id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-019101 | Number Theory: Euler Totient (Variant A) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=885391=13^4\cdot 31^1$.
(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 ... | [
{
"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 $q=31$.",
"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: $\\boxed{790920}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=31$ yields the same integer 790920.",
"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... | 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 $\boxed{790920}$.) |
math-019102 | Number Theory: Euler Totient (Core) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=1020099721=19^2\cdot 41^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=19$ and $q=41$ 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: $\\boxed{942839280}$.\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 942839280.",
"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-019103 | Number Theory: Euler Totient (Variant A) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=2385443281=17^4\cdot 13^4$.
(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 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 $p=17$ 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: $\\boxed{2072421312}$.\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 2072421312.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2072421312}$.) |
math-019104 | Number Theory: Euler Totient (Variant A) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=1176067544=43^5\cdot 2^3$.
(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 $p=43$ and $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{574358568}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=2$ yields the same integer 574358568.",
"robustness_analysis": "Robustness note: Both m... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{574358568}$.) |
math-019105 | Number Theory: Euler Totient (Variant B) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=35=5^1\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{24}$.\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 24.",
"robustness_analysis": "If the probl... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{24}$.) |
math-019106 | Number Theory: Euler Totient (Variant C) | 10 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=425=5^2\cdot 17^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explici... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=5$ and $q=17$ 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: $\\boxed{320}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=17$ yields the same integer 320.",
"robustness_analysis": "Generality 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 $n(1-1/p)(1-1/q)$. |
math-019107 | Number Theory: Euler Totient (Variant C) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=172=43^1\cdot 2^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"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 $q=2$.",
"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: $\\boxed{84}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=2$ yields the same integer 84.",
"robustness_analysis": "If the problem were ... | [
{
"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-019108 | Number Theory: Euler Totient (Variant A) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=25589884853=13^5\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=13$ and $q=41$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23045299680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=41$ yields the same integer 23045299680.",
"robustness_analysis": "Sensitivity analysis: 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 $\boxed{23045299680}$.) |
math-019109 | Number Theory: Euler Totient (Core) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=36501=3^1\cdot 23^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multiplic... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{23276}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=23$ yields the same integer 23276.",
"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... | 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 $\boxed{23276}$.) |
math-019110 | Number Theory: Euler Totient (Core) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=248897=11^4\cdot 17^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "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 $q=17$ 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: $\\boxed{212960}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=17$ yields the same integer 212960.",
"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 $\boxed{212960}$.) |
math-019111 | Number Theory: Euler Totient (Core) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=180625=5^4\cdot 17^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\v... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{136000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=17$ yields the same integer 136000.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{136000}$.) |
math-019112 | Number Theory: Euler Totient (Variant C) | 10 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=134456=7^5\cdot 2^3$.
(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 multipli... | [
{
"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 $q=2$.",
"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: $\\boxed{57624}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=2$ yields the same integer 57624.",
"robustness_analysis": "If the problem ... | [
{
"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 $\boxed{57624}$.) |
math-019113 | Number Theory: Euler Totient (Core) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=991431169=23^2\cdot 37^4$.
(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.
Be explicit about why m... | [
{
"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 $q=37$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"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: $\\boxed{922695048}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=37$ yields the same integer 922695048.",
"robustness_analysis": "If the problem were pe... | [
{
"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-019114 | Number Theory: Euler Totient (Core) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=912247=7^1\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit abo... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=19$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{740772}$.\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 740772.",
"robustness_analysis": "If the problem were perturbed: Both... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019115 | Number Theory: Euler Totient (Core) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=108=3^3\cdot 2^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multiplicat... | [
{
"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 $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{36}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=2$ yields the same integer 36.",
"robustness_analysis": "Generality note: Both methods rely on knowin... | [
{
"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-019116 | Number Theory: Euler Totient (Variant C) | 10 | Compute the requested quantity: Compute Euler's totient function $\varphi(n)$.
Here $n=13357=19^2\cdot 37^1$.
(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 exp... | [
{
"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, $\\gcd(p^2,q^1)=1$.",
"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: $\\boxed{12312}$.\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 12312.",
"robustness_analysis": "Sensitivity a... | [
{
"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 $\boxed{12312}$.) |
math-019117 | Number Theory: Euler Totient (Variant B) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=3211=19^1\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit 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 $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2808}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=13$ yields the same integer 2808.",
"robustness_analysis": "If the problem were perturbed: Both me... | [
{
"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 $\boxed{2808}$.) |
math-019118 | Number Theory: Euler Totient (Variant C) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=2205472=2^5\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit 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 $p=2$ 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: $\\boxed{1075840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=41$ yields the same integer 1075840.",
"robustness_analysis": "Sensitivity analysis: Both ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019119 | Number Theory: Euler Totient (Variant C) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=4121741=29^3\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=29$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3673488}$.\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 3673488.",
"robustness_analysis": "Sensitivity analysis: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019120 | Number Theory: Euler Totient (Core) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=3604711=31^3\cdot 11^2$.
(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... | [
{
"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 $q=11$ are distinct primes, $\\gcd(p^3,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3171300}$.\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 3171300.",
"robustness_analysis": "Sensitivity analysis: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019121 | Number Theory: Euler Totient (Variant B) | 10 | State any required conditions first: Compute Euler's totient function $\varphi(n)$.
Here $n=287=41^1\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=7$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=7$ yields the same integer 240.",
"robustness_analysis": "If the problem were perturbed: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{240}$.) |
math-019122 | Number Theory: Euler Totient (Core) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=874503125=5^5\cdot 23^4$.
(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 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 $p=5$ and $q=23$.",
"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: $\\boxed{669185000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=23$ yields the same integer 669185000.",
"robustness_analysis": "Robust... | [
{
"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-019123 | Number Theory: Euler Totient (Core) | 10 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=9583=37^2\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=37$ and $q=7$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{7992}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=7$ yields the same integer 7992.",
"robustness_analysis": "If the problem were perturbed: Both met... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{7992}$.) |
math-019124 | Number Theory: Euler Totient (Core) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=77=7^1\cdot 11^1$.
(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 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 $p=7$ and $q=11$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{60}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=11$ yields the same integer 60.",
"robustness_analysis": "Sensitivity... | [
{
"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 $\boxed{60}$.) |
math-019125 | Number Theory: Euler Totient (Variant A) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=1147=37^1\cdot 31^1$.
(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 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 $p=37$ and $q=31$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1080}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=31$ yields the same integer 1080.",
"robustness_analysis": "Generality note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019126 | Number Theory: Euler Totient (Core) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=31213=13^1\cdot 7^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=13$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24696}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=7$ yields the same integer 24696.",
"robustness_analysis": "Generality note: Both methods rely on... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{24696}$.) |
math-019127 | Number Theory: Euler Totient (Variant A) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=36501=23^3\cdot 3^1$.
(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 $\varphi(n)$ 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 $q=3$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{23276}$.\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 23276.",
"robustness_analysis": "If th... | [
{
"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-019128 | Number Theory: Euler Totient (Variant A) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=322102=11^5\cdot 2^1$.
(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 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 $p=11$ and $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{146410}$.\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 146410.",
"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 $\boxed{146410}$.) |
math-019129 | Number Theory: Euler Totient (Variant C) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=57289761=29^4\cdot 3^4$.
(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 e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=29$ and $q=3$ are distinct primes, $\\gcd(p^4,q^4)=1$.",
"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: $\\boxed{36876168}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=3$ yields the same integer 36876168.",
"robustness_analysis": "Robustness note: Both met... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{36876168}$.) |
math-019130 | Number Theory: Euler Totient (Core) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=1085773=13^1\cdot 17^4$.
(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 $\varphi(n)$ coun... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=13$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{943296}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=17$ yields the same integer 943296.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{943296}$.) |
math-019131 | Number Theory: Euler Totient (Variant C) | 10 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=1445=17^2\cdot 5^1$.
(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 $\varphi(n)$ counts.
Be explicit about why multipli... | [
{
"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 $q=5$.",
"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: $\\boxed{1088}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=5$ yields the same integer 1088.",
"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... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1088}$.) |
math-019132 | Number Theory: Euler Totient (Variant C) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=8539739=13^5\cdot 23^1$.
(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 a... | [
{
"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 $q=23$.",
"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: $\\boxed{7540104}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=23$ yields the same integer 7540104.",
"robustness_analysis": "... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{7540104}$.) |
math-019133 | Number Theory: Euler Totient (Variant B) | 10 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=5375=5^3\cdot 43^1$.
