id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-019301 | Number Theory: Euler Totient (Variant B) | 10 | Prompt: Compute Euler's totient function $\varphi(n)$.
Here $n=12005773=13^1\cdot 31^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why 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 $p=13$ and $q=31$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{10724760}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=31$ yields the same integer 10724760.",
"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-019302 | Number Theory: Euler Totient (Core) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=329623=7^3\cdot 31^2$.
(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 $\varphi(n)$ co... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{273420}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=31$ yields the same integer 273420.",
"robustness_analysis": "Rob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019303 | Number Theory: Euler Totient (Variant B) | 10 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=24=3^1\cdot 2^3$.
(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 why... | [
{
"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": "The two methods are consistent and must coincide. Final answer: $\\boxed{8}$.\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 8.",
"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 $\boxed{8}$.) |
math-019304 | Number Theory: Euler Totient (Variant A) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=5658248=2^3\cdot 29^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.... | [
{
"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 $q=29$ are distinct primes, $\\gcd(p^3,q^4)=1$.",
"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: $\\boxed{2731568}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=29$ yields the same integer 2731568.",
"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 $n(1-1/p)(1-1/q)$. |
math-019305 | Number Theory: Euler Totient (Core) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=162409=31^2\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ and $q=13$ are distinct primes, $\\gcd(p^2,q^2)=1$.",
"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: $\\boxed{145080}$.\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 145080.",
"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... | 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 $\boxed{145080}$.) |
math-019306 | Number Theory: Euler Totient (Core) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=521284=19^4\cdot 2^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=19$ 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: $\\boxed{246924}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=2$ yields the same integer 246924.",
"robustness_analysis": "Generality note: Both methods... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{246924}$.) |
math-019307 | Number Theory: Euler Totient (Variant C) | 10 | Track units/moduli carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=1038951230297=29^5\cdot 37^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "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 $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{976013830512}$.\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 976013830512.",
"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... | 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 $\boxed{976013830512}$.) |
math-019308 | Number Theory: Euler Totient (Variant C) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=4170272=2^5\cdot 19^4$.
(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 $\varphi(n)$ 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 $p=2$ 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": "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 $p=2,q=19$ yields the same integer 1975392.",
"robustness_analysis": "Generality... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019309 | Number Theory: Euler Totient (Variant A) | 10 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=889249=41^2\cdot 23^2$.
(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 $p=41$ and $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{829840}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=23$ yields the same integer 829840.",
"robustness_analysis": "Sensitivity... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019310 | Number Theory: Euler Totient (Variant C) | 10 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=1943784233=17^5\cdot 37^2$.
(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 w... | [
{
"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, $\\gcd(p^5,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1779999552}$.\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 1779999552.",
"robustness_analysis": "Generality note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1779999552}$.) |
math-019311 | Number Theory: Euler Totient (Core) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=301=43^1\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=43$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{252}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=7$ yields the same integer 252.",
"robustness_analysis": "Generality note: Both methods rely on kno... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{252}$.) |
math-019312 | Number Theory: Euler Totient (Variant B) | 10 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=1413648900229=29^5\cdot 41^3$.
(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.
... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=29$ and $q=41$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1331612084320}$.\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 1331612084320.",
"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 $n(1-1/p)(1-1/q)$. |
math-019313 | Number Theory: Euler Totient (Variant B) | 10 | Compute the requested quantity: Compute Euler's totient function $\varphi(n)$.
Here $n=352967271643=43^5\cdot 7^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{295507483236}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=7$ yields the same integer 295507483236.",
"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 $\boxed{295507483236}$.) |
math-019314 | Number Theory: Euler Totient (Core) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=4107=3^1\cdot 37^2$.
(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.
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 $p=3$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2664}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=37$ yields the same integer 2664.",
"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... | 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 $\boxed{2664}$.) |
math-019315 | Number Theory: Euler Totient (Core) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=390963=3^1\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multip... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=19$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{246924}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=19$ yields the same integer 246924.",
"robustness_analysis": "Sensitivity analysis: Both methods ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019316 | Number Theory: Euler Totient (Variant A) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=1399489=13^4\cdot 7^2$.
(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": "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 $q=7$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1107288}$.\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 1107288.",
"robustness_analysis": "Robustness... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019317 | Number Theory: Euler Totient (Core) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=82572791=7^5\cdot 17^3$.
(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 $\varphi(n)$ co... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=17$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{66613344}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=17$ yields the same integer 66613344.",
"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-019318 | Number Theory: Euler Totient (Core) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=2215457=19^4\cdot 17^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1975392}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=17$ yields the same integer 1975392.",
"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... | 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 $\boxed{1975392}$.) |
math-019319 | Number Theory: Euler Totient (Variant A) | 10 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=12691=37^1\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=7$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{10584}$.\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 10584.",
"robustness_analysis": "Generality note: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019320 | Number Theory: Euler Totient (Variant C) | 10 | Show all reasoning: Compute Euler's totient function $\varphi(n)$.
Here $n=219501=3^2\cdot 29^3$.
(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 ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=29$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"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: $\\boxed{141288}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=29$ yields the same integer 141288.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{141288}$.) |
math-019321 | Number Theory: Euler Totient (Variant B) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=31010762653=13^5\cdot 17^4$.
(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)... | [
{
"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 $q=17$ 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: $\\boxed{26941477056}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=17$ yields the same integer 26941477056.",
"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 $n(1-1/p)(1-1/q)$. |
math-019322 | Number Theory: Euler Totient (Variant A) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=219069601=19^4\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about 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 $p=19$ and $q=41$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{202477680}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=41$ yields the same integer 202477680.",
"robustness_analysis": "Sensitivity analysis: Both m... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019323 | Number Theory: Euler Totient (Variant A) | 10 | Determine the requested value: Compute Euler's totient function $\varphi(n)$.
Here $n=368255999281=41^4\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $q=19$ are distinct primes, $\\gcd(p^4,q^4)=1$.",
"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: $\\boxed{340364980080}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=19$ yields the same integer 340364980080.",
"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 $\boxed{340364980080}$.) |
math-019324 | Number Theory: Euler Totient (Variant C) | 10 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=2875=23^1\cdot 5^3$.
(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 explicit about why multiplic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=23$ and $q=5$ 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: $\\boxed{2200}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=5$ yields the same integer 2200.",
"robustness_analysis": "Sensitivity anal... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{2200}$.) |
math-019325 | Number Theory: Euler Totient (Variant B) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=143133271933=41^4\cdot 37^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"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 $q=37$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{135868102560}$.\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 135868102560.",
"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-019326 | Number Theory: Euler Totient (Variant A) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=634933=13^3\cdot 17^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=13$ and $q=17$ 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: $\\boxed{551616}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=17$ yields the same integer 551616.",
"robustness_analysis": "Sensitivity analysis: Both methods... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{551616}$.) |
math-019327 | Number Theory: Euler Totient (Variant A) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=20577=3^1\cdot 19^3$.
(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 $\varphi(n)$ 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 $p=3$ and $q=19$ 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: $\\boxed{12996}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=19$ yields the same integer 12996.",
"robustness_analysis": "Sensi... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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 $\boxed{12996}$.) |
math-019328 | Number Theory: Euler Totient (Core) | 10 | Complete the analysis: Compute Euler's totient function $\varphi(n)$.
Here $n=912247=19^4\cdot 7^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $p=19$ and $q=7$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{740772}$.\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 740772.",
"robustness_analysis": "Rob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019329 | Number Theory: Euler Totient (Variant B) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=1413721=29^2\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $p=29$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1331680}$.\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 1331680.",
"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-019330 | Number Theory: Euler Totient (Variant B) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=31668003=3^5\cdot 19^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $q=19$.",
"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: $\\boxed{20000844}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=19$ yields the same integer 20000844.",
"robustness_analysis": "If the p... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{20000844}$.) |
math-019331 | Number Theory: Euler Totient (Variant C) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=11900798207=23^5\cdot 43^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.... | [
{
"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, $\\gcd(p^5,q^2)=1$.",
"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: $\\boxed{11118642612}$.\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 11118642612.",
"robustness_anal... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019332 | Number Theory: Euler Totient (Core) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=2951069=29^3\cdot 11^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ab... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2590280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=11$ yields the same integer 2590280.",
"robustness_analysis": "Generality note: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{2590280}$.) |
math-019333 | Number Theory: Euler Totient (Variant A) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=667=29^1\cdot 23^1$.
(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... | [
{
"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: $\\boxed{616}$.\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 616.",
"robustness_analysis": "Generality note: Both methods rely... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{616}$.) |
math-019334 | Number Theory: Euler Totient (Variant B) | 10 | Work this out carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=3604711=11^2\cdot 31^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=11$ and $q=31$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{3171300}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=31$ yields the same integer 3171300.",
"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 $\boxed{3171300}$.) |
math-019335 | Number Theory: Euler Totient (Variant C) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=2025=5^2\cdot 3^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ... | [
{
"method_name": "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 $q=3$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1080}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=3$ yields the same integer 1080.",
"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... | 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 $\boxed{1080}$.) |
math-019336 | Number Theory: Euler Totient (Variant C) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=637=7^2\cdot 13^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $q=13$.",
"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: $\\boxed{504}$.\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 504.",
"robustness_analysis": "If the problem wer... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019337 | Number Theory: Euler Totient (Variant C) | 10 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=22977523=43^3\cdot 17^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"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 $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{21122976}$.\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 21122976.",
"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... | 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-019338 | Number Theory: Euler Totient (Variant B) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=91733851=41^3\cdot 11^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mult... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=11$ are distinct primes, $\\gcd(p^3,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{81360400}$.\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 81360400.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{81360400}$.) |
math-019339 | Number Theory: Euler Totient (Variant C) | 10 | Explain what is being counted/optimized: Compute Euler's totient function $\varphi(n)$.
Here $n=1801068721=31^2\cdot 37^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=31$ and $q=37$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1695862440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=37$ yields the same integer 1695862440.",
"robustness_analysis": "Robustness note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1695862440}$.) |
math-019340 | Number Theory: Euler Totient (Core) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=137842=41^3\cdot 2^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\v... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=41$ 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: $\\boxed{67240}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=2$ yields the same integer 67240.",
"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... | 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-019341 | Number Theory: Euler Totient (Variant C) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=42599173=29^2\cdot 37^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=29$ and $q=37$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{40018608}$.\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 40018608.",
"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-019342 | Number Theory: Euler Totient (Core) | 10 | Derive the result step-by-step: Compute Euler's totient function $\varphi(n)$.
Here $n=6845=5^1\cdot 37^2$.
(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 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 $p=5$ and $q=37$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5328}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=37$ yields the same integer 5328.",
"robustness_analysis": "Generality note: Both methods rel... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019343 | Number Theory: Euler Totient (Core) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=268119=31^3\cdot 3^2$.
(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 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 $p=31$ and $q=3$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{172980}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=3$ yields the same integer 172980.",
"robustness_analysis": "Robustness note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{172980}$.) |
math-019344 | Number Theory: Euler Totient (Variant B) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=1685159=7^3\cdot 17^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=7$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{1359456}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=17$ yields the same integer 1359456.",
"robustness_analysis": "Generality... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{1359456}$.) |
math-019345 | Number Theory: Euler Totient (Core) | 10 | Solve and sanity-check: Compute Euler's totient function $\varphi(n)$.
Here $n=576583=41^2\cdot 7^3$.
(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 ab... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ 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: $\\boxed{482160}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=7$ yields the same integer 482160.",
"robustness_analysis": "Generality note: Both methods... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{482160}$.) |
math-019346 | Number Theory: Euler Totient (Variant A) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=34375=5^5\cdot 11^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about 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 $p=5$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{25000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=11$ yields the same integer 25000.",
"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-019347 | Number Theory: Euler Totient (Variant A) | 10 | Try to avoid pattern-matching; explain why: Compute Euler's totient function $\varphi(n)$.
Here $n=151959=3^1\cdot 37^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=3$ and $q=37$ 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: $\\boxed{98568}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=37$ yields the same integer 98568.",
"robustness_analysis": "Generality note: Both methods rely on... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019348 | