
"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 $u=-280,v=389$ satisfies $u1763+v1269=1$, 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.",
... | 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-015112 | Number Theory: Units mod m — Existence Condition | 8 | Track units/moduli carefully: Find the multiplicative inverse of $64$ modulo $181$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{181}$).
(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 $x=99$ and compute $64x=6336$.",
"Step 2: Reduce: $6336\\equiv 1\\pmod{181}$ (since $6335=6335$ is divisible b... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{99}$.\nMethod 1 constructs an inverse via Bézout, producing $x=99$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast a... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{99}$.) |
math-015113 | Computational Number Theory: Inverses and Certificates | 8 | Give a theorem-based solution: Find the multiplicative inverse of $283$ modulo $988$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{988}$).
(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 $x=199$ and compute $283x=56317$.",
"Step 2: Reduce: $56317\\equiv 1\\pmod{988}$ (since $56316=56316$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{199}$.\nMethod 1 constructs an inverse via Bézout, producing $x=199$. 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-015114 | Number Theory: Units mod m — Existence Condition | 8 | Explain each transformation: Find the multiplicative inverse of $130$ modulo $279$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{279}$).
(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 $\\gcd(130,279)=1$, 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: $\\boxed{88}$.\nMethod 1 constructs an inverse via Bézout, producing $x=88$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast a... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015115 | Number Theory: Divisibility — Linear Combinations | 8 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(1426,911)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1426+v\cdot 911=\gcd(1426,911)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear b... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1426,911)$ to compute $g=\\gcd(1426,911)$.",
"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 $u=-444,v=695$ satisfies $u1426+v911=1$, 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.",
... | 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-015116 | Number Theory: Units mod m — Existence Condition | 8 | Show all reasoning: Find the multiplicative inverse of $817$ modulo $1191$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1191}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=328$ and compute $817x=267976$.",
"Step 2: Reduce: $267976\\equiv 1\\pmod{1191}$ (since $267975=267975$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{328}$.\nMethod 1 constructs an inverse via Bézout, producing $x=328$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{328}$.) |
math-015117 | Number Theory: Units mod m — Existence Condition | 8 | Complete the analysis: Find the multiplicative inverse of $52$ modulo $737$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{737}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=326$ and compute $52x=16952$.",
"Step 2: Reduce: $16952\\equiv 1\\pmod{737}$ (since $16951=16951$ is divisi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{326}$.\nMethod 1 constructs an inverse via Bézout, producing $x=326$. 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 $\boxed{326}$.) |
math-015118 | Number Theory: Units mod m — Existence Condition | 8 | Show all reasoning: Find the multiplicative inverse of $735$ modulo $1889$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1889}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "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 $\\gcd(735,1889)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1082}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1082$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{1082}$.) |
math-015119 | Number Theory: Units mod m — Existence Condition | 8 | Explain what is being counted/optimized: Find the multiplicative inverse of $1120$ modulo $1413$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1413}$).
(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 $x=1114$ and compute $1120x=1247680$.",
"Step 2: Reduce: $1247680\\equiv 1\\pmod{1413}$ (since $1247679=124767... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1114}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1114$. 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 $\boxed{1114}$.) |
math-015120 | Number Theory: Units mod m — Existence Condition | 8 | Solve and then verify: Find the multiplicative inverse of $577$ modulo $784$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{784}$).
(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": "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 $x=481$ and compute $577x=277537$.",
"Step 2: Reduce: $277537\\equiv 1\\pmod{784}$ (since $277536=277536$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "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. (Here the result is $\boxed{481}$.) |
math-015121 | Number Theory: Divisibility — Linear Combinations | 8 | Work carefully and justify each inference: (a) Compute $\gcd(518,394)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 518+v\cdot 394=\gcd(518,394)$.
