id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-017301 | Number Theory: Units mod m — Existence Condition | 9 | Warm-up: Find the multiplicative inverse of $1467$ modulo $1528$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1528}$).
(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 $x=1027$ and compute $1467x=1506609$.",
"Step 2: Reduce: $1506609\\equiv 1\\pmod{1528}$ (since $1506608=150660... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1027}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1027$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robust... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, 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-017302 | Number Theory: Euler Totient (Variant B) | 9 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=184877=7^5\cdot 11^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{144060}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=11$ yields the same integer 144060.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{144060}$.) |
math-017303 | Number Theory: Modular Inverses — Extended Euclid | 9 | Write the solution set clearly: Find the multiplicative inverse of $479$ modulo $1658$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1658}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditio... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(479,1658)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{45}$.\nMethod 1 constructs an inverse via Bézout, producing $x=45$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Exten... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{45}$.) |
math-017304 | Number Theory: Euler Totient (Core) | 9 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=39738676943=41^5\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ c... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=41$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{33230949360}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=7$ yields the same integer 33230949360.",
"robustness_analy... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017305 | Number Theory: Modular Inverses — Extended Euclid | 9 | Track units/moduli carefully: Find the multiplicative inverse of $665$ modulo $901$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{901}$).
(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 $x=859$ and compute $665x=571235$.",
"Step 2: Reduce: $571235\\equiv 1\\pmod{901}$ (since $571234=571234$ is d... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{859}$.\nMethod 1 constructs an inverse via Bézout, producing $x=859$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017306 | Number Theory: Euler Totient (Variant B) | 9 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=60835=5^1\cdot 23^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multip... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=5$ and $q=23$ are distinct primes, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{46552}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=23$ yields the same integer 46552.",
"robustness_analysis": "If the problem were perturbed: ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017307 | Computational Number Theory: Inverses and Certificates | 9 | Solve and then verify: Find the multiplicative inverse of $336$ modulo $503$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{503}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an in... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(336,503)=1$, 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: $\\boxed{253}$.\nMethod 1 constructs an inverse via Bézout, producing $x=253$. 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 $\boxed{253}$.) |
math-017308 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Checkpoint: Find the multiplicative inverse of $251$ modulo $553$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{553}$).
(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 $\\gcd(251,553)=1$, 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: $\\boxed{412}$.\nMethod 1 constructs an inverse via Bézout, producing $x=412$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{412}$.) |
math-017309 | Number Theory: Euler Totient (Variant A) | 9 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=259=37^1\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{216}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=7$ yields the same integer 216.",
"robustness_analysis": "Robustness note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{216}$.) |
math-017310 | Number Theory: Euler Totient (Core) | 9 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=4375=7^1\cdot 5^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=5$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=5$ yields the same integer 3000.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017311 | Number Theory: Modular Inverses — Extended Euclid | 9 | Problem: Find the multiplicative inverse of $390$ modulo $451$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{451}$).
(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 $\\gcd(390,451)=1$, 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: $\\boxed{207}$.\nMethod 1 constructs an inverse via Bézout, producing $x=207$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{207}$.) |
math-017312 | Number Theory: Euler Totient (Variant C) | 9 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=4230589213=17^4\cdot 37^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ coun... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=17$ and $q=37$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3874116672}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=37$ yields the same integer 3874116672.",
"robustness_analysis": "Generality note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{3874116672}$.) |
math-017313 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Determine the requested value: Find the multiplicative inverse of $436$ modulo $1769$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1769}$).
(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 $x=1132$ and compute $436x=493552$.",
"Step 2: Reduce: $493552\\equiv 1\\pmod{1769}$ (since $493551=493551$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1132}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1132$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017314 | Computational Number Theory: Inverses and Certificates | 9 | Work this out carefully: Find the multiplicative inverse of $518$ modulo $1973$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1973}$).
