id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-019201 | Number Theory: Euler Totient (Variant A) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=3518667=19^4\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=19$ 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: $\\boxed{2222316}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=3$ yields the same integer 2222316.",
"robustness_analysis": "S... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{2222316}$.) |
math-019202 | Number Theory: Euler Totient (Variant A) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=11303044=2^2\cdot 41^4$.
(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 explicit about why multi... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{5513680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=41$ yields the same integer 5513680.",
"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-019203 | Number Theory: Euler Totient (Variant A) | 10 | Do not skip justification steps: Compute Euler's totient function $\varphi(n)$.
Here $n=967361669=23^3\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=43$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{903783804}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=43$ yields the same integer 903783804.",
"robustness_analysis... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{903783804}$.) |
math-019204 | Number Theory: Euler Totient (Variant C) | 10 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=717409=11^4\cdot 7^2$.
(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 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 $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{559020}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=7$ yields the same integer 559020.",
"robustness_analysis": "Robustness note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{559020}$.) |
math-019205 | Number Theory: Euler Totient (Core) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=74=37^1\cdot 2^1$.
(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 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 $p=37$ and $q=2$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"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: $\\boxed{36}$.\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 36.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{36}$.) |
math-019206 | Number Theory: Euler Totient (Variant B) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=53527912321=37^4\cdot 13^4$.
(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 $\varphi(n)$ ... | [
{
"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 $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{48074964912}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=13$ yields the same integer 48074964912.",
"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... | 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 $\boxed{48074964912}$.) |
math-019207 | Number Theory: Euler Totient (Variant A) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=648=3^4\cdot 2^3$.
(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 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 $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{216}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=2$ yields the same integer 216.",
"robustness_analysis": "Robustness note: Both methods rely on know... | [
{
"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 $\boxed{216}$.) |
math-019208 | Number Theory: Euler Totient (Core) | 10 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=160=2^5\cdot 5^1$.
(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... | [
{
"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 $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{64}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=5$ yields the same integer 64.",
"robustness_analysis": "Robustness note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{64}$.) |
math-019209 | Number Theory: Euler Totient (Variant A) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=31117=29^2\cdot 37^1$.
(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 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 $q=37$ 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: $\\boxed{29232}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=37$ yields the same integer 29232.",
"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-019210 | Number Theory: Euler Totient (Core) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=68574961=13^4\cdot 7^4$.
(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 explicit about why multip... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=13$ 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: $\\boxed{54257112}$.\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 54257112.",
"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 $\boxed{54257112}$.) |
math-019211 | Number Theory: Euler Totient (Core) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=285541678321=17^4\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts... | [
{
"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 $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{262495222752}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=43$ yields the same integer 262495222752.",
"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... | 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 $\boxed{262495222752}$.) |
math-019212 | Number Theory: Euler Totient (Variant A) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=2840383=7^5\cdot 13^2$.
(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 $\varphi(n)$ c... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2247336}$.\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 2247336.",
"robustness_analysis": "Robustness note: Both methods rel... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2247336}$.) |
math-019213 | Number Theory: Euler Totient (Core) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=62197=37^1\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit abou... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=37$ and $q=41$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{59040}$.\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 59040.",
"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... | 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 $\boxed{59040}$.) |
math-019214 | Number Theory: Euler Totient (Variant A) | 10 | Task: Compute Euler's totient function $\varphi(n)$.
Here $n=6407383500961=37^4\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{6089229323352}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=43$ yields the same integer 6089229323352.",
"robustness_analysis"... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019215 | Number Theory: Euler Totient (Variant C) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=3993=11^3\cdot 3^1$.
(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.
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 $p=11$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{2420}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=3$ yields the same integer 2420.",
"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 $n(1-1/p)(1-1/q)$. |
math-019216 | Number Theory: Euler Totient (Variant A) | 10 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=323433=3^5\cdot 11^3$.
(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 about why multipl... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{196020}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=11$ yields the same integer 196020.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019217 | Number Theory: Euler Totient (Variant B) | 10 | Do not skip justification steps: Compute Euler's totient function $\varphi(n)$.
Here $n=31499023927=7^5\cdot 37^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{26269456248}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=37$ yields the same integer 26269456248.",
"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... | 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 $\boxed{26269456248}$.) |
math-019218 | Number Theory: Euler Totient (Variant C) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=1813=37^1\cdot 7^2$.
(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 $\varphi(n)$ 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 $p=37$ and $q=7$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1512}$.\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 1512.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1512}$.) |
math-019219 | Number Theory: Euler Totient (Variant B) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=9084137=43^2\cdot 17^3$.
(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 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 $p=43$ and $q=17$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{8350944}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=17$ yields the same integer 8350944.",
"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-019220 | Number Theory: Euler Totient (Variant B) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=64992503=23^1\cdot 41^4$.
(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 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 $p=23$ 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: $\\boxed{60650480}$.\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 60650480.",
"robustness_analysis":... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{60650480}$.) |
math-019221 | Number Theory: Euler Totient (Core) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=123634100563=43^5\cdot 29^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence 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 $p=43$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{116594789304}$.\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 116594789304.",
"robustness_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... | 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 $\boxed{116594789304}$.) |
math-019222 | Number Theory: Euler Totient (Variant A) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=58=29^1\cdot 2^1$.
(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 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 $p=29$ 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: $\\boxed{28}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=2$ yields the same integer 28.",
"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... | 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 $\boxed{28}$.) |
math-019223 | Number Theory: Euler Totient (Variant A) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=39100009=37^2\cdot 13^4$.
(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 $\varphi(n)$ ... | [
{
"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 $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{35116848}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=13$ yields the same integer 35116848.",
"robustness_analysis": "Sensitivity analysis: 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... | 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 $\boxed{35116848}$.) |
math-019224 | Number Theory: Euler Totient (Variant C) | 10 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=2527=7^1\cdot 19^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $q=19$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2052}$.\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 2052.",
"robustness_analysis": "Sensitivity analysis: 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 $\boxed{2052}$.) |
math-019225 | Number Theory: Euler Totient (Variant A) | 10 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=72283=41^2\cdot 43^1$.
(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 explicit about ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=43$ 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: $\\boxed{68880}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=43$ yields the same integer 68880.",
"robustness_analysis": "Generality no... | [
{
"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 $\boxed{68880}$.) |
math-019226 | Number Theory: Euler Totient (Variant A) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=697=17^1\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=17$ and $q=41$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{640}$.\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 640.",
"robustness_analysis": "Generality note: Both methods rely on kn... | [
{
"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 $\boxed{640}$.) |
math-019227 | Number Theory: Euler Totient (Variant C) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=22627=17^1\cdot 11^3$.
(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 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 $p=17$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{19360}$.\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 19360.",
"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... | 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-019228 | Number Theory: Euler Totient (Core) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=216=3^3\cdot 2^3$.
(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.... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=2$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{72}$.\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 72.",
"robustness_analysis": "Generality note: Both methods rely on ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019229 | Number Theory: Euler Totient (Variant A) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=560947=29^3\cdot 23^1$.
(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 $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ 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: $\\boxed{518056}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=23$ yields the same integer 518056.",
"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 $\boxed{518056}$.) |
math-019230 | Number Theory: Euler Totient (Variant C) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=232=29^1\cdot 2^3$.
(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 $\varphi(n)$ count... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=2$.",
"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: $\\boxed{112}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=2$ yields the same integer 112.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{112}$.) |
math-019231 | Number Theory: Euler Totient (Core) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=374151649=29^4\cdot 23^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"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 $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{345543352}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=23$ yields the same integer 345543352.",
"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... | 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 $\boxed{345543352}$.) |
math-019232 | Number Theory: Euler Totient (Variant B) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=17689=19^2\cdot 7^2$.
(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 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 $p=19$ 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: $\\boxed{14364}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=7$ yields the same integer 14364.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{14364}$.) |
math-019233 | Number Theory: Euler Totient (Variant A) | 10 | Question: Compute Euler's totient function $\varphi(n)$.
Here $n=104553157=41^4\cdot 37^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=41$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{99246240}$.\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 99246240.",
"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... | 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-019234 | Number Theory: Euler Totient (Variant A) | 10 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=63869=17^3\cdot 13^1$.
(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 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 $p=17$ and $q=13$ 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: $\\boxed{55488}$.\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 55488.",
"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-019235 | Number Theory: Euler Totient (Core) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=36459209=23^2\cdot 41^3$.
