id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-011101 | Number Theory: Divisibility — Linear Combinations | 6 | Explain what is being counted/optimized: (a) Compute $\gcd(1354,442)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1354+v\cdot 442=\gcd(1354,442)$.
(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 $u=79$ and $v=-242$ with $u1354... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=79,v=-242$ satisfies $u1354+v442=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid s... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011102 | Number Theory: Modular Inverses — Extended Euclid | 6 | Work this out carefully: Find the multiplicative inverse of $62$ modulo $749$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{749}$).
(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 $x=447$ and compute $62x=27714$.",
"Step 2: Reduce: $27714\\equiv 1\\pmod{749}$ (since $27713=27713$ is divisi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{447}$.\nMethod 1 constructs an inverse via Bézout, producing $x=447$. 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. (Here the result is $\boxed{447}$.) |
math-011103 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve and include a self-check: Find the multiplicative inverse of $525$ modulo $1037$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1037}$).
(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 $x=399$ and compute $525x=209475$.",
"Step 2: Reduce: $209475\\equiv 1\\pmod{1037}$ (since $209474=209474$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{399}$.\nMethod 1 constructs an inverse via Bézout, producing $x=399$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{399}$.) |
math-011104 | Computational Number Theory: Inverses and Certificates | 6 | Track units/moduli carefully: Find the multiplicative inverse of $483$ modulo $1966$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1966}$).
(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 $x=289$ and compute $483x=139587$.",
"Step 2: Reduce: $139587\\equiv 1\\pmod{1966}$ (since $139586=139586$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{289}$.\nMethod 1 constructs an inverse via Bézout, producing $x=289$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011105 | Number Theory: gcd — Back Substitution | 6 | Warm-up: (a) Compute $\gcd(638,1191)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 638+v\cdot 1191=\gcd(638,1191)$.
(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 $u=-28$ and $v=15$ with $u638+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-28,v=15$ satisfies $u638+v1191=1$, 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.",
... | 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-011106 | Number Theory: Modular Inverses — Extended Euclid | 6 | Track units/moduli carefully: Find the multiplicative inverse of $319$ modulo $326$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{326}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=93$ and compute $319x=29667$.",
"Step 2: Reduce: $29667\\equiv 1\\pmod{326}$ (since $29666=29666$ is divisi... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{93}$.\nMethod 1 constructs an inverse via Bézout, producing $x=93$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is f... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{93}$.) |
math-011107 | Number Theory: Modular Inverses — Extended Euclid | 6 | Challenge: Find the multiplicative inverse of $749$ modulo $1195$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1195}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an 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 $\\gcd(749,1195)=1$, 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: $\\boxed{844}$.\nMethod 1 constructs an inverse via Bézout, producing $x=844$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{844}$.) |
math-011108 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $269$ modulo $432$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{432}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessar... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(269,432)=1$, 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: $\\boxed{53}$.\nMethod 1 constructs an inverse via Bézout, producing $x=53$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Ex... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{53}$.) |
math-011109 | Number Theory: Divisibility — Linear Combinations | 6 | Checkpoint: (a) Compute $\gcd(1403,998)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1403+v\cdot 998=\gcd(1403,998)$.
(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 $u=69$ and $v=-97$ with $u1403+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=69,v=-97$ satisfies $u1403+v998=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: E... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011110 | Number Theory: gcd — Back Substitution | 6 | Solve and sanity-check: (a) Compute $\gcd(1925,715)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1925+v\cdot 715=\gcd(1925,715)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Includ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1925,715)$ to compute $g=\\gcd(1925,715)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{55}$.\nThe Euclidean algorithm computes $g=55$. The Bézout certificate $u=3,v=-8$ satisfies $u1925+v715=55$, 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 $\boxed{55}$.) |
math-011111 | Number Theory: Units mod m — Existence Condition | 6 | Solve with verification: Find the multiplicative inverse of $5$ modulo $1594$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1594}$).
(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 $x=319$ and compute $5x=1595$.",
"Step 2: Reduce: $1595\\equiv 1\\pmod{1594}$ (since $1594=1594$ is divisible ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{319}$.\nMethod 1 constructs an inverse via Bézout, producing $x=319$. 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-011112 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Keep the final answer in boxed form: Find the multiplicative inverse of $965$ modulo $1457$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1457}$).
(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 $\\gcd(965,1457)=1$, 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: $\\boxed{690}$.\nMethod 1 constructs an inverse via Bézout, producing $x=690$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{690}$.) |
math-011113 | Number Theory: Units mod m — Existence Condition | 6 | Proceed methodically: Find the multiplicative inverse of $709$ modulo $1192$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1192}$).