(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 $\\gcd(518,1973)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{857}$.\nMethod 1 constructs an inverse via Bézout, producing $x=857$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{857}$.) |
math-017315 | Number Theory: Units mod m — Existence Condition | 9 | Exercise: Find the multiplicative inverse of $757$ modulo $855$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{855}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(757,855)=1$, 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: $\\boxed{253}$.\nMethod 1 constructs an inverse via Bézout, producing $x=253$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{253}$.) |
math-017316 | Number Theory: Units mod m — Existence Condition | 9 | Solve (and briefly cross-validate): Find the multiplicative inverse of $37$ modulo $1348$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1348}$).
(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 $\\gcd(37,1348)=1$, 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: $\\boxed{1093}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1093$. 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... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017317 | Number Theory: Euler Totient (Core) | 9 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=5488=2^4\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ex... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=2$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2352}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=7$ yields the same integer 2352.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2352}$.) |
math-017318 | Number Theory: Euler Totient (Variant B) | 9 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=3159=3^5\cdot 13^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1944}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=13$ yields the same integer 1944.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017319 | Number Theory: Euler Totient (Variant C) | 9 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=2673=3^5\cdot 11^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1620}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=11$ yields the same integer 1620.",
"robustness_analysis": "Sensiti... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017320 | Computational Number Theory: Inverses and Certificates | 9 | Use two approaches if possible: Find the multiplicative inverse of $394$ modulo $1271$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1271}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditio... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(394,1271)=1$, 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: $\\boxed{1171}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1171$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017321 | Number Theory: Euler Totient (Variant A) | 9 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=1083=3^1\cdot 19^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\var... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=19$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{684}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=19$ yields the same integer 684.",
"robustness_analysis": "Sensitivity analysis: Both methods rely o... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{684}$.) |
math-017322 | Number Theory: Euler Totient (Variant A) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=140170841=43^4\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{133571760}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=41$ yields the same integer 133571760.",
"robustness_analysis... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{133571760}$.) |
math-017323 | Number Theory: Euler Totient (Variant A) | 9 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=42527=23^1\cdot 43^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ count... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=43$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{39732}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=43$ yields the same integer 39732.",
"robustness_analysis": "Sens... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017324 | Number Theory: Euler Totient (Variant B) | 9 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=131769=11^4\cdot 3^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expl... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=11$ and $q=3$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{79860}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=3$ yields the same integer 79860.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{79860}$.) |
math-017325 | Number Theory: Euler Totient (Variant B) | 9 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=48125602831=31^5\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=31$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{45437233200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=41$ yields the same integer 45437233200.",
"robustness_analysis": "If the problem wer... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{45437233200}$.) |
math-017326 | Number Theory: Euler Totient (Variant B) | 9 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=11191=19^2\cdot 31^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=19$ and $q=31$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{10260}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=31$ yields the same integer 10260.",
"robustness_analysis": "Sensitivity analysis: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017327 | Number Theory: Euler Totient (Variant B) | 9 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=15979=29^2\cdot 19^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ count... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=19$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{14616}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=19$ yields the same integer 14616.",
"robustness_analysis": "Robustness note: Both methods rely o... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{14616}$.) |
math-017328 | Number Theory: Euler Totient (Core) | 9 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=1264300249=37^2\cdot 31^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=31$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1190448360}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=31$ yields the same integer 1190448360.",
"robustness_analys... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1190448360}$.) |
math-017329 | Number Theory: Euler Totient (Core) | 9 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=3377129=7^2\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2824080}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=41$ yields the same integer 2824080.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017330 | Number Theory: Units mod m — Existence Condition | 9 | Try to avoid pattern-matching; explain why: Find the multiplicative inverse of $882$ modulo $1459$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1459}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffici... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=177$ and compute $882x=156114$.",
"Step 2: Reduce: $156114\\equiv 1\\pmod{1459}$ (since $156113=156113$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{177}$.\nMethod 1 constructs an inverse via Bézout, producing $x=177$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{177}$.) |
math-017331 | Number Theory: Euler Totient (Core) | 9 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=90424352=2^5\cdot 41^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=41$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{44109440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=41$ yields the same integer 44109440.",
"robustness_analysis": "Robustness note: Both met... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{44109440}$.) |
math-017332 | Computational Number Theory: Inverses and Certificates | 9 | Give a fully justified solution: Find the multiplicative inverse of $716$ modulo $779$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{779}$).