(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 ... | [
{
"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, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{34023440}$.\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 34023440.",
"robustness_analysis": "Generality 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... | 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 $\boxed{34023440}$.) |
math-019236 | Number Theory: Euler Totient (Variant A) | 10 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=3286969=7^4\cdot 37^2$.
(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 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 $p=7$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2741256}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=37$ yields the same integer 2741256.",
"robustness_analysis": "If the problem were perturbed: Bo... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{2741256}$.) |
math-019237 | Number Theory: Euler Totient (Variant B) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=18458141=17^5\cdot 13^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts... | [
{
"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 $q=13$ are distinct primes, $\\gcd(p^5,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16036032}$.\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 16036032.",
"robustness_analysis": "General... | [
{
"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 $\boxed{16036032}$.) |
math-019238 | Number Theory: Euler Totient (Variant C) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=2896405025=41^5\cdot 5^2$.
(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.
... | [
{
"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 $q=5$ are distinct primes, $\\gcd(p^5,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2260608800}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=5$ yields the same integer 2260608800.",
"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... | 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-019239 | Number Theory: Euler Totient (Variant C) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=2121843=29^4\cdot 3^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=29$ and $q=3$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1365784}$.\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 1365784.",
"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 $\boxed{1365784}$.) |
math-019240 | Number Theory: Euler Totient (Variant C) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=2305703=29^1\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be... | [
{
"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 $q=43$ 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: $\\boxed{2174424}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=43$ yields the same integer 2174424.",
"robustness_analysis": "Robustness note: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019241 | Number Theory: Euler Totient (Core) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=161=23^1\cdot 7^1$.
(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 $\varphi(n)$ count... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=7$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{132}$.\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 132.",
"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... | 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-019242 | Number Theory: Euler Totient (Variant A) | 10 | Checkpoint: Compute Euler's totient function $\varphi(n)$.
Here $n=15768841=11^2\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why m... | [
{
"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 $q=19$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13580820}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=19$ yields the same integer 13580820.",
"robustness_analysis": "If the problem were pert... | [
{
"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 $\boxed{13580820}$.) |
math-019243 | Number Theory: Euler Totient (Variant C) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=3048625=5^3\cdot 29^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2354800}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=29$ yields the same integer 2354800.",
"robustness_analysis": "If the pro... | [
{
"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 $\boxed{2354800}$.) |
math-019244 | Number Theory: Euler Totient (Core) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=4121741=13^2\cdot 29^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why 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 $p=13$ 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: $\\boxed{3673488}$.\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 3673488.",
"robustness_analysis": "If the problem were pertur... | [
{
"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 $\boxed{3673488}$.) |
math-019245 | Number Theory: Euler Totient (Variant C) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=107811=3^4\cdot 11^3$.
(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 $p=3$ and $q=11$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{65340}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=11$ yields the same integer 65340.",
"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... | 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 $\boxed{65340}$.) |
math-019246 | Number Theory: Euler Totient (Core) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=29584=43^2\cdot 2^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{14448}$.\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 14448.",
"robustness_analysis": "Generality note: Both methods rely on... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019247 | Number Theory: Euler Totient (Variant B) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=19320201=3^5\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{12580596}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=43$ yields the same integer 12580596.",
"robustness_analysis": "Generality 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... | 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 $\boxed{12580596}$.) |
math-019248 | Number Theory: Euler Totient (Variant B) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=424589=11^4\cdot 29^1$.
(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 explicit about why multipl... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ 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: $\\boxed{372680}$.\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 372680.",
"robustness_analysis": "If the problem were perturbe... | [
{
"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 $\boxed{372680}$.) |
math-019249 | Number Theory: Euler Totient (Variant B) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=206763=41^3\cdot 3^1$.
(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.
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 $q=3$ are distinct primes, $\\gcd(p^3,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{134480}$.\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 134480.",
"robustness_analysis": "Robustness note: Both methods... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019250 | Number Theory: Euler Totient (Core) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=80706559921=41^4\cdot 13^4$.
(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 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 $p=41$ and $q=13$ 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: $\\boxed{72681329760}$.\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 72681329760.",
"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... | 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 $\boxed{72681329760}$.) |
math-019251 | Number Theory: Euler Totient (Variant A) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=76840601=37^4\cdot 41^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
B... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{72940320}$.\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 72940320.",
"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 $\boxed{72940320}$.) |
math-019252 | Number Theory: Euler Totient (Variant B) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=6250=5^5\cdot 2^1$.
(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.... | [
{
"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 $q=2$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{2500}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=2$ yields the same integer 2500.",
"robustness_analysis": "Robustness note: Both methods rely... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2500}$.) |
math-019253 | Number Theory: Euler Totient (Core) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=969236023=7^3\cdot 41^4$.
(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 explicit... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=41$ are distinct primes, $\\gcd(p^3,q^4)=1$.",
"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: $\\boxed{810510960}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=41$ yields the same integer 810510960.",
"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... | 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-019254 | Number Theory: Euler Totient (Core) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=154204603531=41^5\cdot 11^3$.
(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.
... | [
{
"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 $q=11$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{136766832400}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=11$ yields the same integer 136766832400.",
"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 $\boxed{136766832400}$.) |
math-019255 | Number Theory: Euler Totient (Variant B) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=3090277=17^4\cdot 37^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=17$ and $q=37$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"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: $\\boxed{2829888}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=37$ yields the same integer 2829888.",
"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 $n(1-1/p)(1-1/q)$. |
math-019256 | Number Theory: Euler Totient (Core) | 10 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=7267=13^2\cdot 43^1$.
(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 explicit about... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=13$ and $q=43$ 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: $\\boxed{6552}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=43$ yields the same integer 6552.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{6552}$.) |
math-019257 | Number Theory: Euler Totient (Core) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=1863=3^4\cdot 23^1$.
(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... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1188}$.\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 1188.",
"robustness_analysis": "If the ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1188}$.) |
math-019258 | Number Theory: Euler Totient (Core) | 10 | Task: Compute Euler's totient function $\varphi(n)$.
Here $n=129=3^1\cdot 43^1$.
(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 explicit about why multiplicativ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "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 $p=3,q=43$ yields the same integer 84.",
"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... | 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 $\boxed{84}$.) |
math-019259 | Number Theory: Euler Totient (Core) | 10 | Compute the requested quantity: Compute Euler's totient function $\varphi(n)$.
Here $n=4625=37^1\cdot 5^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3600}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=5$ yields the same integer 3600.",
"robustness_analysis": "Sensitivity analysis: 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 $\boxed{3600}$.) |
math-019260 | Number Theory: Euler Totient (Variant C) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=278179=19^1\cdot 11^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=19$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{239580}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=11$ yields the same integer 239580.",
"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... | 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 $\boxed{239580}$.) |
math-019261 | Number Theory: Euler Totient (Variant B) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=22802916887=37^4\cdot 23^3$.
(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 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 $p=37$ and $q=23$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{21221986104}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=23$ yields the same integer 21221986104.",
"robustness_anal... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{21221986104}$.) |
math-019262 | Number Theory: Euler Totient (Variant B) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=184=23^1\cdot 2^3$.
(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 ... | [
{
"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, $\\gcd(p^1,q^3)=1$.",
"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: $\\boxed{88}$.\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 88.",
"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 $n(1-1/p)(1-1/q)$. |
math-019263 | Number Theory: Euler Totient (Variant B) | 10 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=2673033448829=29^5\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)... | [
{
"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 $q=19$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2445025151016}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=19$ yields the same integer 2445025151016.",
"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 $\boxed{2445025151016}$.) |
math-019264 | Number Theory: Euler Totient (Variant A) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=178137047=11^4\cdot 23^3$.
(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... | [
{
"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 $q=23$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{154901780}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=23$ yields the same integer 154901780.",
"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 $n(1-1/p)(1-1/q)$. |
math-019265 | Number Theory: Euler Totient (Variant C) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=62898244747=31^5\cdot 13^3$.
(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)... | [
{
"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 $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{56187017640}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=13$ yields the same integer 56187017640.",
"robustness_analysis": "Sensitivity 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... | 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 $\boxed{56187017640}$.) |
math-019266 | Number Theory: Euler Totient (Core) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=4761=3^2\cdot 23^2$.
(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 $\varphi(n)$ 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 $p=3$ 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: $\\boxed{3036}$.\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 3036.",
"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 $\boxed{3036}$.) |
math-019267 | Number Theory: Euler Totient (Variant B) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=456533=11^3\cdot 7^3$.
(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 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 $p=11$ 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: $\\boxed{355740}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=7$ yields the same integer 355740.",
"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... | 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 $\boxed{355740}$.) |
math-019268 | Number Theory: Euler Totient (Variant A) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=29986576=2^4\cdot 37^4$.
(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 $\varphi(n)$ 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 $p=2$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{14588064}$.\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 14588064.",
"robustness_analysis": "If the problem were pertu... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{14588064}$.) |
math-019269 | Number Theory: Euler Totient (Core) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=31487=37^2\cdot 23^1$.
(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 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 $p=37$ and $q=23$ 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: $\\boxed{29304}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=37,q=23$ yields the same integer 29304.",
"robustness_analysis": "Generality note: Both methods rely o... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{29304}$.) |
math-019270 | Number Theory: Euler Totient (Variant C) | 10 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=3971=11^1\cdot 19^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=11$ and $q=19$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{3420}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=19$ yields the same integer 3420.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{3420}$.) |
math-019271 | Number Theory: Euler Totient (Variant A) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=2215457=17^1\cdot 19^4$.
(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.
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 $q=19$.",
"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: $\\boxed{1975392}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=19$ yields the same integer 1975392.",
"robustness_analysis": "Sensitivi... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{1975392}$.) |
math-019272 | Number Theory: Euler Totient (Variant C) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=20295603=3^5\cdot 17^4$.
(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 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 $p=3$ and $q=17$.",
"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: $\\boxed{12734496}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=17$ yields the same integer 12734496.",
"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 $n(1-1/p)(1-1/q)$. |
math-019273 | Number Theory: Euler Totient (Variant C) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=4779254860397=37^5\cdot 41^3$.
(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 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 $p=37$ 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": "The two methods are consistent and must coincide. Final answer: $\\boxed{4536669083040}$.\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 4536669083040.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019274 | Number Theory: Euler Totient (Variant A) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=6=2^1\cdot 3^1$.
(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(... | [
{
"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: $\\boxed{2}$.\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 2.",
"robustness_analysis": "Generality note: Both methods rely on knowing ... | [
{
"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 $\boxed{2}$.) |
math-019275 | Number Theory: Euler Totient (Variant B) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=6185325623=23^5\cdot 31^2$.
(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.
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 $p=23$ and $q=31$.",
"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: $\\boxed{5725546860}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=31$ yields the same integer 5725546860.",
"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... | 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-019276 | Number Theory: Euler Totient (Core) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=1680914269=29^3\cdot 41^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=29$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1583367520}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=41$ yields the same integer 1583367520.",
"robustness_analysis": "Generality note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019277 | Number Theory: Euler Totient (Core) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=559682=2^1\cdot 23^4$.
(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 $\varphi(n)$ 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 $p=2$ and $q=23$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{267674}$.\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 267674.",
"robustness_analysis": "If the problem were perturbed... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019278 | Number Theory: Euler Totient (Variant A) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=7511105797=13^3\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.... | [
{
"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 $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{6772088232}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=43$ yields the same integer 6772088232.",
"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... | 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 $\boxed{6772088232}$.) |
math-019279 | Number Theory: Euler Totient (Variant C) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=1113879=13^5\cdot 3^1$.
(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 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 $p=13$ 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: $\\boxed{685464}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=3$ yields the same integer 685464.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{685464}$.) |
math-019280 | Number Theory: Euler Totient (Variant A) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=178409449=19^4\cdot 37^2$.
(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 $\varphi(n)$ c... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=19$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{164451384}$.\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 164451384.",
"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... | 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-019281 | Number Theory: Euler Totient (Variant B) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=39711403=31^4\cdot 43^1$.
(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 explicit about why mul... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=43$ 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: $\\boxed{37536660}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=43$ yields the same integer 37536660.",
"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-019282 | Number Theory: Euler Totient (Variant A) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=19773=13^3\cdot 3^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=13$ and $q=3$ are distinct primes, $\\gcd(p^3,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{12168}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=3$ yields the same integer 12168.",
"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 $n(1-1/p)(1-1/q)$. |
math-019283 | Number Theory: Euler Totient (Variant B) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=2628125=5^5\cdot 29^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=5$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2030000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=29$ yields the same integer 2030000.",
"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... | 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 $\boxed{2030000}$.) |
math-019284 | Number Theory: Euler Totient (Variant A) | 10 | Checkpoint: Compute Euler's totient function $\varphi(n)$.