(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 $u=-54$ and $v=71$ with $u518+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-54,v=71$ satisfies $u518+v394=2$, 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 $ua+vb=g$. |
math-015122 | Number Theory: gcd — Euclidean Algorithm | 8 | Solve with verification: (a) Compute $\gcd(1161,1116)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1161+v\cdot 1116=\gcd(1161,1116)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
In... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=25$ and $v=-26$ with $u1161+... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{9}$.\nThe Euclidean algorithm computes $g=9$. The Bézout certificate $u=25,v=-26$ satisfies $u1161+v1116=9$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{9}$.) |
math-015123 | Computational Number Theory: Inverses and Certificates | 8 | Carefully track domains: Find the multiplicative inverse of $991$ modulo $1968$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1968}$).
(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 $x=703$ and compute $991x=696673$.",
"Step 2: Reduce: $696673\\equiv 1\\pmod{1968}$ (since $696672=696672$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{703}$.\nMethod 1 constructs an inverse via Bézout, producing $x=703$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{703}$.) |
math-015124 | Number Theory: Bézout Identity — Certificates | 8 | Where appropriate, name the theorem you use: (a) Compute $\gcd(162,1370)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 162+v\cdot 1370=\gcd(162,1370)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substi... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=296$ and $v=-35$ with $u162+... | {
"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 $u=296,v=-35$ satisfies $u162+v1370=2$, 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 $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-015125 | Number Theory: Units mod m — Existence Condition | 8 | Find the exact value: Find the multiplicative inverse of $46$ modulo $343$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{343}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inve... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(46,343)=1$, 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: $\\boxed{261}$.\nMethod 1 constructs an inverse via Bézout, producing $x=261$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{261}$.) |
math-015126 | Number Theory: Divisibility — Linear Combinations | 8 | Challenge: (a) Compute $\gcd(840,1300)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 840+v\cdot 1300=\gcd(840,1300)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief ver... | [
{
"method_name": "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 $u=-17$ and $v=11$ with $u840+v... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{20}$.\nThe Euclidean algorithm computes $g=20$. The Bézout certificate $u=-17,v=11$ satisfies $u840+v1300=20$, 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.",
... | 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-015127 | Number Theory: gcd — Back Substitution | 8 | Where appropriate, name the theorem you use: (a) Compute $\gcd(1931,1873)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1931+v\cdot 1873=\gcd(1931,1873)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-sub... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1931,1873)$ to compute $g=\\gcd(1931,1873)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=549,v=-566$ satisfies $u1931+v1873=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scale... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-015128 | Number Theory: gcd — Back Substitution | 8 | Checkpoint: (a) Compute $\gcd(1421,554)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1421+v\cdot 554=\gcd(1421,554)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief ve... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=177$ and $v=-454$ with $u142... | {
"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 $u=177,v=-454$ satisfies $u1421+v554=1$, 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.",
... | 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-015129 | Number Theory: Bézout Identity — Certificates | 8 | Find the exact value: (a) Compute $\gcd(97,936)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 97+v\cdot 936=\gcd(97,936)$.
(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 brie... | [
{
"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 $(97,936)$ to compute $g=\\gcd(97,936)$.",
"Step 2: Record the remainder equ... | {
"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 $u=193,v=-20$ satisfies $u97+v936=1$, 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-015130 | Number Theory: Bézout Identity — Certificates | 8 | Solve and justify each step: (a) Compute $\gcd(694,946)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 694+v\cdot 946=\gcd(694,946)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Incl... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-229$ and $v=168$ with $u694... | {
"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 $u=-229,v=168$ satisfies $u694+v946=2$, 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.",
... | 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-015131 | Number Theory: Modular Inverses — Extended Euclid | 8 | Track quantifiers carefully: Find the multiplicative inverse of $1358$ modulo $1471$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1471}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the 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 $x=742$ and compute $1358x=1007636$.",
"Step 2: Reduce: $1007636\\equiv 1\\pmod{1471}$ (since $1007635=1007635... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{742}$.\nMethod 1 constructs an inverse via Bézout, producing $x=742$. 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 $\boxed{742}$.) |
math-015132 | Number Theory: Units mod m — Existence Condition | 8 | Explain what is being counted/optimized: Find the multiplicative inverse of $733$ modulo $1448$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1448}$).
(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 $x=885$ and compute $733x=648705$.",
"Step 2: Reduce: $648705\\equiv 1\\pmod{1448}$ (since $648704=648704$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{885}$.\nMethod 1 constructs an inverse via Bézout, producing $x=885$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{885}$.) |
math-015133 | Computational Number Theory: Inverses and Certificates | 8 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $1169$ modulo $1346$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1346}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessa... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1019$ and compute $1169x=1191211$.",
"Step 2: Reduce: $1191211\\equiv 1\\pmod{1346}$ (since $1191210=119121... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1019}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1019$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness 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 $\boxed{1019}$.) |
math-015134 | Number Theory: Bézout Identity — Certificates | 8 | Checkpoint: (a) Compute $\gcd(1136,1187)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1136+v\cdot 1187=\gcd(1136,1187)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1136,1187)$ to compute $g=\\gcd(1136,1187)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=256,v=-245$ satisfies $u1136+v1187=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scale... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015135 | Number Theory: Modular Inverses — Extended Euclid | 8 | Explain why your operations are valid: Find the multiplicative inverse of $419$ modulo $1688$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1688}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(419,1688)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1547}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1547$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 $\boxed{1547}$.) |
math-015136 | Number Theory: Divisibility — Linear Combinations | 8 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(904,603)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 904+v\cdot 603=\gcd(904,603)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substituti... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-2$ and $v=3$ with $u904+v60... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-2,v=3$ satisfies $u904+v603=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales effi... | [
{
"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-015137 | Number Theory: Units mod m — Existence Condition | 8 | Explain what is being counted/optimized: Find the multiplicative inverse of $247$ modulo $743$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{743}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(247,743)=1$, 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: $\\boxed{370}$.\nMethod 1 constructs an inverse via Bézout, producing $x=370$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{370}$.) |
math-015138 | Number Theory: Bézout Identity — Certificates | 8 | Track units/moduli carefully: (a) Compute $\gcd(189,868)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 189+v\cdot 868=\gcd(189,868)$.
(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 $u=23$ and $v=-5$ with $u189+v8... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{7}$.\nThe Euclidean algorithm computes $g=7$. The Bézout certificate $u=23,v=-5$ satisfies $u189+v868=7$, 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 $\boxed{7}$.) |
math-015139 | Number Theory: Divisibility — Linear Combinations | 8 | Explain each transformation: (a) Compute $\gcd(1291,691)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1291+v\cdot 691=\gcd(1291,691)$.
(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 $u=-243$ and $v=454$ with $u129... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-243,v=454$ satisfies $u1291+v691=1$, 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.",
... | 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-015140 | Number Theory: Modular Inverses — Extended Euclid | 8 | Problem: Find the multiplicative inverse of $432$ modulo $577$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{577}$).
(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 $x=191$ and compute $432x=82512$.",
"Step 2: Reduce: $82512\\equiv 1\\pmod{577}$ (since $82511=82511$ is divis... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{191}$.\nMethod 1 constructs an inverse via Bézout, producing $x=191$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{191}$.) |
math-015141 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Give a fully justified solution: Find the multiplicative inverse of $126$ modulo $163$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{163}$).
(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 $\\gcd(126,163)=1$, 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: $\\boxed{22}$.\nMethod 1 constructs an inverse via Bézout, producing $x=22$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the pro... | [
{
"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-015142 | Number Theory: gcd — Euclidean Algorithm | 8 | Start by stating any domain restrictions: (a) Compute $\gcd(229,1018)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 229+v\cdot 1018=\gcd(229,1018)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitut... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(229,1018)$ to compute $g=\\gcd(229,1018)$.",
"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 $u=489,v=-110$ satisfies $u229+v1018=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015143 | Number Theory: Modular Inverses — Extended Euclid | 8 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $307$ modulo $356$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{356}$).
(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 $\\gcd(307,356)=1$, 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: $\\boxed{247}$.\nMethod 1 constructs an inverse via Bézout, producing $x=247$. 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 $\boxed{247}$.) |
math-015144 | Number Theory: Divisibility — Linear Combinations | 8 | Do not skip justification steps: (a) Compute $\gcd(1453,1401)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1453+v\cdot 1401=\gcd(1453,1401)$.
(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 $(1453,1401)$ to compute $g=\\gcd(1453,1401)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-458,v=475$ satisfies $u1453+v1401=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-015145 | Number Theory: gcd — Back Substitution | 8 | Complete the analysis: (a) Compute $\gcd(1008,200)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1008+v\cdot 200=\gcd(1008,200)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1008,200)$ to compute $g=\\gcd(1008,200)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{8}$.\nThe Euclidean algorithm computes $g=8$. The Bézout certificate $u=1,v=-5$ satisfies $u1008+v200=8$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{8}$.) |
math-015146 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Explain each transformation: Find the multiplicative inverse of $140$ modulo $887$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{887}$).
(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 $\\gcd(140,887)=1$, 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: $\\boxed{868}$.\nMethod 1 constructs an inverse via Bézout, producing $x=868$. 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 $\boxed{868}$.) |
math-015147 | Number Theory: gcd — Euclidean Algorithm | 8 | Task: (a) Compute $\gcd(941,596)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 941+v\cdot 596=\gcd(941,596)$.
(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 verificatio... | [
{
"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 $(941,596)$ to compute $g=\\gcd(941,596)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-19,v=30$ satisfies $u941+v596=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-015148 | Computational Number Theory: Inverses and Certificates | 8 | Challenge: Find the multiplicative inverse of $445$ modulo $1928$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1928}$).
(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 e... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(445,1928)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{13}$.\nMethod 1 constructs an inverse via Bézout, producing $x=13$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended Eu... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015149 | Number Theory: Bézout Identity — Certificates | 8 | Find the exact value: (a) Compute $\gcd(811,1123)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 811+v\cdot 1123=\gcd(811,1123)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(811,1123)$ to compute $g=\\gcd(811,1123)$.",
"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 $u=-18,v=13$ satisfies $u811+v1123=1$, 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-015150 | Number Theory: Bézout Identity — Certificates | 8 | Solve (and briefly cross-validate): (a) Compute $\gcd(944,225)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 944+v\cdot 225=\gcd(944,225)$.
(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 $(944,225)$ to compute $g=\\gcd(944,225)$.",
"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 $u=-46,v=193$ satisfies $u944+v225=1$, 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 $\boxed{1}$.) |
math-015151 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Problem: Find the multiplicative inverse of $542$ modulo $633$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{633}$).
(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 $\\gcd(542,633)=1$, 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: $\\boxed{473}$.\nMethod 1 constructs an inverse via Bézout, producing $x=473$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015152 | Number Theory: Divisibility — Linear Combinations | 8 | Exercise: (a) Compute $\gcd(1706,1737)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1706+v\cdot 1737=\gcd(1706,1737)$.
(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": "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 $u=56$ and $v=-55$ with $u1706+... | {
"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 $u=56,v=-55$ satisfies $u1706+v1737=1$, 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.",
... | 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-015153 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Exercise: Find the multiplicative inverse of $667$ modulo $950$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{950}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(667,950)=1$, 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: $\\boxed{903}$.\nMethod 1 constructs an inverse via Bézout, producing $x=903$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{903}$.) |
math-015154 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Prompt: Find the multiplicative inverse of $161$ modulo $1718$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1718}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=843$ and compute $161x=135723$.",
"Step 2: Reduce: $135723\\equiv 1\\pmod{1718}$ (since $135722=135722$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{843}$.\nMethod 1 constructs an inverse via Bézout, producing $x=843$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{843}$.) |
math-015155 | Number Theory: gcd — Euclidean Algorithm | 8 | Do not skip justification steps: (a) Compute $\gcd(1984,1859)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1984+v\cdot 1859=\gcd(1984,1859)$.
(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 $(1984,1859)$ to compute $g=\\gcd(1984,1859)$.",
"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 $u=-580,v=619$ satisfies $u1984+v1859=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015156 | Number Theory: gcd — Euclidean Algorithm | 8 | Prompt: (a) Compute $\gcd(999,652)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 999+v\cdot 652=\gcd(999,652)$.
(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 verificat... | [
{
"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 $(999,652)$ to compute $g=\\gcd(999,652)$.",
"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 $u=295,v=-452$ satisfies $u999+v652=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015157 | Computational Number Theory: Inverses and Certificates | 8 | Keep the final answer in boxed form: Find the multiplicative inverse of $104$ modulo $215$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{215}$).
(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 $\\gcd(104,215)=1$, 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: $\\boxed{184}$.\nMethod 1 constructs an inverse via Bézout, producing $x=184$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015158 | Number Theory: gcd — Back Substitution | 8 | Give reasoning, not just computation: (a) Compute $\gcd(265,1779)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 265+v\cdot 1779=\gcd(265,1779)$.
(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 $(265,1779)$ to compute $g=\\gcd(265,1779)$.",
"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 $u=-866,v=129$ satisfies $u265+v1779=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015159 | Number Theory: Units mod m — Existence Condition | 8 | Give a theorem-based solution: Find the multiplicative inverse of $251$ modulo $793$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{793}$).
(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 $x=218$ and compute $251x=54718$.",
"Step 2: Reduce: $54718\\equiv 1\\pmod{793}$ (since $54717=54717$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{218}$.\nMethod 1 constructs an inverse via Bézout, producing $x=218$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{218}$.) |
math-015160 | Number Theory: Units mod m — Existence Condition | 8 | Solve and sanity-check: Find the multiplicative inverse of $585$ 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 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 $x=547$ and compute $585x=319995$.",
"Step 2: Reduce: $319995\\equiv 1\\pmod{829}$ (since $319994=319994$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{547}$.\nMethod 1 constructs an inverse via Bézout, producing $x=547$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{547}$.) |
math-015161 | Number Theory: Units mod m — Existence Condition | 8 | Give an answer and a quick verification: Find the multiplicative inverse of $133$ modulo $769$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{769}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=451$ and compute $133x=59983$.",
"Step 2: Reduce: $59983\\equiv 1\\pmod{769}$ (since $59982=59982$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{451}$.\nMethod 1 constructs an inverse via Bézout, producing $x=451$. 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... | 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 $\boxed{451}$.) |
math-015162 | Number Theory: Divisibility — Linear Combinations | 8 | Indicate where a theorem is used: (a) Compute $\gcd(445,837)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 445+v\cdot 837=\gcd(445,837)$.
(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 $u=79$ and $v=-42$ with $u445+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 $u=79,v=-42$ satisfies $u445+v837=1$, 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.",
... | 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-015163 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Complete the analysis: Find the multiplicative inverse of $100$ modulo $293$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{293}$).
(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": "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 $x=126$ and compute $100x=12600$.",
"Step 2: Reduce: $12600\\equiv 1\\pmod{293}$ (since $12599=12599$ is divis... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{126}$.\nMethod 1 constructs an inverse via Bézout, producing $x=126$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{126}$.) |
math-015164 | Number Theory: Units mod m — Existence Condition | 8 | Prompt: Find the multiplicative inverse of $801$ modulo $1039$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1039}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=633$ and compute $801x=507033$.",
"Step 2: Reduce: $507033\\equiv 1\\pmod{1039}$ (since $507032=507032$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{633}$.\nMethod 1 constructs an inverse via Bézout, producing $x=633$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{633}$.) |
math-015165 | Number Theory: Modular Inverses — Extended Euclid | 8 | Solve with verification: Find the multiplicative inverse of $651$ modulo $1531$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1531}$).
(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 $\\gcd(651,1531)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1063}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1063$. 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... | 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 $\boxed{1063}$.) |
math-015166 | Number Theory: gcd — Back Substitution | 8 | Track units/moduli carefully: (a) Compute $\gcd(687,1249)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 687+v\cdot 1249=\gcd(687,1249)$.
(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 $u=20$ and $v=-11$ with $u687+v... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=20,v=-11$ satisfies $u687+v1249=1$, 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-015167 | Number Theory: gcd — Back Substitution | 8 | Make each step logically reversible (or explain if not): (a) Compute $\gcd(1013,180)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1013+v\cdot 180=\gcd(1013,180)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear bac... | [
{
"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 $u=-43$ and $v=242$ with $u1013... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-43,v=242$ satisfies $u1013+v180=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertu... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015168 | Computational Number Theory: Inverses and Certificates | 8 | Show all reasoning: Find the multiplicative inverse of $902$ modulo $1875$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1875}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=713$ and compute $902x=643126$.",
"Step 2: Reduce: $643126\\equiv 1\\pmod{1875}$ (since $643125=643125$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{713}$.\nMethod 1 constructs an inverse via Bézout, producing $x=713$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015169 | Number Theory: Units mod m — Existence Condition | 8 | Carefully track domains: Find the multiplicative inverse of $215$ modulo $342$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{342}$).
(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 $x=35$ and compute $215x=7525$.",
"Step 2: Reduce: $7525\\equiv 1\\pmod{342}$ (since $7524=7524$ is divisible ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{35}$.\nMethod 1 constructs an inverse via Bézout, producing $x=35$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{35}$.) |
math-015170 | Number Theory: Bézout Identity — Certificates | 8 | Determine the requested value: (a) Compute $\gcd(1056,1711)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1056+v\cdot 1711=\gcd(1056,1711)$.
(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 $(1056,1711)$ to compute $g=\\gcd(1056,1711)$.",
"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 $u=-128,v=79$ satisfies $u1056+v1711=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-015171 | Number Theory: Units mod m — Existence Condition | 8 | Carefully track domains: Find the multiplicative inverse of $1397$ modulo $1887$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1887}$).
(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 $x=1706$ and compute $1397x=2383282$.",
"Step 2: Reduce: $2383282\\equiv 1\\pmod{1887}$ (since $2383281=238328... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1706}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1706$. 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... | 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 $\boxed{1706}$.) |
math-015172 | Number Theory: gcd — Back Substitution | 8 | Keep the final answer in boxed form: (a) Compute $\gcd(663,147)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 663+v\cdot 147=\gcd(663,147)$.
(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 $(663,147)$ to compute $g=\\gcd(663,147)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=2,v=-9$ satisfies $u663+v147=3$, 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.",
... | 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-015173 | Computational Number Theory: Inverses and Certificates | 8 | Give a theorem-based solution: Find the multiplicative inverse of $1418$ modulo $1469$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1469}$).
(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 $x=144$ and compute $1418x=204192$.",
"Step 2: Reduce: $204192\\equiv 1\\pmod{1469}$ (since $204191=204191$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{144}$.\nMethod 1 constructs an inverse via Bézout, producing $x=144$. 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-015174 | Number Theory: Units mod m — Existence Condition | 8 | Find the exact value: Find the multiplicative inverse of $391$ modulo $1225$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1225}$).
(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 $\\gcd(391,1225)=1$, 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: $\\boxed{636}$.\nMethod 1 constructs an inverse via Bézout, producing $x=636$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{636}$.) |
math-015175 | Computational Number Theory: Inverses and Certificates | 8 | Find the exact value: Find the multiplicative inverse of $555$ modulo $1634$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1634}$).
(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 $\\gcd(555,1634)=1$, 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: $\\boxed{1107}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1107$. 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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015176 | Number Theory: Divisibility — Linear Combinations | 8 | Prompt: (a) Compute $\gcd(1129,1754)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1129+v\cdot 1754=\gcd(1129,1754)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief ver... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1129,1754)$ to compute $g=\\gcd(1129,1754)$.",
"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 $u=-87,v=56$ satisfies $u1129+v1754=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertu... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015177 | Number Theory: Bézout Identity — Certificates | 8 | Explain why your operations are valid: (a) Compute $\gcd(1695,1730)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1695+v\cdot 1730=\gcd(1695,1730)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitut... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1695,1730)$ to compute $g=\\gcd(1695,1730)$.",
"Step 2: Record the remaind... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5}$.\nThe Euclidean algorithm computes $g=5$. The Bézout certificate $u=-99,v=97$ satisfies $u1695+v1730=5$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid s... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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 $\boxed{5}$.) |
math-015178 | Number Theory: Divisibility — Linear Combinations | 8 | Indicate where a theorem is used: (a) Compute $\gcd(415,1842)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 415+v\cdot 1842=\gcd(415,1842)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chai... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=253$ and $v=-57$ with $u415+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=253,v=-57$ satisfies $u415+v1842=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were pertu... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-015179 | Number Theory: Divisibility — Linear Combinations | 8 | Give an answer and a quick verification: (a) Compute $\gcd(1657,1950)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1657+v\cdot 1950=\gcd(1657,1950)$.
(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 $(1657,1950)$ to compute $g=\\gcd(1657,1950)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=193,v=-164$ satisfies $u1657+v1950=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensit... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015180 | Computational Number Theory: Inverses and Certificates | 8 | Solve and justify each step: Find the multiplicative inverse of $731$ modulo $1197$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1197}$).
(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 $\\gcd(731,1197)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{131}$.\nMethod 1 constructs an inverse via Bézout, producing $x=131$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{131}$.) |
math-015181 | Number Theory: Units mod m — Existence Condition | 8 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $49$ modulo $1609$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1609}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffici... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1412$ and compute $49x=69188$.",
"Step 2: Reduce: $69188\\equiv 1\\pmod{1609}$ (since $69187=69187$ is divi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1412}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1412$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness 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... | 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 $\boxed{1412}$.) |
math-015182 | Computational Number Theory: Extended Euclid | 8 | Proceed methodically: (a) Compute $\gcd(753,1828)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 753+v\cdot 1828=\gcd(753,1828)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(753,1828)$ to compute $g=\\gcd(753,1828)$.",
"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 $u=369,v=-152$ satisfies $u753+v1828=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-015183 | Number Theory: gcd — Euclidean Algorithm | 8 | Problem: (a) Compute $\gcd(577,1119)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 577+v\cdot 1119=\gcd(577,1119)$.
(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 $(577,1119)$ to compute $g=\\gcd(577,1119)$.",
"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 $u=64,v=-33$ satisfies $u577+v1119=1$, 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-015184 | Number Theory: Modular Inverses — Extended Euclid | 8 | Keep the final answer in boxed form: Find the multiplicative inverse of $575$ modulo $1098$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1098}$).
(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 $\\gcd(575,1098)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{359}$.\nMethod 1 constructs an inverse via Bézout, producing $x=359$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015185 | Number Theory: Modular Inverses — Extended Euclid | 8 | Task: Find the multiplicative inverse of $145$ modulo $397$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{397}$).
(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 $\\gcd(145,397)=1$, 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: $\\boxed{115}$.\nMethod 1 constructs an inverse via Bézout, producing $x=115$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015186 | Number Theory: Modular Inverses — Extended Euclid | 8 | Warm-up: Find the multiplicative inverse of $819$ modulo $1139$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1139}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exi... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(819,1139)=1$, 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: $\\boxed{210}$.\nMethod 1 constructs an inverse via Bézout, producing $x=210$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the 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... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015187 | Number Theory: Divisibility — Linear Combinations | 8 | Track units/moduli carefully: (a) Compute $\gcd(1698,676)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1698+v\cdot 676=\gcd(1698,676)$.
(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 $u=127$ and $v=-319$ with $u169... | {
"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 $u=127,v=-319$ satisfies $u1698+v676=2$, 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.",
... | 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-015188 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Give reasoning, not just computation: Find the multiplicative inverse of $1154$ modulo $1183$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1183}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient c... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=979$ and compute $1154x=1129766$.",
"Step 2: Reduce: $1129766\\equiv 1\\pmod{1183}$ (since $1129765=1129765... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{979}$.\nMethod 1 constructs an inverse via Bézout, producing $x=979$. 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-015189 | Number Theory: Congruences — Solving $ax\equiv 1$ | 8 | Challenge: Find the multiplicative inverse of $223$ modulo $1186$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1186}$).
(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 e... | [
{
"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 $x=1069$ and compute $223x=238387$.",
"Step 2: Reduce: $238387\\equiv 1\\pmod{1186}$ (since $238386=238386$ is... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1069}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1069$. 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-015190 | Computational Number Theory: Extended Euclid | 8 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(663,1054)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 663+v\cdot 1054=\gcd(663,1054)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear b... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(663,1054)$ to compute $g=\\gcd(663,1054)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{17}$.\nThe Euclidean algorithm computes $g=17$. The Bézout certificate $u=-27,v=17$ satisfies $u663+v1054=17$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scale... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{17}$.) |
math-015191 | Number Theory: Units mod m — Existence Condition | 8 | Checkpoint: Find the multiplicative inverse of $347$ modulo $1363$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1363}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=927$ and compute $347x=321669$.",
"Step 2: Reduce: $321669\\equiv 1\\pmod{1363}$ (since $321668=321668$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{927}$.\nMethod 1 constructs an inverse via Bézout, producing $x=927$. 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. (Here the result is $\boxed{927}$.) |
math-015192 | Number Theory: Modular Inverses — Extended Euclid | 8 | Work carefully and justify each inference: Find the multiplicative inverse of $549$ modulo $677$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{677}$).
(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 $x=238$ and compute $549x=130662$.",
"Step 2: Reduce: $130662\\equiv 1\\pmod{677}$ (since $130661=130661$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{238}$.\nMethod 1 constructs an inverse via Bézout, producing $x=238$. 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... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-015193 | Number Theory: gcd — Euclidean Algorithm | 8 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(1884,1993)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1884+v\cdot 1993=\gcd(1884,1993)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clea... | [
{
"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 $(1884,1993)$ to compute $g=\\gcd(1884,1993)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-128,v=121$ satisfies $u1884+v1993=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euclid ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-015194 | Number Theory: Units mod m — Existence Condition | 8 | Checkpoint: Find the multiplicative inverse of $304$ 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 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 $\\gcd(304,1255)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1094}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1094$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem ... | [
{
"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 $\boxed{1094}$.) |
math-015195 | Number Theory: Bézout Identity — Certificates | 8 | Explain what is being counted/optimized: (a) Compute $\gcd(1405,1503)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1405+v\cdot 1503=\gcd(1405,1503)$.
(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 $u=46$ and $v=-43$ with $u1405+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=46,v=-43$ satisfies $u1405+v1503=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-015196 | Computational Number Theory: Inverses and Certificates | 8 | Track quantifiers carefully: Find the multiplicative inverse of $339$ modulo $527$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{527}$).
(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 $x=356$ and compute $339x=120684$.",
"Step 2: Reduce: $120684\\equiv 1\\pmod{527}$ (since $120683=120683$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{356}$.\nMethod 1 constructs an inverse via Bézout, producing $x=356$. 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... | 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 $\boxed{356}$.) |
math-015197 | Number Theory: Bézout Identity — Certificates | 8 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(1298,426)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1298+v\cdot 426=\gcd(1298,426)$.
(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 $(1298,426)$ to compute $g=\\gcd(1298,426)$.",
"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 $u=64,v=-195$ satisfies $u1298+v426=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-015198 | Number Theory: Modular Inverses — Extended Euclid | 8 | Challenge: Find the multiplicative inverse of $1016$ modulo $1671$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1671}$).
(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 $\\gcd(1016,1671)=1$, 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: $\\boxed{449}$.\nMethod 1 constructs an inverse via Bézout, producing $x=449$. 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 $\boxed{449}$.) |
math-015199 | Number Theory: Bézout Identity — Certificates | 8 | Solve and then verify: (a) Compute $\gcd(1498,301)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1498+v\cdot 301=\gcd(1498,301)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1498,301)$ to compute $g=\\gcd(1498,301)$.",
"Step 2: Record the remainder... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{7}$.\nThe Euclidean algorithm computes $g=7$. The Bézout certificate $u=-1,v=5$ satisfies $u1498+v301=7$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: 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 $ua+vb=g$. (Here the result is $\boxed{7}$.) |
math-015200 | Computational Number Theory: Inverses and Certificates | 8 | Warm-up: Find the multiplicative inverse of $1610$ modulo $1691$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1691}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ex... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(1610,1691)=1$, 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: $\\boxed{167}$.\nMethod 1 constructs an inverse via Bézout, producing $x=167$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
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