(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 $x=136$ and compute $716x=97376$.",
"Step 2: Reduce: $97376\\equiv 1\\pmod{779}$ (since $97375=97375$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{136}$.\nMethod 1 constructs an inverse via Bézout, producing $x=136$. 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-017333 | Number Theory: Euler Totient (Variant B) | 9 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=772987077=31^5\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=3$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{498701340}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=3$ yields the same integer 498701340.",
"robustness_analysis": "Genera... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017334 | Computational Number Theory: Inverses and Certificates | 9 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $173$ modulo $277$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{277}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficie... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(173,277)=1$, 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: $\\boxed{269}$.\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: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{269}$.) |
math-017335 | Number Theory: Euler Totient (Core) | 9 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=148=37^1\cdot 2^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=37$ and $q=2$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{72}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=2$ yields the same integer 72.",
"robustness_analysis": "Sensitivity analysis: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017336 | Number Theory: Euler Totient (Core) | 9 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=27783=3^4\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{15876}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=7$ yields the same integer 15876.",
"robustness_analysis": "Generality note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{15876}$.) |
math-017337 | Number Theory: Modular Inverses — Extended Euclid | 9 | Solve and include a self-check: Find the multiplicative inverse of $1698$ modulo $1847$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1847}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $x=595$ and compute $1698x=1010310$.",
"Step 2: Reduce: $1010310\\equiv 1\\pmod{1847}$ (since $1010309=1010309... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{595}$.\nMethod 1 constructs an inverse via Bézout, producing $x=595$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{595}$.) |
math-017338 | Number Theory: Euler Totient (Variant A) | 9 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=57289761=3^4\cdot 29^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{36876168}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=29$ yields the same integer 36876168.",
"robustness_analysis": ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{36876168}$.) |
math-017339 | Computational Number Theory: Inverses and Certificates | 9 | Work this out carefully: Find the multiplicative inverse of $173$ modulo $654$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{654}$).
(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 $x=155$ and compute $173x=26815$.",
"Step 2: Reduce: $26815\\equiv 1\\pmod{654}$ (since $26814=26814$ is divis... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. 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": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017340 | Number Theory: Euler Totient (Core) | 9 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=281219=19^3\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multi... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=19$ and $q=41$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{259920}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=41$ yields the same integer 259920.",
"robustness_analysis": "Se... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{259920}$.) |
math-017341 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Complete the analysis: Find the multiplicative inverse of $332$ modulo $553$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{553}$).