Here $n=319=11^1\cdot 29^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multip... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ and $q=29$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{280}$.\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 280.",
"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-019285 | Number Theory: Euler Totient (Core) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=3087=7^3\cdot 3^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=7$ and $q=3$ 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: $\\boxed{1764}$.\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 1764.",
"robustness_analysis": "If the problem were perturbed: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1764}$.) |
math-019286 | Number Theory: Euler Totient (Core) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=62=2^1\cdot 31^1$.
(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 $\varph... | [
{
"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 $q=31$.",
"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: $\\boxed{30}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=31$ yields the same integer 30.",
"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... | 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-019287 | Number Theory: Euler Totient (Variant C) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=32166277=11^4\cdot 13^3$.
(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 $\varphi(n)$ c... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ 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: $\\boxed{26992680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=13$ yields the same integer 26992680.",
"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 $\boxed{26992680}$.) |
math-019288 | Number Theory: Euler Totient (Variant A) | 10 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=28252567=7^5\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence 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 $p=7$ 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: $\\boxed{23625840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=41$ yields the same integer 23625840.",
"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 $n(1-1/p)(1-1/q)$. |
math-019289 | Number Theory: Euler Totient (Core) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=1272112=2^4\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=2$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{621264}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=43$ yields the same integer 621264.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019290 | Number Theory: Euler Totient (Core) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=1102736=2^4\cdot 41^3$.
(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 $\varphi(n)$ co... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{537920}$.\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 537920.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{537920}$.) |
math-019291 | Number Theory: Euler Totient (Variant B) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=28702027=19^2\cdot 43^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=19$ and $q=43$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{26559036}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=43$ yields the same integer 26559036.",
"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-019292 | Number Theory: Euler Totient (Variant A) | 10 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=200564019=19^5\cdot 3^4$.
(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 $\varphi(n)$ 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 $p=19$ 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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{126672012}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=3$ yields the same integer 126672012.",
"robustness_analysis": "If the... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{126672012}$.) |
math-019293 | Number Theory: Euler Totient (Variant A) | 10 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=41323=43^1\cdot 31^2$.
(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 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 $p=43$ and $q=31$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{39060}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=31$ yields the same integer 39060.",
"robustness_analysis": "Sens... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019294 | Number Theory: Euler Totient (Variant C) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=48=3^1\cdot 2^4$.
(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... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=2$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{16}$.\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 16.",
"robustness_analysis": "Robustness 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... | 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-019295 | Number Theory: Euler Totient (Variant B) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=33698267=17^3\cdot 19^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=17$ 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: $\\boxed{30046752}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=19$ yields the same integer 30046752.",
"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-019296 | Number Theory: Euler Totient (Core) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=215=43^1\cdot 5^1$.
(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 ... | [
{
"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 $q=5$ 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: $\\boxed{168}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=5$ yields the same integer 168.",
"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... | 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-019297 | Number Theory: Euler Totient (Variant A) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=12=3^1\cdot 2^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{4}$.\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 4.",
"robustness_analysis": "Sensitivity analysis: 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-019298 | Number Theory: Euler Totient (Variant A) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=469567=37^2\cdot 7^3$.
(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 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 $p=37$ and $q=7$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{391608}$.\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 391608.",
"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... | 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 $\boxed{391608}$.) |
math-019299 | Number Theory: Euler Totient (Variant B) | 10 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=76406227=43^3\cdot 31^2$.
(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 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 $p=43$ and $q=31$ are distinct primes, $\\gcd(p^3,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{72221940}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=31$ yields the same integer 72221940.",
"robustness_analysis": "General... | [
{
"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-019300 | Number Theory: Euler Totient (Variant C) | 10 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=1161=43^1\cdot 3^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=43$ and $q=3$ 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: $\\boxed{756}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=3$ yields the same integer 756.",
"robustness_analysis": "Robustness note: B... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{756}$.) |
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