(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 $x=269$ and compute $709x=190721$.",
"Step 2: Reduce: $190721\\equiv 1\\pmod{1192}$ (since $190720=190720$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{269}$.\nMethod 1 constructs an inverse via Bézout, producing $x=269$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{269}$.) |
math-011114 | Number Theory: Units mod m — Existence Condition | 6 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $45$ modulo $61$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{61}$).
(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 $\\gcd(45,61)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{19}$.\nMethod 1 constructs an inverse via Bézout, producing $x=19$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is f... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{19}$.) |
math-011115 | Number Theory: gcd — Euclidean Algorithm | 6 | Challenge: (a) Compute $\gcd(723,1458)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 723+v\cdot 1458=\gcd(723,1458)$.
(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 $(723,1458)$ to compute $g=\\gcd(723,1458)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=121,v=-60$ satisfies $u723+v1458=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011116 | Number Theory: Bézout Identity — Certificates | 6 | Use two approaches if possible: (a) Compute $\gcd(1329,1716)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1329+v\cdot 1716=\gcd(1329,1716)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution cha... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1329,1716)$ to compute $g=\\gcd(1329,1716)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=235,v=-182$ satisfies $u1329+v1716=3$, 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.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-011117 | Number Theory: Modular Inverses — Extended Euclid | 6 | State any required conditions first: Find the multiplicative inverse of $610$ modulo $1301$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1301}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient con... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=787$ and compute $610x=480070$.",
"Step 2: Reduce: $480070\\equiv 1\\pmod{1301}$ (since $480069=480069$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{787}$.\nMethod 1 constructs an inverse via Bézout, producing $x=787$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011118 | Number Theory: Units mod m — Existence Condition | 6 | Solve and sanity-check: Find the multiplicative inverse of $1609$ modulo $1987$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1987}$).
(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 $\\gcd(1609,1987)=1$, 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: $\\boxed{1393}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1393$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the... | [
{
"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 $\boxed{1393}$.) |
math-011119 | Computational Number Theory: Inverses and Certificates | 6 | Be explicit about assumptions: Find the multiplicative inverse of $122$ modulo $387$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{387}$).
(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 $x=92$ and compute $122x=11224$.",
"Step 2: Reduce: $11224\\equiv 1\\pmod{387}$ (since $11223=11223$ is divisi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{92}$.\nMethod 1 constructs an inverse via Bézout, producing $x=92$. 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... | 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 $\boxed{92}$.) |
math-011120 | Number Theory: Bézout Identity — Certificates | 6 | Solve and then verify: (a) Compute $\gcd(126,103)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 126+v\cdot 103=\gcd(126,103)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a ... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=9$ and $v=-11$ with $u126+v1... | {
"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 $u=9,v=-11$ satisfies $u126+v103=1$, 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 $ua+vb=g$. |
math-011121 | Number Theory: gcd — Back Substitution | 6 | Where appropriate, name the theorem you use: (a) Compute $\gcd(535,559)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 535+v\cdot 559=\gcd(535,559)$.
(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 $(535,559)$ to compute $g=\\gcd(535,559)$.",
"Step 2: Record the remainder e... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=163,v=-156$ satisfies $u535+v559=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011122 | Number Theory: Modular Inverses — Extended Euclid | 6 | Explain each transformation: Find the multiplicative inverse of $865$ modulo $1421$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1421}$).
(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 $\\gcd(865,1421)=1$, 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: $\\boxed{23}$.\nMethod 1 constructs an inverse via Bézout, producing $x=23$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysi... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{23}$.) |
math-011123 | Computational Number Theory: Extended Euclid | 6 | Work this out carefully: (a) Compute $\gcd(1939,1618)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1939+v\cdot 1618=\gcd(1939,1618)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
In... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1939,1618)$ to compute $g=\\gcd(1939,1618)$.",
"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 $u=373,v=-447$ satisfies $u1939+v1618=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: E... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Core principle: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011124 | Number Theory: Bézout Identity — Certificates | 6 | Find the exact value: (a) Compute $\gcd(791,410)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 791+v\cdot 410=\gcd(791,410)$.
(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 b... | [
{
"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 $u=-99$ and $v=191$ with $u791+... | {
"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 $u=-99,v=191$ satisfies $u791+v410=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011125 | Number Theory: Divisibility — Linear Combinations | 6 | Track units/moduli carefully: (a) Compute $\gcd(1461,744)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1461+v\cdot 744=\gcd(1461,744)$.
(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 $u=55$ and $v=-108$ with $u1461... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=55,v=-108$ satisfies $u1461+v744=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011126 | Number Theory: Bézout Identity — Certificates | 6 | Show all reasoning: (a) Compute $\gcd(315,1191)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 315+v\cdot 1191=\gcd(315,1191)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(315,1191)$ to compute $g=\\gcd(315,1191)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=121,v=-32$ satisfies $u315+v1191=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011127 | Number Theory: gcd — Back Substitution | 6 | Explain what is being counted/optimized: (a) Compute $\gcd(314,805)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 314+v\cdot 805=\gcd(314,805)$.
(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 $(314,805)$ to compute $g=\\gcd(314,805)$.",
"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 $u=-141,v=55$ satisfies $u314+v805=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales e... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011128 | Computational Number Theory: Extended Euclid | 6 | Complete the analysis: (a) Compute $\gcd(602,131)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 602+v\cdot 131=\gcd(602,131)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a ... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=42$ and $v=-193$ with $u602+... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=42,v=-193$ satisfies $u602+v131=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid sc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011129 | Number Theory: Divisibility — Linear Combinations | 6 | Answer using clear logical steps: (a) Compute $\gcd(538,641)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 538+v\cdot 641=\gcd(538,641)$.
(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 $(538,641)$ to compute $g=\\gcd(538,641)$.",
"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 $u=56,v=-47$ satisfies $u538+v641=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid sca... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011130 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Challenge: Find the multiplicative inverse of $607$ modulo $1023$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1023}$).
(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 $x=391$ and compute $607x=237337$.",
"Step 2: Reduce: $237337\\equiv 1\\pmod{1023}$ (since $237336=237336$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{391}$.\nMethod 1 constructs an inverse via Bézout, producing $x=391$. 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 $\boxed{391}$.) |
math-011131 | Number Theory: Divisibility — Linear Combinations | 6 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(494,1867)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 494+v\cdot 1867=\gcd(494,1867)$.
(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 $u=-805$ and $v=213$ with $u494... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-805,v=213$ satisfies $u494+v1867=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011132 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve with verification: Find the multiplicative inverse of $1180$ modulo $1557$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1557}$).
(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 $\\gcd(1180,1557)=1$, 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: $\\boxed{1144}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1144$. 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 $\boxed{1144}$.) |
math-011133 | Number Theory: gcd — Euclidean Algorithm | 6 | Exercise: (a) Compute $\gcd(1788,1562)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1788+v\cdot 1562=\gcd(1788,1562)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief v... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1788,1562)$ to compute $g=\\gcd(1788,1562)$.",
"Step 2: Record the remaind... | {
"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 $u=235,v=-269$ satisfies $u1788+v1562=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robust... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{2}$.) |
math-011134 | Number Theory: Modular Inverses — Extended Euclid | 6 | Carefully track domains: Find the multiplicative inverse of $403$ modulo $413$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{413}$).
(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 $x=289$ and compute $403x=116467$.",
"Step 2: Reduce: $116467\\equiv 1\\pmod{413}$ (since $116466=116466$ is d... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{289}$.\nMethod 1 constructs an inverse via Bézout, producing $x=289$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{289}$.) |
math-011135 | Computational Number Theory: Inverses and Certificates | 6 | Give a theorem-based solution: Find the multiplicative inverse of $221$ modulo $330$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{330}$).
(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 $\\gcd(221,330)=1$, 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: $\\boxed{221}$.\nMethod 1 constructs an inverse via Bézout, producing $x=221$. 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 $\boxed{221}$.) |
math-011136 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $732$ 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 condit... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=103$ and compute $732x=75396$.",
"Step 2: Reduce: $75396\\equiv 1\\pmod{887}$ (since $75395=75395$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{103}$.\nMethod 1 constructs an inverse via Bézout, producing $x=103$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011137 | Number Theory: Bézout Identity — Certificates | 6 | Work this out carefully: (a) Compute $\gcd(377,275)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 377+v\cdot 275=\gcd(377,275)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include ... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=-62$ and $v=85$ with $u377+v... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=-62,v=85$ satisfies $u377+v275=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid scales ef... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011138 | Number Theory: gcd — Back Substitution | 6 | Provide both a computational and a conceptual explanation: (a) Compute $\gcd(538,477)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 538+v\cdot 477=\gcd(538,477)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear back... | [
{
"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 $(538,477)$ to compute $g=\\gcd(538,477)$.",
"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 $u=-86,v=97$ satisfies $u538+v477=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the pro... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Takeaway: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011139 | Number Theory: Units mod m — Existence Condition | 6 | Give a theorem-based solution: Find the multiplicative inverse of $49$ modulo $1343$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1343}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"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 $\\gcd(49,1343)=1$, 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: $\\boxed{603}$.\nMethod 1 constructs an inverse via Bézout, producing $x=603$. 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... | 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 $\boxed{603}$.) |
math-011140 | Computational Number Theory: Extended Euclid | 6 | Warm-up: (a) Compute $\gcd(1631,1593)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1631+v\cdot 1593=\gcd(1631,1593)$.
(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 $u=545$ and $v=-558$ with $u163... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=545,v=-558$ satisfies $u1631+v1593=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011141 | Number Theory: gcd — Back Substitution | 6 | Answer with a short justification: (a) Compute $\gcd(1303,1682)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1303+v\cdot 1682=\gcd(1303,1682)$.
(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 $(1303,1682)$ to compute $g=\\gcd(1303,1682)$.",
"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 $u=-537,v=416$ satisfies $u1303+v1682=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Generality note: Euclid... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011142 | Number Theory: gcd — Euclidean Algorithm | 6 | Exercise: (a) Compute $\gcd(1743,823)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1743+v\cdot 823=\gcd(1743,823)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Include a brief veri... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=280$ and $v=-593$ with $u174... | {
"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 $u=280,v=-593$ satisfies $u1743+v823=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011143 | Computational Number Theory: Inverses and Certificates | 6 | Warm-up: Find the multiplicative inverse of $1054$ 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 an inverse to ex... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1332$ and compute $1054x=1403928$.",
"Step 2: Reduce: $1403928\\equiv 1\\pmod{1531}$ (since $1403927=140392... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1332}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1332$. 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... | 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 $\boxed{1332}$.) |
math-011144 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Task: Find the multiplicative inverse of $293$ modulo $975$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{975}$).
(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 $x=782$ and compute $293x=229126$.",
"Step 2: Reduce: $229126\\equiv 1\\pmod{975}$ (since $229125=229125$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{782}$.\nMethod 1 constructs an inverse via Bézout, producing $x=782$. 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 $\boxed{782}$.) |
math-011145 | Number Theory: Units mod m — Existence Condition | 6 | Give a theorem-based solution: Find the multiplicative inverse of $329$ modulo $428$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{428}$).
(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 $x=281$ and compute $329x=92449$.",
"Step 2: Reduce: $92449\\equiv 1\\pmod{428}$ (since $92448=92448$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{281}$.\nMethod 1 constructs an inverse via Bézout, producing $x=281$. 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... | 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 $\boxed{281}$.) |
math-011146 | Number Theory: Units mod m — Existence Condition | 6 | Give a fully justified solution: Find the multiplicative inverse of $454$ modulo $1359$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1359}$).
(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 $\\gcd(454,1359)=1$, 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: $\\boxed{907}$.\nMethod 1 constructs an inverse via Bézout, producing $x=907$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{907}$.) |
math-011147 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Give reasoning, not just computation: Find the multiplicative inverse of $1095$ modulo $1357$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1357}$).
(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 $x=1155$ and compute $1095x=1264725$.",
"Step 2: Reduce: $1264725\\equiv 1\\pmod{1357}$ (since $1264724=126472... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1155}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1155$. 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... | 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 $\boxed{1155}$.) |
math-011148 | Number Theory: Units mod m — Existence Condition | 6 | Exercise: Find the multiplicative inverse of $725$ modulo $1613$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1613}$).
(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 $\\gcd(725,1613)=1$, 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: $\\boxed{663}$.\nMethod 1 constructs an inverse via Bézout, producing $x=663$. 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 $\boxed{663}$.) |
math-011149 | Number Theory: gcd — Back Substitution | 6 | Give reasoning, not just computation: (a) Compute $\gcd(613,1323)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 613+v\cdot 1323=\gcd(613,1323)$.
(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 $(613,1323)$ to compute $g=\\gcd(613,1323)$.",
"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 $u=-341,v=158$ satisfies $u613+v1323=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011150 | Computational Number Theory: Inverses and Certificates | 6 | Work carefully and justify each inference: Find the multiplicative inverse of $1187$ modulo $1293$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1293}$).
(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": "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 $\\gcd(1187,1293)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1232}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1232$. 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... | 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 $\boxed{1232}$.) |
math-011151 | Computational Number Theory: Inverses and Certificates | 6 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $1163$ modulo $1419$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1419}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the neces... | [
{
"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 $x=194$ and compute $1163x=225622$.",
"Step 2: Reduce: $225622\\equiv 1\\pmod{1419}$ (since $225621=225621$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{194}$.\nMethod 1 constructs an inverse via Bézout, producing $x=194$. 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 $\boxed{194}$.) |
math-011152 | Number Theory: Modular Inverses — Extended Euclid | 6 | Prompt: Find the multiplicative inverse of $837$ modulo $1141$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1141}$).
(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 $x=289$ and compute $837x=241893$.",
"Step 2: Reduce: $241893\\equiv 1\\pmod{1141}$ (since $241892=241892$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{289}$.\nMethod 1 constructs an inverse via Bézout, producing $x=289$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011153 | Computational Number Theory: Inverses and Certificates | 6 | Give a theorem-based solution: Find the multiplicative inverse of $377$ modulo $1462$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1462}$).
(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 $\\gcd(377,1462)=1$, 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: $\\boxed{159}$.\nMethod 1 constructs an inverse via Bézout, producing $x=159$. 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-011154 | Number Theory: Units mod m — Existence Condition | 6 | Do not skip justification steps: Find the multiplicative inverse of $346$ modulo $755$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{755}$).