(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 $x=5$ and compute $332x=1660$.",
"Step 2: Reduce: $1660\\equiv 1\\pmod{553}$ (since $1659=1659$ is divisible b... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5}$.\nMethod 1 constructs an inverse via Bézout, producing $x=5$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended Eucl... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, 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 $\boxed{5}$.) |
math-017342 | Number Theory: Euler Totient (Variant A) | 9 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=115625=5^5\cdot 37^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ coun... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{90000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=37$ yields the same integer 90000.",
"robustness_analysis": "If the problem were perturbed: Both m... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017343 | Number Theory: Euler Totient (Variant A) | 9 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=1445=5^1\cdot 17^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1088}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=17$ yields the same integer 1088.",
"robustness_analysis": "Sensitivity anal... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1088}$.) |
math-017344 | Number Theory: Euler Totient (Variant A) | 9 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=58557989=29^3\cdot 7^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ c... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{48461784}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=7$ yields the same integer 48461784.",
"robustness_analysis": ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017345 | Computational Number Theory: Inverses and Certificates | 9 | Question: Find the multiplicative inverse of $269$ modulo $938$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{938}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(269,938)=1$, 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: $\\boxed{537}$.\nMethod 1 constructs an inverse via Bézout, producing $x=537$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017346 | Number Theory: Euler Totient (Variant B) | 9 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=21853=13^1\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=13$ and $q=41$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{19680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=41$ yields the same integer 19680.",
"robustness_analysis": "Robustness note: Both methods ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{19680}$.) |
math-017347 | Computational Number Theory: Inverses and Certificates | 9 | Solve and sanity-check: Find the multiplicative inverse of $295$ 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 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 $\\gcd(295,1421)=1$, 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: $\\boxed{1079}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1079$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1079}$.) |
math-017348 | Number Theory: Euler Totient (Core) | 9 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=1071875=5^5\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit a... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=5$ and $q=7$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{735000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=7$ yields the same integer 735000.",
"robustness_analysis": "If the problem were perturbed:... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{735000}$.) |
math-017349 | Number Theory: Euler Totient (Variant A) | 9 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=10648=2^3\cdot 11^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=11$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{4840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=11$ yields the same integer 4840.",
"robustness_analysis": "Sensiti... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{4840}$.) |
math-017350 | Number Theory: Euler Totient (Variant C) | 9 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=5043=3^1\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=41$ yields the same integer 3280.",
"robustness_analysis": "If the problem w... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{3280}$.) |
math-017351 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Work this out carefully: Find the multiplicative inverse of $193$ modulo $795$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{795}$).
(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 $x=622$ and compute $193x=120046$.",
"Step 2: Reduce: $120046\\equiv 1\\pmod{795}$ (since $120045=120045$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{622}$.\nMethod 1 constructs an inverse via Bézout, producing $x=622$. 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 $\boxed{622}$.) |
math-017352 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | State any required conditions first: Find the multiplicative inverse of $257$ modulo $488$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{488}$).
(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 $x=169$ and compute $257x=43433$.",
"Step 2: Reduce: $43433\\equiv 1\\pmod{488}$ (since $43432=43432$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{169}$.\nMethod 1 constructs an inverse via Bézout, producing $x=169$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017353 | Number Theory: Modular Inverses — Extended Euclid | 9 | Track quantifiers carefully: Find the multiplicative inverse of $240$ modulo $947$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{947}$).
(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 $\\gcd(240,947)=1$, 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: $\\boxed{146}$.\nMethod 1 constructs an inverse via Bézout, producing $x=146$. 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 $\boxed{146}$.) |
math-017354 | Number Theory: Euler Totient (Variant C) | 9 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=48749=29^1\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=29$ and $q=41$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{45920}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=41$ yields the same integer 45920.",
"robustness_analysis": "Sensitivity analysis: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{45920}$.) |
math-017355 | Number Theory: Euler Totient (Variant C) | 9 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=290521=7^4\cdot 11^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit abo... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{226380}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=11$ yields the same integer 226380.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017356 | Number Theory: Euler Totient (Variant C) | 9 | Do not skip justification steps: Compute Euler's totient function $\varphi(n)$.
Here $n=325=5^2\cdot 13^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=13$ yields the same integer 240.",
"robustness_analysis": "If the problem were perturbed: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017357 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Compute the requested quantity: Find the multiplicative inverse of $906$ modulo $973$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{973}$).
(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 $\\gcd(906,973)=1$, 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: $\\boxed{334}$.\nMethod 1 constructs an inverse via Bézout, producing $x=334$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{334}$.) |
math-017358 | Number Theory: Euler Totient (Variant C) | 9 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=4039951=31^1\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ex... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=19$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3703860}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=19$ yields the same integer 3703860.",
"robustness_analysis": "Robustness note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{3703860}$.) |
math-017359 | Number Theory: Euler Totient (Variant C) | 9 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=689087=7^5\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=41$ are distinct primes, $\\gcd(p^5,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{576240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=41$ yields the same integer 576240.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{576240}$.) |
math-017360 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Work carefully and justify each inference: Find the multiplicative inverse of $63$ modulo $134$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{134}$).
(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 $\\gcd(63,134)=1$, 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: $\\boxed{117}$.\nMethod 1 constructs an inverse via Bézout, producing $x=117$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017361 | Number Theory: Units mod m — Existence Condition | 9 | Problem: Find the multiplicative inverse of $23$ modulo $105$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{105}$).
(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 $\\gcd(23,105)=1$, 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: $\\boxed{32}$.\nMethod 1 constructs an inverse via Bézout, producing $x=32$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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. (Here the result is $\boxed{32}$.) |
math-017362 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Solve and justify each step: Find the multiplicative inverse of $815$ modulo $1918$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1918}$).
(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 $x=313$ and compute $815x=255095$.",
"Step 2: Reduce: $255095\\equiv 1\\pmod{1918}$ (since $255094=255094$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{313}$.\nMethod 1 constructs an inverse via Bézout, producing $x=313$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{313}$.) |
math-017363 | Number Theory: Euler Totient (Variant C) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=68107=31^1\cdot 13^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=13$ are distinct primes, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{60840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=13$ yields the same integer 60840.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017364 | Number Theory: Euler Totient (Core) | 9 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=125238481=31^2\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=19$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{114819660}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=19$ yields the same integer 114819660.",
"robustness_analysis... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{114819660}$.) |
math-017365 | Computational Number Theory: Inverses and Certificates | 9 | Explain what is being counted/optimized: Find the multiplicative inverse of $529$ modulo $1135$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1135}$).
(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 $x=339$ and compute $529x=179331$.",
"Step 2: Reduce: $179331\\equiv 1\\pmod{1135}$ (since $179330=179330$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{339}$.\nMethod 1 constructs an inverse via Bézout, producing $x=339$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{339}$.) |
math-017366 | Number Theory: Euler Totient (Core) | 9 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=231125=43^2\cdot 5^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ coun... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{180600}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=5$ yields the same integer 180600.",
"robustness_analysis": "If the probl... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017367 | Number Theory: Euler Totient (Variant C) | 9 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=126495637=43^4\cdot 37^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{120214584}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=37$ yields the same integer 120214584.",
"robustness_analysis": "Generality note: Both ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{120214584}$.) |
math-017368 | Computational Number Theory: Inverses and Certificates | 9 | Explain why your operations are valid: Find the multiplicative inverse of $316$ modulo $1921$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1921}$).
(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 $x=845$ and compute $316x=267020$.",
"Step 2: Reduce: $267020\\equiv 1\\pmod{1921}$ (since $267019=267019$ is ... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{845}$.\nMethod 1 constructs an inverse via Bézout, producing $x=845$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017369 | Number Theory: Units mod m — Existence Condition | 9 | Task: Find the multiplicative inverse of $1170$ modulo $1187$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1187}$).
(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 $\\gcd(1170,1187)=1$, 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: $\\boxed{768}$.\nMethod 1 constructs an inverse via Bézout, producing $x=768$. 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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017370 | Number Theory: Euler Totient (Variant B) | 9 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=3675211075=43^5\cdot 5^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2871792840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=5$ yields the same integer 2871792840.",
"robustness_analysis": "Sensitivity analysis:... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017371 | Number Theory: Euler Totient (Variant C) | 9 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=2056223=23^3\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=23$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1815528}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=13$ yields the same integer 1815528.",
"robustness_analysis": "Generality note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1815528}$.) |
math-017372 | Number Theory: Modular Inverses — Extended Euclid | 9 | Track units/moduli carefully: Find the multiplicative inverse of $207$ modulo $377$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{377}$).