(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": "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 $q=43$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"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: $\\boxed{4200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=43$ yields the same integer 4200.",
"robustness_analysis": "If the problem w... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{4200}$.) |
math-019134 | Number Theory: Euler Totient (Variant A) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=8477283=3^1\cdot 41^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $p=3$ 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: $\\boxed{5513680}$.\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 5513680.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{5513680}$.) |
math-019135 | Number Theory: Euler Totient (Variant C) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=6027346163=43^5\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ coun... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=41$.",
"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: $\\boxed{5743585680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=41$ yields the same integer 5743585680.",
"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... | 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 $\boxed{5743585680}$.) |
math-019136 | Number Theory: Euler Totient (Variant B) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=592=37^1\cdot 2^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit abou... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{288}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=2$ yields the same integer 288.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019137 | Number Theory: Euler Totient (Core) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=3464127271=31^5\cdot 11^2$.
(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 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3047619300}$.\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 3047619300.",
"robustness_analysis": "Sen... | [
{
"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-019138 | Number Theory: Euler Totient (Variant C) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=470412721=41^2\cdot 23^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "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 $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{438985360}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=23$ yields the same integer 438985360.",
"robustness_analysis": "Generality note: Both ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{438985360}$.) |
math-019139 | Number Theory: Euler Totient (Variant C) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=8954912=2^5\cdot 23^4$.
(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 multipl... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=23$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4282784}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=23$ yields the same integer 4282784.",
"robustness_analysis": "Sensitivit... | [
{
"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 $\boxed{4282784}$.) |
math-019140 | Number Theory: Euler Totient (Core) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=48037937=41^4\cdot 17^1$.
(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 $p=41$ and $q=17$ 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: $\\boxed{44109440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=17$ yields the same integer 44109440.",
"robustness_analysis":... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{44109440}$.) |
math-019141 | Number Theory: Euler Totient (Variant B) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=21296=2^4\cdot 11^3$.
(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 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 $p=2$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{9680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=11$ yields the same integer 9680.",
"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 $\boxed{9680}$.) |
math-019142 | Number Theory: Euler Totient (Core) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=71875=5^5\cdot 23^1$.
(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 $q=23$.",
"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: $\\boxed{55000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=23$ yields the same integer 55000.",
"robustness_analysis": "If the problem... | [
{
"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-019143 | Number Theory: Euler Totient (Variant A) | 10 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=5295931061521=41^4\cdot 37^4$.
(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 $\varphi(n)$ 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 $p=41$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5027119794720}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=37$ yields the same integer 5027119794720.",
"robustness_analysis": "Generality note: Bot... | [
{
"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 $\boxed{5027119794720}$.) |
math-019144 | Number Theory: Euler Totient (Variant A) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=2550077=37^1\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=37$ and $q=41$ are distinct primes, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2420640}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=41$ yields the same integer 2420640.",
"robustness_analysis": "Generalit... | [
{
"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-019145 | Number Theory: Euler Totient (Variant A) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=1920983=17^4\cdot 23^1$.
(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 mult... | [
{
"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 $q=23$ 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: $\\boxed{1729376}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=23$ yields the same integer 1729376.",
"robustness_analysis": "... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1729376}$.) |
math-019146 | Number Theory: Euler Totient (Variant A) | 10 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=1029059101=43^5\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{861537852}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=7$ yields the same integer 861537852.",
"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-019147 | Number Theory: Euler Totient (Core) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=38336139=17^5\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ ... | [
{
"method_name": "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 $q=3$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{24054048}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=3$ yields the same integer 24054048.",
"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 $n(1-1/p)(1-1/q)$. |
math-019148 | Number Theory: Euler Totient (Core) | 10 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=2888=19^2\cdot 2^3$.
(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 $\v... | [
{
"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 $q=2$.",
"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: $\\boxed{1368}$.\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 1368.",
"robustness_analysis": "If the problem w... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{1368}$.) |
math-019149 | Number Theory: Euler Totient (Variant C) | 10 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=8575=5^2\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about w... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{5880}$.\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 5880.",
"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... | 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-019150 | Number Theory: Euler Totient (Core) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=7496644=2^2\cdot 37^4$.
(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": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=37$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3647016}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=37$ yields the same integer 3647016.",
"robustness_analysis": "R... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019151 | Number Theory: Euler Totient (Variant A) | 10 | State any required conditions first: Compute Euler's totient function $\varphi(n)$.
Here $n=1591=43^1\cdot 37^1$.
(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... | [
{
"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": "Both approaches agree after simplification. Final answer: $\\boxed{1512}$.\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 1512.",
"robustness_analysis": "If the problem were perturbed: Both me... | [
{
"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-019152 | Number Theory: Euler Totient (Variant C) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=614125=5^3\cdot 17^3$.
(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": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{462400}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=17$ yields the same integer 462400.",
"robustness_analysis": "If the probl... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{462400}$.) |
math-019153 | Number Theory: Euler Totient (Variant A) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=4131833=29^2\cdot 17^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3754688}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=17$ yields the same integer 3754688.",
"robustness_analysis": "Sensitivity analysis: Both... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{3754688}$.) |
math-019154 | Number Theory: Euler Totient (Core) | 10 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=29212967=23^3\cdot 7^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $q=7$ are distinct primes, $\\gcd(p^3,q^4)=1$.",
"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: $\\boxed{23951004}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=7$ yields the same integer 23951004.",
"robustness_analysis": ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{23951004}$.) |
math-019155 | Number Theory: Euler Totient (Variant A) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=3474871553=17^3\cdot 29^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=17$ 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: $\\boxed{3157692608}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=29$ yields the same integer 3157692608.",
"robustness_analysis": "Robustness note: Bot... | [
{
"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-019156 | Number Theory: Euler Totient (Core) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=559682=23^4\cdot 2^1$.
(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 $\varphi(n)$ coun... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=2$ 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: $\\boxed{267674}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=2$ yields the same integer 267674.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019157 | Number Theory: Euler Totient (Variant B) | 10 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=596183=23^3\cdot 7^2$.
(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": "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 $q=7$ are distinct primes, $\\gcd(p^3,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{488796}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=7$ yields the same integer 488796.",
"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... | 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-019158 | Number Theory: Euler Totient (Variant B) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=157757=19^3\cdot 23^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ coun... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=19$ and $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{142956}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=23$ yields the same integer 142956.",
"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 $n(1-1/p)(1-1/q)$. |
math-019159 | Number Theory: Euler Totient (Variant C) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=1361367=7^5\cdot 3^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ab... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=3$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"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: $\\boxed{777924}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=3$ yields the same integer 777924.",
"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... | 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 $\boxed{777924}$.) |
math-019160 | Number Theory: Euler Totient (Core) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=2494508291=37^4\cdot 11^3$.
(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 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 $p=37$ 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: $\\boxed{2206444680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=11$ yields the same integer 2206444680.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2206444680}$.) |
math-019161 | Number Theory: Euler Totient (Variant A) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=405769=7^4\cdot 13^2$.
(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 explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=13$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{321048}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=13$ yields the same integer 321048.",
"robustness_analysis": "Rob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019162 | Number Theory: Euler Totient (Variant A) | 10 | Task: Compute Euler's totient function $\varphi(n)$.
Here $n=162=3^4\cdot 2^1$.
(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 about why multiplicativi... | [
{
"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 $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{54}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=2$ yields the same integer 54.",
"robustness_analysis": "Robustness n... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{54}$.) |
math-019163 | Number Theory: Euler Totient (Variant C) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=34381034087=41^4\cdot 23^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be ... | [
{
"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 $q=23$.",
"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: $\\boxed{32084103920}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=23$ yields the same integer 32084103920.",
"robustness_analysis": "G... | [
{
"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 $\boxed{32084103920}$.) |
math-019164 | Number Theory: Euler Totient (Variant A) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=36=2^2\cdot 3^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=2$ and $q=3$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{12}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=3$ yields the same integer 12.",
"robustness_analysis": "Robustness note: Both methods rely on knowin... | [
{
"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 $\boxed{12}$.) |
math-019165 | Number Theory: Euler Totient (Core) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=1685159=17^3\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ab... | [
{
"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 $q=7$ 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: $\\boxed{1359456}$.\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 1359456.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{1359456}$.) |
math-019166 | Number Theory: Euler Totient (Variant A) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=5043=41^2\cdot 3^1$.
(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 $\varphi(n)$ counts.
Be explicit about why multipli... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=3$ 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: $\\boxed{3280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=3$ yields the same integer 3280.",
"robustness_analysis": "Sensiti... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019167 | Number Theory: Euler Totient (Core) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=1311267756497=17^5\cdot 31^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varph... | [
{
"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 $q=31$ 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: $\\boxed{1194323573280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=31$ yields the same integer 1194323573280.",
"robustness_... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{1194323573280}$.) |
math-019168 | Number Theory: Euler Totient (Variant A) | 10 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=1119371=11^3\cdot 29^2$.
(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 ex... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{982520}$.\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 982520.",
"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... | 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 $\boxed{982520}$.) |
math-019169 | Number Theory: Euler Totient (Variant A) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=187=17^1\cdot 11^1$.
(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 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 $p=17$ and $q=11$ 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: $\\boxed{160}$.\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 160.",
"robustness_analysis": "If the p... | [
{
"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 $\boxed{160}$.) |
math-019170 | Number Theory: Euler Totient (Core) | 10 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=19918169=17^2\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=17$ and $q=41$ 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: $\\boxed{18289280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=41$ yields the same integer 18289280.",
"robustness_analysis":... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{18289280}$.) |
math-019171 | Number Theory: Euler Totient (Variant A) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=315380807=23^5\cdot 7^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=23$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{258573084}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=7$ yields the same integer 258573084.",
"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 $n(1-1/p)(1-1/q)$. |
math-019172 | Number Theory: Euler Totient (Variant C) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=21853=41^2\cdot 13^1$.
(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 $\varphi(n)$ cou... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=13$ yields the same integer 19680.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019173 | Number Theory: Euler Totient (Variant B) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=10952=2^3\cdot 37^2$.
(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": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=2$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{5328}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=37$ yields the same integer 5328.",
"robustness_analysis": "Robustness 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... | 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-019174 | Number Theory: Euler Totient (Variant A) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=57233=43^1\cdot 11^3$.
(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 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 $p=43$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{50820}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=11$ yields the same integer 50820.",
"robustness_analysis": "Sensitivity analysis: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{50820}$.) |
math-019175 | Number Theory: Euler Totient (Variant B) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=158171=13^1\cdot 23^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"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 $q=23$.",
"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: $\\boxed{139656}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=23$ yields the same integer 139656.",
"robustness_analysis": "If... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{139656}$.) |
math-019176 | Number Theory: Euler Totient (Variant C) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=59582=31^3\cdot 2^1$.
(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 $\varphi(n)$ 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 $p=31$ and $q=2$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{28830}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=2$ yields the same integer 28830.",
"robustness_analysis": "Sensitivity analysis: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019177 | Number Theory: Euler Totient (Variant A) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=147=3^1\cdot 7^2$.
(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 a... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{84}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=7$ yields the same integer 84.",
"robustness_analysis": "Generality note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{84}$.) |
math-019178 | Number Theory: Euler Totient (Variant B) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=158171=23^3\cdot 13^1$.
(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": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=23$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{139656}$.\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 139656.",
"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... | 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 $\boxed{139656}$.) |
math-019179 | Number Theory: Euler Totient (Variant A) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=2941759=43^3\cdot 37^1$.
(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 $\varphi(n)$ c... | [
{
"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 $q=37$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"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: $\\boxed{2795688}$.\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 2795688.",
"robustness_analysis": "Robustnes... | [
{
"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-019180 | Number Theory: Euler Totient (Variant C) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=77767466347=43^5\cdot 23^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)... | [
{
"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 $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{72656358852}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=23$ yields the same integer 72656358852.",
"robustness_analysis": "Robustness note: Both me... | [
{
"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-019181 | Number Theory: Euler Totient (Variant A) | 10 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=99145229=43^4\cdot 29^1$.
(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 explici... | [
{
"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 $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{93500232}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=29$ yields the same integer 93500232.",
"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... | 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 $\boxed{93500232}$.) |
math-019182 | Number Theory: Euler Totient (Variant A) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=928=2^5\cdot 29^1$.
(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 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 $p=2$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{448}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=29$ yields the same integer 448.",
"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... | 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 $\boxed{448}$.) |
math-019183 | Number Theory: Euler Totient (Variant A) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=17576=2^3\cdot 13^3$.
(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 $\varphi(n)$ counts.
Be explici... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=2$ and $q=13$.",
"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: $\\boxed{8112}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=13$ yields the same integer 8112.",
"robustness_analysis": "Robustness note:... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019184 | Number Theory: Euler Totient (Variant A) | 10 | State any required conditions first: Compute Euler's totient function $\varphi(n)$.
Here $n=8365427=29^3\cdot 7^3$.
