id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-017201 | Number Theory: Euler Totient (Core) | 9 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=76759069=19^5\cdot 31^1$.
(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.
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 $p=19$ 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: $\\boxed{70373340}$.\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 70373340.",
"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 $\boxed{70373340}$.) |
math-017202 | Number Theory: Euler Totient (Variant A) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=214375=5^4\cdot 7^3$.
(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... | [
{
"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 $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{147000}$.\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 147000.",
"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... | 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 $\boxed{147000}$.) |
math-017203 | Number Theory: Euler Totient (Variant C) | 9 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=111166451=17^4\cdot 11^3$.
(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 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 $p=17$ and $q=11$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"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: $\\boxed{95115680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=11$ yields the same integer 95115680.",
"robustness_analysis": "Robustn... | [
{
"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-017204 | Number Theory: Euler Totient (Variant C) | 9 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=275517490541323=43^5\cdot 37^4$.
(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": "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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{261836860904136}$.\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 261836860904136.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{261836860904136}$.) |
math-017205 | Number Theory: Modular Inverses — Extended Euclid | 9 | Explain each transformation: Find the multiplicative inverse of $673$ modulo $1234$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1234}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition f... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(673,1234)=1$, 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: $\\boxed{1223}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1223$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017206 | Number Theory: Euler Totient (Variant A) | 9 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=10256403=43^4\cdot 3^1$.
(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 $... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=43$ and $q=3$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6678588}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=3$ yields the same integer 6678588.",
"robustness_analysis": "Generality note: Both methods rel... | [
{
"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 $\boxed{6678588}$.) |
math-017207 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Try to avoid pattern-matching; explain why: Find the multiplicative inverse of $577$ modulo $1165$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1165}$).
(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 $x=953$ and compute $577x=549881$.",
"Step 2: Reduce: $549881\\equiv 1\\pmod{1165}$ (since $549880=549880$ is ... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{953}$.\nMethod 1 constructs an inverse via Bézout, producing $x=953$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{953}$.) |
math-017208 | Number Theory: Euler Totient (Variant B) | 9 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=2511=31^1\cdot 3^4$.
(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 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 $p=31$ and $q=3$ 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: $\\boxed{1620}$.\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 1620.",
"robustness_analysis": "If the problem were perturbed: Bo... | [
{
"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 $\boxed{1620}$.) |
math-017209 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Explain why your operations are valid: Find the multiplicative inverse of $47$ modulo $1554$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1554}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient co... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1025$ and compute $47x=48175$.",
"Step 2: Reduce: $48175\\equiv 1\\pmod{1554}$ (since $48174=48174$ is divi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1025}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1025$. 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... | 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 $\boxed{1025}$.) |
math-017210 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Complete the analysis: Find the multiplicative inverse of $259$ modulo $906$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{906}$).
(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 $\\gcd(259,906)=1$, 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: $\\boxed{7}$.\nMethod 1 constructs an inverse via Bézout, producing $x=7$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fast and... | [
{
"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-017211 | Number Theory: Units mod m — Existence Condition | 9 | Where appropriate, name the theorem you use: Find the multiplicative inverse of $216$ modulo $1879$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1879}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and suffic... | [
{
"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 $\\gcd(216,1879)=1$, 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: $\\boxed{983}$.\nMethod 1 constructs an inverse via Bézout, producing $x=983$. 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 $\boxed{983}$.) |
math-017212 | Number Theory: Units mod m — Existence Condition | 9 | Compute the requested quantity: Find the multiplicative inverse of $370$ modulo $1649$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1649}$).
(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 $\\gcd(370,1649)=1$, 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: $\\boxed{361}$.\nMethod 1 constructs an inverse via Bézout, producing $x=361$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: 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 $\boxed{361}$.) |
math-017213 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Answer with a short justification: Find the multiplicative inverse of $311$ modulo $395$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{395}$).
(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 $x=221$ and compute $311x=68731$.",
"Step 2: Reduce: $68731\\equiv 1\\pmod{395}$ (since $68730=68730$ is divis... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{221}$.\nMethod 1 constructs an inverse via Bézout, producing $x=221$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid 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-017214 | Number Theory: Euler Totient (Variant C) | 9 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=145=5^1\cdot 29^1$.
(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 explic... | [
{
"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 $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{112}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=29$ yields the same integer 112.",
"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... | 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 $\boxed{112}$.) |
math-017215 | Number Theory: Units mod m — Existence Condition | 9 | Complete the analysis: Find the multiplicative inverse of $1084$ modulo $1307$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1307}$).
(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 $x=1096$ and compute $1084x=1188064$.",
"Step 2: Reduce: $1188064\\equiv 1\\pmod{1307}$ (since $1188063=118806... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1096}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1096$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017216 | Number Theory: Units mod m — Existence Condition | 9 | Give a fully justified solution: Find the multiplicative inverse of $951$ modulo $1760$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1760}$).
(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 $x=1351$ and compute $951x=1284801$.",
"Step 2: Reduce: $1284801\\equiv 1\\pmod{1760}$ (since $1284800=1284800... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1351}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1351$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid is fa... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1351}$.) |
math-017217 | Number Theory: Units mod m — Existence Condition | 9 | Derive the result step-by-step: Find the multiplicative inverse of $1262$ modulo $1317$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1317}$).
(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 $\\gcd(1262,1317)=1$, 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: $\\boxed{431}$.\nMethod 1 constructs an inverse via Bézout, producing $x=431$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017218 | Number Theory: Euler Totient (Core) | 9 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=4102893=3^4\cdot 37^3$.
(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 $\varphi(n)$ 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 $p=3$ 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: $\\boxed{2661336}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=37$ yields the same integer 2661336.",
"robustness_analysis": "Generality note: Both methods rel... | [
{
"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 $\boxed{2661336}$.) |
math-017219 | Number Theory: Euler Totient (Variant C) | 9 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=240737=17^3\cdot 7^2$.
(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.
Be explicit about why multipl... | [
{
"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 $p=17$ 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: $\\boxed{194208}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=7$ yields the same integer 194208.",
"robustness_analysis": "Generality 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... | 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 $\boxed{194208}$.) |
math-017220 | Number Theory: Euler Totient (Variant C) | 9 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=779=19^1\cdot 41^1$.
(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 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 $p=19$ 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: $\\boxed{720}$.\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 720.",
"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 $\boxed{720}$.) |
math-017221 | Number Theory: Euler Totient (Variant B) | 9 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=5776=19^2\cdot 2^4$.
(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 $\varphi(n)$ counts.
Be explicit about why multiplic... | [
{
"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 $q=2$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"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: $\\boxed{2736}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=2$ yields the same integer 2736.",
"robustness_analysis": "Generality 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... | 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 $\boxed{2736}$.) |
math-017222 | Number Theory: Euler Totient (Core) | 9 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=10158731=31^4\cdot 11^1$.
(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 why mul... | [
{
"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 $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{8937300}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=11$ yields the same integer 8937300.",
"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... | 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-017223 | Number Theory: Euler Totient (Variant C) | 9 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=449307=3^5\cdot 43^2$.
(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 $\varphi(n)$ 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 $p=3$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{292572}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=43$ yields the same integer 292572.",
"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... | 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 $\boxed{292572}$.) |
math-017224 | Number Theory: Euler Totient (Variant B) | 9 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=53582633=13^3\cdot 29^3$.
(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 $\varphi(n)$ cou... | [
{
"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 $p=13$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{47755344}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=29$ yields the same integer 47755344.",
"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 $n(1-1/p)(1-1/q)$. |
math-017225 | Number Theory: Euler Totient (Variant A) | 9 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=533=41^1\cdot 13^1$.
(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 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 $p=41$ 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: $\\boxed{480}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=13$ yields the same integer 480.",
"robustness_analysis": "Generality 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... | 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 $\boxed{480}$.) |
math-017226 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Give an answer and a quick verification: Find the multiplicative inverse of $592$ modulo $1523$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1523}$).
(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 $x=301$ and compute $592x=178192$.",
"Step 2: Reduce: $178192\\equiv 1\\pmod{1523}$ (since $178191=178191$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{301}$.\nMethod 1 constructs an inverse via Bézout, producing $x=301$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{301}$.) |
math-017227 | Number Theory: Modular Inverses — Extended Euclid | 9 | Find the exact value: Find the multiplicative inverse of $1402$ modulo $1483$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1483}$).
(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 $x=238$ and compute $1402x=333676$.",
"Step 2: Reduce: $333676\\equiv 1\\pmod{1483}$ (since $333675=333675$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{238}$.\nMethod 1 constructs an inverse via Bézout, producing $x=238$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustne... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{238}$.) |
math-017228 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Give an answer and a quick verification: Find the multiplicative inverse of $147$ modulo $247$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{247}$).
(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 $\\gcd(147,247)=1$, 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: $\\boxed{205}$.\nMethod 1 constructs an inverse via Bézout, producing $x=205$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017229 | Number Theory: Euler Totient (Variant B) | 9 | Compute the requested quantity: Compute Euler's totient function $\varphi(n)$.
Here $n=1715=5^1\cdot 7^3$.
(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 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 $p=5$ and $q=7$ 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: $\\boxed{1176}$.\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 1176.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 $\boxed{1176}$.) |
math-017230 | Number Theory: Units mod m — Existence Condition | 9 | Checkpoint: Find the multiplicative inverse of $149$ modulo $1071$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1071}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to ... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(149,1071)=1$, 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: $\\boxed{956}$.\nMethod 1 constructs an inverse via Bézout, producing $x=956$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the p... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017231 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Determine the requested value: Find the multiplicative inverse of $374$ modulo $895$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{895}$).
(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 $x=414$ and compute $374x=154836$.",
"Step 2: Reduce: $154836\\equiv 1\\pmod{895}$ (since $154835=154835$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{414}$.\nMethod 1 constructs an inverse via Bézout, producing $x=414$. 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. |
math-017232 | Number Theory: Units mod m — Existence Condition | 9 | Question: Find the multiplicative inverse of $111$ modulo $158$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{158}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=121$ and compute $111x=13431$.",
"Step 2: Reduce: $13431\\equiv 1\\pmod{158}$ (since $13430=13430$ is divis... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{121}$.\nMethod 1 constructs an inverse via Bézout, producing $x=121$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{121}$.) |
math-017233 | Number Theory: Euler Totient (Variant A) | 9 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=123=3^1\cdot 41^1$.
(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... | [
{
"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: $\\boxed{80}$.\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 80.",
"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... | 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 $\boxed{80}$.) |
math-017234 | Number Theory: Euler Totient (Variant B) | 9 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=261=3^2\cdot 29^1$.
(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": "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 $q=29$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"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: $\\boxed{168}$.\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 168.",
"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 $n(1-1/p)(1-1/q)$. |
math-017235 | Number Theory: Euler Totient (Core) | 9 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=242=11^2\cdot 2^1$.
(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 abou... | [
{
"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 $q=2$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"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: $\\boxed{110}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=2$ yields the same integer 110.",
"robustness_analysis": "Robustness note: 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... | 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 $\boxed{110}$.) |
math-017236 | Computational Number Theory: Inverses and Certificates | 9 | Work this out carefully: Find the multiplicative inverse of $1643$ modulo $1775$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1775}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for ... | [
{
"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 $\\gcd(1643,1775)=1$, 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: $\\boxed{632}$.\nMethod 1 constructs an inverse via Bézout, producing $x=632$. 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. (Here the result is $\boxed{632}$.) |
math-017237 | Number Theory: Euler Totient (Core) | 9 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=1085773=17^4\cdot 13^1$.
(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 ... | [
{
"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 $p=17$ 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: $\\boxed{943296}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=13$ yields the same integer 943296.",
"robustness_analysis": "Generality note: 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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{943296}$.) |
math-017238 | Number Theory: Modular Inverses — Extended Euclid | 9 | Solve and then verify: Find the multiplicative inverse of $593$ modulo $783$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{783}$).
(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 $x=647$ and compute $593x=383671$.",
"Step 2: Reduce: $383671\\equiv 1\\pmod{783}$ (since $383670=383670$ is d... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{647}$.\nMethod 1 constructs an inverse via Bézout, producing $x=647$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euc... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{647}$.) |
math-017239 | Number Theory: Modular Inverses — Extended Euclid | 9 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $169$ modulo $1197$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1197}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessar... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(169,1197)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{85}$.\nMethod 1 constructs an inverse via Bézout, producing $x=85$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: Ex... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017240 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Answer with a short justification: Find the multiplicative inverse of $1039$ modulo $1358$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1358}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient cond... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(1039,1358)=1$, 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: $\\boxed{1209}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1209$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{1209}$.) |
math-017241 | Number Theory: Euler Totient (Variant C) | 9 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=1235376017=37^3\cdot 29^3$.
(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": "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 $q=29$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"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: $\\boxed{1160539632}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=29$ yields the same integer 1160539632.",
"robustness_analysis": "Gen... | [
{
"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-017242 | Number Theory: Euler Totient (Core) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=866761=7^4\cdot 19^2$.
(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 $\varphi(n)$ 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 $p=7$ and $q=19$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"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: $\\boxed{703836}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=19$ yields the same integer 703836.",
"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 $\boxed{703836}$.) |
math-017243 | Number Theory: Euler Totient (Variant C) | 9 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=14018600321=41^5\cdot 11^2$.
(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 expl... | [
{
"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 $q=11$.",
"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: $\\boxed{12433348400}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=11$ yields the same integer 12433348400.",
"robustness_analysis": "S... | [
{
"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-017244 | Number Theory: Euler Totient (Core) | 9 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=23125=5^4\cdot 37^1$.
(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 ab... | [
{
"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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{18000}$.\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 18000.",
"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... | 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 $\boxed{18000}$.) |
math-017245 | Number Theory: Modular Inverses — Extended Euclid | 9 | Exercise: Find the multiplicative inverse of $179$ modulo $1455$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1455}$).
(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 $\\gcd(179,1455)=1$, 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: $\\boxed{569}$.\nMethod 1 constructs an inverse via Bézout, producing $x=569$. 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-017246 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Prompt: Find the multiplicative inverse of $167$ modulo $1233$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1233}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=443$ and compute $167x=73981$.",
"Step 2: Reduce: $73981\\equiv 1\\pmod{1233}$ (since $73980=73980$ is divi... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{443}$.\nMethod 1 constructs an inverse via Bézout, producing $x=443$. 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 $\boxed{443}$.) |
math-017247 | Number Theory: Euler Totient (Variant C) | 9 | Checkpoint: Compute Euler's totient function $\varphi(n)$.
Here $n=377=29^1\cdot 13^1$.
(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 multip... | [
{
"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 $q=13$.",
"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: $\\boxed{336}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=13$ yields the same integer 336.",
"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... | 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-017248 | Computational Number Theory: Inverses and Certificates | 9 | Solve and include a self-check: Find the multiplicative inverse of $1490$ modulo $1703$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1703}$).
(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 $x=1695$ and compute $1490x=2525550$.",
"Step 2: Reduce: $2525550\\equiv 1\\pmod{1703}$ (since $2525549=252554... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1695}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1695$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1695}$.) |
math-017249 | Number Theory: Units mod m — Existence Condition | 9 | Determine the requested value: Find the multiplicative inverse of $189$ 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 condition f... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(189,488)=1$, 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: $\\boxed{173}$.\nMethod 1 constructs an inverse via Bézout, producing $x=173$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generali... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{173}$.) |
math-017250 | Number Theory: Units mod m — Existence Condition | 9 | Solve and include a self-check: Find the multiplicative inverse of $686$ modulo $731$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{731}$).
(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 $\\gcd(686,731)=1$, 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: $\\boxed{666}$.\nMethod 1 constructs an inverse via Bézout, producing $x=666$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Ext... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017251 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Show all reasoning: Find the multiplicative inverse of $88$ modulo $257$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{257}$).
(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 invers... | [
{
"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 $x=184$ and compute $88x=16192$.",
"Step 2: Reduce: $16192\\equiv 1\\pmod{257}$ (since $16191=16191$ is divisi... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{184}$.\nMethod 1 constructs an inverse via Bézout, producing $x=184$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017252 | Number Theory: Units mod m — Existence Condition | 9 | Solve and justify each step: Find the multiplicative inverse of $522$ modulo $701$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{701}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=654$ and compute $522x=341388$.",
"Step 2: Reduce: $341388\\equiv 1\\pmod{701}$ (since $341387=341387$ is d... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{654}$.\nMethod 1 constructs an inverse via Bézout, producing $x=654$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017253 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Solve and then verify: Find the multiplicative inverse of $347$ modulo $515$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{515}$).
(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 $\\gcd(347,515)=1$, 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: $\\boxed{328}$.\nMethod 1 constructs an inverse via Bézout, producing $x=328$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "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 $\boxed{328}$.) |
math-017254 | Number Theory: Units mod m — Existence Condition | 9 | Determine the requested value: Find the multiplicative inverse of $250$ modulo $1529$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1529}$).
(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 $\\gcd(250,1529)=1$, so an inverse exists.",
"Step 2: Use the extended Euclidean algorithm to find integers $u,v$ such th... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1107}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1107$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Genera... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1107}$.) |
math-017255 | Number Theory: Euler Totient (Core) | 9 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=7604375=23^3\cdot 5^4$.