(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 $x=51$ and compute $207x=10557$.",
"Step 2: Reduce: $10557\\equiv 1\\pmod{377}$ (since $10556=10556$ is divisi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{51}$.\nMethod 1 constructs an inverse via Bézout, producing $x=51$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{51}$.) |
math-017373 | Number Theory: Euler Totient (Variant A) | 9 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=36252565459=19^5\cdot 11^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ co... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=19$ and $q=11$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{31222305180}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=11$ yields the same integer 31222305180.",
"robustness_analysis": "R... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017374 | Number Theory: Units mod m — Existence Condition | 9 | Do not skip justification steps: Find the multiplicative inverse of $123$ modulo $1043$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1043}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $\\gcd(123,1043)=1$, 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: $\\boxed{212}$.\nMethod 1 constructs an inverse via Bézout, producing $x=212$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017375 | Number Theory: Modular Inverses — Extended Euclid | 9 | Solve and include a self-check: Find the multiplicative inverse of $1109$ modulo $1222$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1222}$).
(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 $\\gcd(1109,1222)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such t... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{465}$.\nMethod 1 constructs an inverse via Bézout, producing $x=465$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{465}$.) |
math-017376 | Number Theory: Euler Totient (Variant A) | 9 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=6877=13^1\cdot 23^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ex... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=13$ and $q=23$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{6072}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=23$ yields the same integer 6072.",
"robustness_analysis": "Generality note: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017377 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Answer using clear logical steps: Find the multiplicative inverse of $394$ modulo $1761$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1761}$).
(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 $\\gcd(394,1761)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{514}$.\nMethod 1 constructs an inverse via Bézout, producing $x=514$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{514}$.) |
math-017378 | Number Theory: Euler Totient (Variant B) | 9 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=496=31^1\cdot 2^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varp... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=31$ and $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=2$ yields the same integer 240.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{240}$.) |
math-017379 | Number Theory: Euler Totient (Core) | 9 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=1808545729=23^2\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=43$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1689682764}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=43$ yields the same integer 1689682764.",
"robustness_analys... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1689682764}$.) |
math-017380 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Explain why your operations are valid: Find the multiplicative inverse of $531$ modulo $641$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{641}$).
(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 $\\gcd(531,641)=1$, 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: $\\boxed{472}$.\nMethod 1 constructs an inverse via Bézout, producing $x=472$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017381 | Number Theory: Euler Totient (Variant C) | 9 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=64827=7^4\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=3$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{37044}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=3$ yields the same integer 37044.",
"robustness_analysis": "Generality note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017382 | Number Theory: Euler Totient (Variant C) | 9 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=84182245951=41^4\cdot 31^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=41$ and $q=31$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{79479697200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=31$ yields the same integer 79479697200.",
"robustness_anal... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017383 | Number Theory: Euler Totient (Variant A) | 9 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=5616860517=37^5\cdot 3^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=37$ and $q=3$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3643368984}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=3$ yields the same integer 3643368984.",
"robustness_analysis": "Sensitivity analysis:... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017384 | Computational Number Theory: Inverses and Certificates | 9 | Give reasoning, not just computation: Find the multiplicative inverse of $349$ modulo $1175$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1175}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(349,1175)=1$, 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: $\\boxed{1074}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1074$. 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... | 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 $\boxed{1074}$.) |
math-017385 | Computational Number Theory: Inverses and Certificates | 9 | Compute the requested quantity: Find the multiplicative inverse of $276$ modulo $421$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{421}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition ... | [
{
"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 $\\gcd(276,421)=1$, 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: $\\boxed{90}$.\nMethod 1 constructs an inverse via Bézout, producing $x=90$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{90}$.) |
math-017386 | Number Theory: Euler Totient (Variant C) | 9 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=4680338569=37^2\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4447939608}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=43$ yields the same integer 4447939608.",
"robustness_analysis": "Rob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{4447939608}$.) |
math-017387 | Number Theory: Modular Inverses — Extended Euclid | 9 | Solve (and briefly cross-validate): Find the multiplicative inverse of $305$ modulo $388$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{388}$).