(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.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6923112}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=7$ yields the same integer 6923112.",
"robustness_analysis": "Generality... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{6923112}$.) |
math-019185 | Number Theory: Euler Totient (Variant B) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=6075=3^5\cdot 5^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=5$ yields the same integer 3240.",
"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... | 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 $\boxed{3240}$.) |
math-019186 | Number Theory: Euler Totient (Variant B) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=52200625=17^4\cdot 5^4$.
(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... | [
{
"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 $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{39304000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=5$ yields the same integer 39304000.",
"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... | 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-019187 | Number Theory: Euler Totient (Core) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=2715556321=31^2\cdot 41^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=41$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2563861200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=41$ yields the same integer 2563861200.",
"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... | 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 $\boxed{2563861200}$.) |
math-019188 | Number Theory: Euler Totient (Variant C) | 10 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=317057=13^1\cdot 29^3$.
(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 $\varphi(n)$ count... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=13$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{282576}$.\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 282576.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{282576}$.) |
math-019189 | Number Theory: Euler Totient (Variant A) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=1585183=23^1\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ cou... | [
{
"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, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1479280}$.\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 1479280.",
"robustness_analysis": "Generality note: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019190 | Number Theory: Euler Totient (Core) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=90077=41^1\cdot 13^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $p=41$ and $q=13$ are distinct primes, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{81120}$.\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 81120.",
"robustness_analysis": "Sensitivity a... | [
{
"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-019191 | Number Theory: Euler Totient (Variant C) | 10 | State any required conditions first: Compute Euler's totient function $\varphi(n)$.
Here $n=70241161=29^2\cdot 17^4$.
(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.... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=29$ and $q=17$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"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: $\\boxed{63829696}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=17$ yields the same integer 63829696.",
"robustness_analysis": "Robustness note: Both me... | [
{
"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-019192 | Number Theory: Euler Totient (Variant B) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=7225=5^2\cdot 17^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"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 $q=17$ are distinct primes, $\\gcd(p^2,q^2)=1$.",
"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: $\\boxed{5440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=17$ yields the same integer 5440.",
"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... | 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 $\boxed{5440}$.) |
math-019193 | Number Theory: Euler Totient (Variant C) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=27447121=13^4\cdot 31^2$.
(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": "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 $q=31$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{24518520}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=31$ yields the same integer 24518520.",
"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... | 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-019194 | Number Theory: Euler Totient (Variant C) | 10 | Compute the requested quantity: Compute Euler's totient function $\varphi(n)$.
Here $n=557183=37^3\cdot 11^1$.
(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 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 $p=37$ and $q=11$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"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: $\\boxed{492840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=11$ yields the same integer 492840.",
"robustness_analysis": "Sensitivity analysis: Both m... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019195 | Number Theory: Euler Totient (Core) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=1919429419=19^3\cdot 23^4$.
(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 $\varphi(n)$ 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 $p=19$ and $q=23$.",
"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: $\\boxed{1739345652}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=23$ yields the same integer 1739345652.",
"robustness_analysis": "Sen... | [
{
"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 $\boxed{1739345652}$.) |
math-019196 | Number Theory: Euler Totient (Variant A) | 10 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=107653=7^2\cdot 13^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit a... | [
{
"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 $q=13$.",
"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: $\\boxed{85176}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=13$ yields the same integer 85176.",
"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... | 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-019197 | Number Theory: Euler Totient (Core) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=2385443281=13^4\cdot 17^4$.
(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 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 $p=13$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2072421312}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=17$ yields the same integer 2072421312.",
"robustness_analysis": "Sensitivity analysis... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2072421312}$.) |
math-019198 | Number Theory: Euler Totient (Core) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=28=2^2\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=2$ and $q=7$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{12}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=7$ yields the same integer 12.",
"robustness_analysis": "Generality note: Both methods rely on knowin... | [
{
"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 $\boxed{12}$.) |
math-019199 | Number Theory: Euler Totient (Core) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=5603803=43^1\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=19$.",
"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: $\\boxed{5185404}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=19$ yields the same integer 5185404.",
"robustness_analysis": "... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{5185404}$.) |
math-019200 | Number Theory: Euler Totient (Variant A) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=3722098081=13^4\cdot 19^4$.
(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 $\varphi(n)$ 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 $q=19$.",
"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: $\\boxed{3254952168}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=19$ yields the same integer 3254952168.",
"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... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
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