(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 $\\gcd(346,755)=1$, 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: $\\boxed{731}$.\nMethod 1 constructs an inverse via Bézout, producing $x=731$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{731}$.) |
math-011155 | Number Theory: gcd — Back Substitution | 6 | Problem: (a) Compute $\gcd(1317,706)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1317+v\cdot 706=\gcd(1317,706)$.
(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 $(1317,706)$ to compute $g=\\gcd(1317,706)$.",
"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 $u=-327,v=610$ satisfies $u1317+v706=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011156 | Number Theory: gcd — Back Substitution | 6 | Do not skip justification steps: (a) Compute $\gcd(1789,1201)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1789+v\cdot 1201=\gcd(1789,1201)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=96$ and $v=-143$ with $u1789... | {
"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 $u=96,v=-143$ satisfies $u1789+v1201=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Key idea: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011157 | Number Theory: gcd — Back Substitution | 6 | State any required conditions first: (a) Compute $\gcd(474,947)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 474+v\cdot 947=\gcd(474,947)$.
(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 $(474,947)$ to compute $g=\\gcd(474,947)$.",
"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 $u=2,v=-1$ satisfies $u474+v947=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the probl... | [
{
"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-011158 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Task: Find the multiplicative inverse of $184$ modulo $361$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{361}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the 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 $x=155$ and compute $184x=28520$.",
"Step 2: Reduce: $28520\\equiv 1\\pmod{361}$ (since $28519=28519$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{155}$.\nMethod 1 constructs an inverse via Bézout, producing $x=155$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011159 | Number Theory: Modular Inverses — Extended Euclid | 6 | Keep the final answer in boxed form: Find the multiplicative inverse of $553$ modulo $698$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{698}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=491$ and compute $553x=271523$.",
"Step 2: Reduce: $271523\\equiv 1\\pmod{698}$ (since $271522=271522$ is d... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{491}$.\nMethod 1 constructs an inverse via Bézout, producing $x=491$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{491}$.) |
math-011160 | Number Theory: Units mod m — Existence Condition | 6 | Prompt: Find the multiplicative inverse of $234$ modulo $863$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{863}$).
(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 $\\gcd(234,863)=1$, 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: $\\boxed{402}$.\nMethod 1 constructs an inverse via Bézout, producing $x=402$. 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-011161 | Number Theory: Modular Inverses — Extended Euclid | 6 | Explain each transformation: Find the multiplicative inverse of $792$ 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 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 $x=126$ and compute $792x=99792$.",
"Step 2: Reduce: $99792\\equiv 1\\pmod{1367}$ (since $99791=99791$ is divi... | {
"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... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011162 | Number Theory: Units mod m — Existence Condition | 6 | Carefully track domains: Find the multiplicative inverse of $827$ modulo $1446$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1446}$).
(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 $\\gcd(827,1446)=1$, 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: $\\boxed{869}$.\nMethod 1 constructs an inverse via Bézout, producing $x=869$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{869}$.) |
math-011163 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Complete the analysis: Find the multiplicative inverse of $245$ modulo $823$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{823}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an 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 $x=346$ and compute $245x=84770$.",
"Step 2: Reduce: $84770\\equiv 1\\pmod{823}$ (since $84769=84769$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{346}$.\nMethod 1 constructs an inverse via Bézout, producing $x=346$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011164 | Number Theory: gcd — Euclidean Algorithm | 6 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(227,1458)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 227+v\cdot 1458=\gcd(227,1458)$.
(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 $u=167$ and $v=-26$ with $u227+... | {
"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 $u=167,v=-26$ satisfies $u227+v1458=1$, 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-011165 | Computational Number Theory: Inverses and Certificates | 6 | Track units/moduli carefully: Find the multiplicative inverse of $598$ modulo $651$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{651}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition fo... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(598,651)=1$, 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: $\\boxed{565}$.\nMethod 1 constructs an inverse via Bézout, producing $x=565$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{565}$.) |
math-011166 | Number Theory: Units mod m — Existence Condition | 6 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $43$ modulo $584$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{584}$).
(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 $x=163$ and compute $43x=7009$.",
"Step 2: Reduce: $7009\\equiv 1\\pmod{584}$ (since $7008=7008$ is divisible ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{163}$.\nMethod 1 constructs an inverse via Bézout, producing $x=163$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011167 | Number Theory: Units mod m — Existence Condition | 6 | Task: Find the multiplicative inverse of $73$ modulo $933$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{933}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exist.
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 $x=409$ and compute $73x=29857$.",
"Step 2: Reduce: $29857\\equiv 1\\pmod{933}$ (since $29856=29856$ is divisi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{409}$.\nMethod 1 constructs an inverse via Bézout, producing $x=409$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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... | 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 $\boxed{409}$.) |
math-011168 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve (and briefly cross-validate): Find the multiplicative inverse of $73$ modulo $224$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{224}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=89$ and compute $73x=6497$.",
"Step 2: Reduce: $6497\\equiv 1\\pmod{224}$ (since $6496=6496$ is divisible b... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{89}$.\nMethod 1 constructs an inverse via Bézout, producing $x=89$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is f... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{89}$.) |
math-011169 | Number Theory: Modular Inverses — Extended Euclid | 6 | Solve and include a self-check: Find the multiplicative inverse of $869$ modulo $1028$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1028}$).
(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 $x=653$ and compute $869x=567457$.",
"Step 2: Reduce: $567457\\equiv 1\\pmod{1028}$ (since $567456=567456$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{653}$.\nMethod 1 constructs an inverse via Bézout, producing $x=653$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{653}$.) |
math-011170 | Number Theory: Divisibility — Linear Combinations | 6 | Derive the result step-by-step: (a) Compute $\gcd(764,1057)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 764+v\cdot 1057=\gcd(764,1057)$.
(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 $u=386$ and $v=-279$ with $u764... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=386,v=-279$ satisfies $u764+v1057=1$, 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-011171 | Computational Number Theory: Inverses and Certificates | 6 | Give an answer and a quick verification: Find the multiplicative inverse of $485$ modulo $569$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{569}$).
(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 $\\gcd(485,569)=1$, 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: $\\boxed{359}$.\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: ... | [
{
"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-011172 | Number Theory: Divisibility — Linear Combinations | 6 | Write the solution set clearly: (a) Compute $\gcd(487,1481)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 487+v\cdot 1481=\gcd(487,1481)$.
(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 $u=222$ and $v=-73$ with $u487+... | {
"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 $u=222,v=-73$ satisfies $u487+v1481=1$, 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-011173 | Number Theory: gcd — Euclidean Algorithm | 6 | Give a fully justified solution: (a) Compute $\gcd(1686,1465)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1686+v\cdot 1465=\gcd(1686,1465)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution ch... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=411$ and $v=-473$ with $u168... | {
"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 $u=411,v=-473$ satisfies $u1686+v1465=1$, 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.",
... | 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-011174 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Give an answer and a quick verification: Find the multiplicative inverse of $3$ modulo $92$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{92}$).
(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 $\\gcd(3,92)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that $... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{31}$.\nMethod 1 constructs an inverse via Bézout, producing $x=31$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Ex... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{31}$.) |
math-011175 | Number Theory: Modular Inverses — Extended Euclid | 6 | Track quantifiers carefully: Find the multiplicative inverse of $510$ modulo $1199$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1199}$).
(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 $x=355$ and compute $510x=181050$.",
"Step 2: Reduce: $181050\\equiv 1\\pmod{1199}$ (since $181049=181049$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{355}$.\nMethod 1 constructs an inverse via Bézout, producing $x=355$. 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 $\boxed{355}$.) |
math-011176 | Number Theory: Divisibility — Linear Combinations | 6 | Work this out carefully: (a) Compute $\gcd(1303,628)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1303+v\cdot 628=\gcd(1303,628)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Inclu... | [
{
"method_name": "Bézout + Divisibility Argument",
"approach": "Use the definition of gcd as the smallest positive linear combination and show any common divisor divides your combination.",
"steps": [
"Step 1: From the extended Euclidean algorithm we obtain integers $u=147$ and $v=-305$ with $u130... | {
"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 $u=147,v=-305$ satisfies $u1303+v628=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011177 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Solve and justify each step: Find the multiplicative inverse of $571$ modulo $1532$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1532}$).
(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 $x=711$ and compute $571x=405981$.",
"Step 2: Reduce: $405981\\equiv 1\\pmod{1532}$ (since $405980=405980$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{711}$.\nMethod 1 constructs an inverse via Bézout, producing $x=711$. 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 $\boxed{711}$.) |
math-011178 | Number Theory: Modular Inverses — Extended Euclid | 6 | Give reasoning, not just computation: Find the multiplicative inverse of $797$ modulo $1574$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1574}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=551$ and compute $797x=439147$.",
"Step 2: Reduce: $439147\\equiv 1\\pmod{1574}$ (since $439146=439146$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{551}$.\nMethod 1 constructs an inverse via Bézout, producing $x=551$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{551}$.) |
math-011179 | Number Theory: Units mod m — Existence Condition | 6 | Solve with verification: Find the multiplicative inverse of $198$ modulo $1019$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1019}$).
(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 $\\gcd(198,1019)=1$, 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: $\\boxed{808}$.\nMethod 1 constructs an inverse via Bézout, producing $x=808$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011180 | Number Theory: Units mod m — Existence Condition | 6 | Warm-up: Find the multiplicative inverse of $17$ modulo $593$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{593}$).
(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 $\\gcd(17,593)=1$, 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: $\\boxed{314}$.\nMethod 1 constructs an inverse via Bézout, producing $x=314$. 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... | 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 $\boxed{314}$.) |
math-011181 | Number Theory: Bézout Identity — Certificates | 6 | Determine the requested value: (a) Compute $\gcd(639,1802)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 639+v\cdot 1802=\gcd(639,1802)$.
(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 $(639,1802)$ to compute $g=\\gcd(639,1802)$.",
"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 $u=-141,v=50$ satisfies $u639+v1802=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scales ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011182 | Number Theory: Units mod m — Existence Condition | 6 | Indicate where a theorem is used: Find the multiplicative inverse of $793$ modulo $1073$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1073}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condit... | [
{
"method_name": "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 $\\gcd(793,1073)=1$, 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: $\\boxed{525}$.\nMethod 1 constructs an inverse via Bézout, producing $x=525$. 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 $\boxed{525}$.) |
math-011183 | Number Theory: Divisibility — Linear Combinations | 6 | Track units/moduli carefully: (a) Compute $\gcd(1600,1336)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1600+v\cdot 1336=\gcd(1600,1336)$.
(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 $(1600,1336)$ to compute $g=\\gcd(1600,1336)$.",
"Step 2: Record the remaind... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{8}$.\nThe Euclidean algorithm computes $g=8$. The Bézout certificate $u=81,v=-97$ satisfies $u1600+v1336=8$, 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 $ua+vb=g$. |
math-011184 | Computational Number Theory: Extended Euclid | 6 | Explain why your operations are valid: (a) Compute $\gcd(1933,1609)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1933+v\cdot 1609=\gcd(1933,1609)$.
(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 $(1933,1609)$ to compute $g=\\gcd(1933,1609)$.",
"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 $u=293,v=-352$ satisfies $u1933+v1609=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness note: Euclid scale... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011185 | Number Theory: Units mod m — Existence Condition | 6 | Work this out carefully: Find the multiplicative inverse of $385$ modulo $818$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{818}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(385,818)=1$, 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: $\\boxed{17}$.\nMethod 1 constructs an inverse via Bézout, producing $x=17$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivit... | [
{
"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-011186 | Number Theory: gcd — Back Substitution | 6 | Try to avoid pattern-matching; explain why: (a) Compute $\gcd(988,842)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 988+v\cdot 842=\gcd(988,842)$.
(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 $u=-173$ and $v=203$ with $u988... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2}$.\nThe Euclidean algorithm computes $g=2$. The Bézout certificate $u=-173,v=203$ satisfies $u988+v842=2$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensitivity analysis: Euc... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | 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-011187 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Derive the result step-by-step: Find the multiplicative inverse of $881$ modulo $1014$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1014}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $x=953$ and compute $881x=839593$.",
"Step 2: Reduce: $839593\\equiv 1\\pmod{1014}$ (since $839592=839592$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{953}$.\nMethod 1 constructs an inverse via Bézout, producing $x=953$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011188 | Number Theory: Bézout Identity — Certificates | 6 | Track quantifiers carefully: (a) Compute $\gcd(1941,1329)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1941+v\cdot 1329=\gcd(1941,1329)$.
(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 $(1941,1329)$ to compute $g=\\gcd(1941,1329)$.",
"Step 2: Record the remaind... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3}$.\nThe Euclidean algorithm computes $g=3$. The Bézout certificate $u=-76,v=111$ satisfies $u1941+v1329=3$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis":... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. (Here the result is $\boxed{3}$.) |
math-011189 | Number Theory: Divisibility — Linear Combinations | 6 | Solve and sanity-check: (a) Compute $\gcd(337,1139)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 337+v\cdot 1139=\gcd(337,1139)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Includ... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(337,1139)$ to compute $g=\\gcd(337,1139)$.",
"Step 2: Record the remainder... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1}$.\nThe Euclidean algorithm computes $g=1$. The Bézout certificate $u=436,v=-129$ satisfies $u337+v1139=1$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Sensiti... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011190 | Number Theory: gcd — Euclidean Algorithm | 6 | Answer with a short justification: (a) Compute $\gcd(90,900)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 90+v\cdot 900=\gcd(90,900)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
I... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(90,900)$ to compute $g=\\gcd(90,900)$.",
"Step 2: Record the remainder equ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{90}$.\nThe Euclidean algorithm computes $g=90$. The Bézout certificate $u=1,v=0$ satisfies $u90+v900=90$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "Robustness 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.",
... | 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-011191 | Number Theory: Units mod m — Existence Condition | 6 | Work carefully and justify each inference: Find the multiplicative inverse of $133$ modulo $645$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{645}$).
(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 $x=97$ and compute $133x=12901$.",
"Step 2: Reduce: $12901\\equiv 1\\pmod{645}$ (since $12900=12900$ is divisi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{97}$.\nMethod 1 constructs an inverse via Bézout, producing $x=97$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Ex... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{97}$.) |
math-011192 | Computational Number Theory: Inverses and Certificates | 6 | Question: Find the multiplicative inverse of $59$ modulo $303$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{303}$).
(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 $\\gcd(59,303)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such that... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{113}$.\nMethod 1 constructs an inverse via Bézout, producing $x=113$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{113}$.) |
math-011193 | Number Theory: gcd — Back Substitution | 6 | Complete the analysis: (a) Compute $\gcd(1787,1492)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 1787+v\cdot 1492=\gcd(1787,1492)$.
(c) Briefly explain why your coefficients certify the gcd.
You must show the Euclidean algorithm remainder steps or a clear backward-substitution chain.
Incl... | [
{
"method_name": "Euclidean Algorithm + Back-Substitution",
"approach": "Compute the gcd by repeated division, then reverse the steps to express the gcd as a linear combination.",
"steps": [
"Step 1: Apply Euclid to $(1787,1492)$ to compute $g=\\gcd(1787,1492)$.",
"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 $u=263,v=-315$ satisfies $u1787+v1492=1$, 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 $ua+vb=g$. (Here the result is $\boxed{1}$.) |
math-011194 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Give reasoning, not just computation: Find the multiplicative inverse of $46$ modulo $201$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{201}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condi... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=118$ and compute $46x=5428$.",
"Step 2: Reduce: $5428\\equiv 1\\pmod{201}$ (since $5427=5427$ is divisible ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{118}$.\nMethod 1 constructs an inverse via Bézout, producing $x=118$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{118}$.) |
math-011195 | Number Theory: Units mod m — Existence Condition | 6 | Solve and justify each step: Find the multiplicative inverse of $881$ modulo $1153$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1153}$).
(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 $x=479$ and compute $881x=421999$.",
"Step 2: Reduce: $421999\\equiv 1\\pmod{1153}$ (since $421998=421998$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{479}$.\nMethod 1 constructs an inverse via Bézout, producing $x=479$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-011196 | Number Theory: Modular Inverses — Extended Euclid | 6 | State any required conditions first: Find the multiplicative inverse of $655$ modulo $1177$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1177}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient con... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=354$ and compute $655x=231870$.",
"Step 2: Reduce: $231870\\equiv 1\\pmod{1177}$ (since $231869=231869$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{354}$.\nMethod 1 constructs an inverse via Bézout, producing $x=354$. 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 $\boxed{354}$.) |
math-011197 | Number Theory: Congruences — Solving $ax\equiv 1$ | 6 | Provide a rigorous solution: Find the multiplicative inverse of $765$ modulo $796$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{796}$).
(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 $\\gcd(765,796)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such tha... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{77}$.\nMethod 1 constructs an inverse via Bézout, producing $x=77$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast a... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{77}$.) |
math-011198 | Number Theory: Divisibility — Linear Combinations | 6 | Where appropriate, name the theorem you use: (a) Compute $\gcd(155,1795)$ using the Euclidean algorithm.
(b) Find integers $u,v$ such that $u\cdot 155+v\cdot 1795=\gcd(155,1795)$.
(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 $u=139$ and $v=-12$ with $u155+... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5}$.\nThe Euclidean algorithm computes $g=5$. The Bézout certificate $u=139,v=-12$ satisfies $u155+v1795=5$, and divisibility shows no larger common divisor can exist.",
"robustness_analysis": "If the problem were perturbed: ... | [
{
"error_description": "Stopped Euclid early and used the last remainder before reaching 0.",
"why_plausible": "The repeated division process is easy to truncate accidentally.",
"why_wrong": "The gcd is the last nonzero remainder; stopping early yields a multiple of the gcd, not necessarily the gcd.",
... | Remember: The Euclidean algorithm computes gcds efficiently, and Bézout coefficients $u,v$ certify the result because every common divisor must divide $ua+vb=g$. |
math-011199 | Number Theory: Units mod m — Existence Condition | 6 | Give a theorem-based solution: Find the multiplicative inverse of $817$ modulo $1640$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1640}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $x=273$ and compute $817x=223041$.",
"Step 2: Reduce: $223041\\equiv 1\\pmod{1640}$ (since $223040=223040$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{273}$.\nMethod 1 constructs an inverse via Bézout, producing $x=273$. 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... | 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 $\boxed{273}$.) |
math-011200 | Computational Number Theory: Inverses and Certificates | 6 | Give reasoning, not just computation: Find the multiplicative inverse of $251$ modulo $603$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{603}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient cond... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(251,603)=1$, 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: $\\boxed{197}$.\nMethod 1 constructs an inverse via Bézout, producing $x=197$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{197}$.) |
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