(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 $\\gcd(305,388)=1$, 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: $\\boxed{201}$.\nMethod 1 constructs an inverse via Bézout, producing $x=201$. 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 $\boxed{201}$.) |
math-017388 | Number Theory: Euler Totient (Core) | 9 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=297=11^1\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ex... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ and $q=3$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{180}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=3$ yields the same integer 180.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{180}$.) |
math-017389 | Number Theory: Euler Totient (Core) | 9 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=507=3^1\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multiplic... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{312}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=13$ yields the same integer 312.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{312}$.) |
math-017390 | Number Theory: Euler Totient (Variant C) | 9 | Task: Compute Euler's totient function $\varphi(n)$.
Here $n=5513329989199=37^5\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5239569417768}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=43$ yields the same integer 5239569417768.",
"robustness_analysis"... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{5239569417768}$.) |
math-017391 | Number Theory: Euler Totient (Core) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=7891499=11^5\cdot 7^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6149220}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=7$ yields the same integer 6149220.",
"robustness_analysis": "Sensitivity analysis: Both method... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017392 | Number Theory: Units mod m — Existence Condition | 9 | Solve and sanity-check: Find the multiplicative inverse of $88$ modulo $249$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{249}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an in... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(88,249)=1$, 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: $\\boxed{133}$.\nMethod 1 constructs an inverse via Bézout, producing $x=133$. 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... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017393 | Number Theory: Modular Inverses — Extended Euclid | 9 | Warm-up: Find the multiplicative inverse of $346$ modulo $1935$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1935}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exi... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(346,1935)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{151}$.\nMethod 1 constructs an inverse via Bézout, producing $x=151$. 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 $\boxed{151}$.) |
math-017394 | Number Theory: Modular Inverses — Extended Euclid | 9 | Determine the requested value: Find the multiplicative inverse of $457$ modulo $1560$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1560}$).
(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 $x=553$ and compute $457x=252721$.",
"Step 2: Reduce: $252721\\equiv 1\\pmod{1560}$ (since $252720=252720$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{553}$.\nMethod 1 constructs an inverse via Bézout, producing $x=553$. 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 $\boxed{553}$.) |
math-017395 | Number Theory: Units mod m — Existence Condition | 9 | Give an answer and a quick verification: Find the multiplicative inverse of $404$ modulo $463$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{463}$).
(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 $\\gcd(404,463)=1$, 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: $\\boxed{102}$.\nMethod 1 constructs an inverse via Bézout, producing $x=102$. 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. |
math-017396 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Give a fully justified solution: Find the multiplicative inverse of $565$ modulo $688$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{688}$).
(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 $\\gcd(565,688)=1$, 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: $\\boxed{509}$.\nMethod 1 constructs an inverse via Bézout, producing $x=509$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem we... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{509}$.) |
math-017397 | Number Theory: Euler Totient (Core) | 9 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=44444413=43^4\cdot 13^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{40071528}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=13$ yields the same integer 40071528.",
"robustness_analysis":... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{40071528}$.) |
math-017398 | Computational Number Theory: Inverses and Certificates | 9 | Provide both a computational and a conceptual explanation: Find the multiplicative inverse of $1352$ modulo $1515$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1515}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the neces... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(1352,1515)=1$, 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: $\\boxed{158}$.\nMethod 1 constructs an inverse via Bézout, producing $x=158$. 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... | 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 $\boxed{158}$.) |
math-017399 | Number Theory: Euler Totient (Variant B) | 9 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=77=11^1\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explici... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=11$ and $q=7$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{60}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=7$ yields the same integer 60.",
"robustness_analysis": "Generality ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-017400 | Number Theory: Euler Totient (Variant A) | 9 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=2963603=43^1\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multi... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2824080}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=41$ yields the same integer 2824080.",
"robustness_analysis": "If the problem were perturbed: B... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
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