(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 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 $p=23$ and $q=5$ are distinct primes, $\\gcd(p^3,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5819000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=5$ yields the same integer 5819000.",
"robustness_analysis": "Robustness note: Both methods rel... | [
{
"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 $\boxed{5819000}$.) |
math-017256 | Number Theory: Euler Totient (Core) | 9 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=600281=11^4\cdot 41^1$.
(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.
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 $p=11$ and $q=41$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"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: $\\boxed{532400}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=41$ yields the same integer 532400.",
"robustness_analysis": "Ro... | [
{
"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 $\boxed{532400}$.) |
math-017257 | Number Theory: Euler Totient (Variant A) | 9 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=42025=5^2\cdot 41^2$.
(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 expl... | [
{
"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 $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{32800}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=41$ yields the same integer 32800.",
"robustness_analysis": "Robustness 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... | 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-017258 | Number Theory: Euler Totient (Variant A) | 9 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=2523=3^1\cdot 29^2$.
(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 exp... | [
{
"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": "Both approaches agree after simplification. Final answer: $\\boxed{1624}$.\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 1624.",
"robustness_analysis": "Generality note: Both methods rely on k... | [
{
"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 $\boxed{1624}$.) |
math-017259 | Computational Number Theory: Inverses and Certificates | 9 | Do not skip justification steps: Find the multiplicative inverse of $269$ modulo $1808$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1808}$).
(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 $x=1109$ and compute $269x=298321$.",
"Step 2: Reduce: $298321\\equiv 1\\pmod{1808}$ (since $298320=298320$ is... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1109}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1109$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Genera... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017260 | Number Theory: Modular Inverses — Extended Euclid | 9 | Solve and then verify: Find the multiplicative inverse of $689$ modulo $1057$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1057}$).
(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 $\\gcd(689,1057)=1$, 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: $\\boxed{922}$.\nMethod 1 constructs an inverse via Bézout, producing $x=922$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017261 | Number Theory: Units mod m — Existence Condition | 9 | Write the solution set clearly: Find the multiplicative inverse of $1281$ modulo $1447$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1447}$).
(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 $x=985$ and compute $1281x=1261785$.",
"Step 2: Reduce: $1261785\\equiv 1\\pmod{1447}$ (since $1261784=1261784... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{985}$.\nMethod 1 constructs an inverse via Bézout, producing $x=985$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017262 | Number Theory: Euler Totient (Variant A) | 9 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=83453453=23^3\cdot 19^3$.
(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 ... | [
{
"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 $q=19$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{75623724}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=19$ yields the same integer 75623724.",
"robustness_analysis": "Generality 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... | 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-017263 | Number Theory: Modular Inverses — Extended Euclid | 9 | Give reasoning, not just computation: Find the multiplicative inverse of $107$ modulo $109$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{109}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient cond... | [
{
"method_name": "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 $x=54$ and compute $107x=5778$.",
"Step 2: Reduce: $5778\\equiv 1\\pmod{109}$ (since $5777=5777$ is divisible ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{54}$.\nMethod 1 constructs an inverse via Bézout, producing $x=54$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Eucli... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{54}$.) |
math-017264 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Answer with a short justification: Find the multiplicative inverse of $377$ modulo $737$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{737}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $x=477$ and compute $377x=179829$.",
"Step 2: Reduce: $179829\\equiv 1\\pmod{737}$ (since $179828=179828$ is d... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{477}$.\nMethod 1 constructs an inverse via Bézout, producing $x=477$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "If the problem were perturbed: Extended ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017265 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Explain what is being counted/optimized: Find the multiplicative inverse of $239$ modulo $342$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{342}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient 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 $\\gcd(239,342)=1$, 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: $\\boxed{83}$.\nMethod 1 constructs an inverse via Bézout, producing $x=83$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | 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 $\boxed{83}$.) |
math-017266 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Compute the requested quantity: Find the multiplicative inverse of $277$ modulo $535$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{535}$).
(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 $\\gcd(277,535)=1$, 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: $\\boxed{338}$.\nMethod 1 constructs an inverse via Bézout, producing $x=338$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{338}$.) |
math-017267 | Number Theory: Euler Totient (Core) | 9 | Checkpoint: Compute Euler's totient function $\varphi(n)$.
Here $n=943=23^1\cdot 41^1$.
(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 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 $p=23$ and $q=41$ 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: $\\boxed{880}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=41$ yields the same integer 880.",
"robustness_analysis": "Generali... | [
{
"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 $\boxed{880}$.) |
math-017268 | Number Theory: Euler Totient (Variant B) | 9 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=83672459=23^5\cdot 13^1$.
(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 ... | [
{
"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 $q=13$ are distinct primes, $\\gcd(p^5,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{73878024}$.\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 73878024.",
"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... | 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 $\boxed{73878024}$.) |
math-017269 | Number Theory: Euler Totient (Core) | 9 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=845=5^1\cdot 13^2$.
(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 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 $p=5$ and $q=13$ 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: $\\boxed{624}$.\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 624.",
"robustness_analysis": "Robustness note: Both methods rely on kno... | [
{
"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-017270 | Number Theory: Modular Inverses — Extended Euclid | 9 | Work this out carefully: Find the multiplicative inverse of $703$ modulo $915$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{915}$).
(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 $\\gcd(703,915)=1$, 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: $\\boxed{82}$.\nMethod 1 constructs an inverse via Bézout, producing $x=82$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid is fast a... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017271 | Number Theory: Euler Totient (Variant A) | 9 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=6877=23^2\cdot 13^1$.
(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.
... | [
{
"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 $q=13$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"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: $\\boxed{6072}$.\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 6072.",
"robustness_analysis": "If the... | [
{
"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 $\boxed{6072}$.) |
math-017272 | Number Theory: Euler Totient (Variant A) | 9 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=4375=5^4\cdot 7^1$.
(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.
... | [
{
"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, $\\gcd(p^4,q^1)=1$.",
"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: $\\boxed{3000}$.\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 3000.",
"robustness_analysis": "Generali... | [
{
"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-017273 | Number Theory: Euler Totient (Variant A) | 9 | Do not skip justification steps: Compute Euler's totient function $\varphi(n)$.
Here $n=125421842647=19^5\cdot 37^3$.
(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 $\varphi(n)$ 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 $q=37$ 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: $\\boxed{115609322952}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=37$ yields the same integer 115609322952.",
"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-017274 | Number Theory: Euler Totient (Variant A) | 9 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=250563=3^1\cdot 17^4$.
(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 abou... | [
{
"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 $q=17$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{157216}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=17$ yields the same integer 157216.",
"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... | 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-017275 | Number Theory: Euler Totient (Variant A) | 9 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=204363=3^5\cdot 29^2$.
(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 multip... | [
{
"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": "The two methods are consistent and must coincide. Final answer: $\\boxed{131544}$.\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 131544.",
"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... | 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 $\boxed{131544}$.) |
math-017276 | Number Theory: Euler Totient (Variant B) | 9 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=2190305047=7^5\cdot 19^4$.
(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 $p=7$ and $q=19$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"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: $\\boxed{1778593572}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=19$ yields the same integer 1778593572.",
"robustness_analysi... | [
{
"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 $\boxed{1778593572}$.) |
math-017277 | Number Theory: Euler Totient (Variant B) | 9 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=60835=23^3\cdot 5^1$.
(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 $\... | [
{
"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 $q=5$.",
"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: $\\boxed{46552}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=5$ yields the same integer 46552.",
"robustness_analysis": "Robus... | [
{
"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-017278 | Number Theory: Euler Totient (Variant B) | 9 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=8157716303=11^5\cdot 37^3$.
(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... | [
{
"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 $q=37$.",
"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: $\\boxed{7215670440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=37$ yields the same integer 7215670440.",
"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 $n(1-1/p)(1-1/q)$. |
math-017279 | Number Theory: Euler Totient (Variant B) | 9 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=34481=41^1\cdot 29^2$.
(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 $p=41$ 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: $\\boxed{32480}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=29$ yields the same integer 32480.",
"robustness_analysis": "Gene... | [
{
"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 $\boxed{32480}$.) |
math-017280 | Number Theory: Euler Totient (Variant A) | 9 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=14283=23^2\cdot 3^3$.
(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 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 $p=23$ and $q=3$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"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: $\\boxed{9108}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=3$ yields the same integer 9108.",
"robustness_analysis": "Robustn... | [
{
"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-017281 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Provide a rigorous solution: Find the multiplicative inverse of $54$ modulo $95$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{95}$).
(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 $x=44$ and compute $54x=2376$.",
"Step 2: Reduce: $2376\\equiv 1\\pmod{95}$ (since $2375=2375$ is divisible by... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{44}$.\nMethod 1 constructs an inverse via Bézout, producing $x=44$. 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... | 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 $\boxed{44}$.) |
math-017282 | Number Theory: Euler Totient (Variant C) | 9 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=23639903=7^3\cdot 41^3$.
(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... | [
{
"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": "The two methods are consistent and must coincide. Final answer: $\\boxed{19768560}$.\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 19768560.",
"robustness_analysis": "If the problem were pertu... | [
{
"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-017283 | Number Theory: Modular Inverses — Extended Euclid | 9 | Indicate where a theorem is used: Find the multiplicative inverse of $311$ modulo $1724$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1724}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condit... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=255$ and compute $311x=79305$.",
"Step 2: Reduce: $79305\\equiv 1\\pmod{1724}$ (since $79304=79304$ is divi... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{255}$.\nMethod 1 constructs an inverse via Bézout, producing $x=255$. 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-017284 | Number Theory: Modular Inverses — Extended Euclid | 9 | Answer using clear logical steps: Find the multiplicative inverse of $251$ modulo $288$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{288}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient conditio... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=179$ and compute $251x=44929$.",
"Step 2: Reduce: $44929\\equiv 1\\pmod{288}$ (since $44928=44928$ is divis... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{179}$.\nMethod 1 constructs an inverse via Bézout, producing $x=179$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Robustness note: ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Takeaway: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017285 | Number Theory: Modular Inverses — Extended Euclid | 9 | Prompt: Find the multiplicative inverse of $397$ modulo $1633$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1633}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inverse to exis... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1016$ and compute $397x=403352$.",
"Step 2: Reduce: $403352\\equiv 1\\pmod{1633}$ (since $403351=403351$ is... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1016}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1016$. 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... | 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 $\boxed{1016}$.) |
math-017286 | Number Theory: Euler Totient (Core) | 9 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=105125=5^3\cdot 29^2$.
(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 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 $p=5$ and $q=29$ are distinct primes, $\\gcd(p^3,q^2)=1$.",
"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: $\\boxed{81200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=29$ yields the same integer 81200.",
"robustness_analysis": "Sensitivity an... | [
{
"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 $\boxed{81200}$.) |
math-017287 | Number Theory: Euler Totient (Variant C) | 9 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=268279=11^1\cdot 29^3$.
(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": "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 $q=29$.",
"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: $\\boxed{235480}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=29$ yields the same integer 235480.",
"robustness_analysis": "Ro... | [
{
"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 $\boxed{235480}$.) |
math-017288 | Number Theory: Euler Totient (Variant B) | 9 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=158290625=5^5\cdot 37^3$.
(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 $\varphi(n)$ counts.
Be explicit about why mul... | [
{
"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 $q=37$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"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: $\\boxed{123210000}$.\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 123210000.",
"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 $\boxed{123210000}$.) |
math-017289 | Computational Number Theory: Inverses and Certificates | 9 | Start by stating any domain restrictions: Find the multiplicative inverse of $1529$ modulo $1856$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1856}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficie... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1033$ and compute $1529x=1579457$.",
"Step 2: Reduce: $1579457\\equiv 1\\pmod{1856}$ (since $1579456=157945... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1033}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1033$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017290 | Number Theory: Euler Totient (Variant C) | 9 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=6765201=3^4\cdot 17^4$.
(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 abo... | [
{
"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 $q=17$ are distinct primes, $\\gcd(p^4,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{4244832}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=17$ yields the same integer 4244832.",
"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... | 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-017291 | Number Theory: Modular Inverses — Extended Euclid | 9 | Show all reasoning: Find the multiplicative inverse of $199$ modulo $1555$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1555}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an inv... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=1219$ and compute $199x=242581$.",
"Step 2: Reduce: $242581\\equiv 1\\pmod{1555}$ (since $242580=242580$ is... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1219}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1219$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended E... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1219}$.) |
math-017292 | Number Theory: Congruences — Solving $ax\equiv 1$ | 9 | Find the exact value: Find the multiplicative inverse of $725$ modulo $1133$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1133}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for an i... | [
{
"method_name": "Uniqueness + Direct Check",
"approach": "Solve by finding any $x$ that works and use uniqueness of inverses modulo $m$ to certify it.",
"steps": [
"Step 1: Propose $x=747$ and compute $725x=541575$.",
"Step 2: Reduce: $541575\\equiv 1\\pmod{1133}$ (since $541574=541574$ is ... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{747}$.\nMethod 1 constructs an inverse via Bézout, producing $x=747$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Generality note: Extended Euclid i... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Remember: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{747}$.) |
math-017293 | Computational Number Theory: Inverses and Certificates | 9 | Show all reasoning: Find the multiplicative inverse of $51$ modulo $233$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{233}$).
(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 invers... | [
{
"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 $x=32$ and compute $51x=1632$.",
"Step 2: Reduce: $1632\\equiv 1\\pmod{233}$ (since $1631=1631$ is divisible b... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. 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": "If the problem were... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017294 | Number Theory: Modular Inverses — Extended Euclid | 9 | Make each step logically reversible (or explain if not): Find the multiplicative inverse of $942$ modulo $1369$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1369}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessar... | [
{
"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 $x=949$ and compute $942x=893958$.",
"Step 2: Reduce: $893958\\equiv 1\\pmod{1369}$ (since $893957=893957$ is ... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{949}$.\nMethod 1 constructs an inverse via Bézout, producing $x=949$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitiv... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Key idea: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. |
math-017295 | Computational Number Theory: Inverses and Certificates | 9 | Answer using clear logical steps: Find the multiplicative inverse of $53$ modulo $447$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{447}$).
(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 $x=194$ and compute $53x=10282$.",
"Step 2: Reduce: $10282\\equiv 1\\pmod{447}$ (since $10281=10281$ is divisi... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{194}$.\nMethod 1 constructs an inverse via Bézout, producing $x=194$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid is... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{194}$.) |
math-017296 | Number Theory: Units mod m — Existence Condition | 9 | Explain each transformation: Find the multiplicative inverse of $993$ modulo $1304$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1304}$).
(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 $x=1065$ and compute $993x=1057545$.",
"Step 2: Reduce: $1057545\\equiv 1\\pmod{1304}$ (since $1057544=1057544... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1065}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1065$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended Euclid ... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1065}$.) |
math-017297 | Number Theory: Euler Totient (Variant C) | 9 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=6601149613=19^4\cdot 37^3$.
(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 $\varphi(n)$ 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 $p=19$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{6084701208}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=37$ yields the same integer 6084701208.",
"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 $\boxed{6084701208}$.) |
math-017298 | Computational Number Theory: Inverses and Certificates | 9 | Carefully track domains: Find the multiplicative inverse of $599$ modulo $1775$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{1775}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition for 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 $\\gcd(599,1775)=1$, 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: $\\boxed{1049}$.\nMethod 1 constructs an inverse via Bézout, producing $x=1049$. Method 2 verifies $ax\\equiv 1$ directly and uses uniqueness, so both necessarily agree.",
"robustness_analysis": "Sensitivity analysis: Extended E... | [
{
"error_description": "Tried to invert even when $\\gcd(a,m)\\ne 1$.",
"why_plausible": "The inverse notation $a^{-1}$ suggests it always exists.",
"why_wrong": "If $d=\\gcd(a,m)>1$, then $ax$ is always divisible by $d$ modulo $m$, so it cannot be congruent to 1.",
"which_method_catches_it": "Exten... | Core principle: A modular inverse exists iff $\gcd(a,m)=1$. Extended Euclid constructs it and also provides a proof certificate via Bézout. (Here the result is $\boxed{1049}$.) |
math-017299 | Number Theory: Units mod m — Existence Condition | 9 | Derive the result step-by-step: Find the multiplicative inverse of $71$ modulo $241$ (i.e., find an integer $x$ such that $ax\equiv 1\pmod{241}$).
(a) Solve using the extended Euclidean algorithm.
(b) Give an independent verification by checking the congruence.
(c) Briefly state the necessary and sufficient condition f... | [
{
"method_name": "Extended Euclidean Algorithm",
"approach": "Use Bézout: find $u,v$ with $au+mv=1$, then $u$ is an inverse of $a$ modulo $m$.",
"steps": [
"Step 1: Check $\\gcd(71,241)=1$, 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: $\\boxed{129}$.\nMethod 1 constructs an inverse via Bézout, producing $x=129$. 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. (Here the result is $\boxed{129}$.) |
math-017300 | Number Theory: Euler Totient (Variant C) | 9 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=26730899=23^3\cdot 13^3$.
(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 $\varphi(n)$ 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 $q=13$ 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: $\\boxed{23601864}$.\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 23601864.",
"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 $\boxed{23601864}$.) |
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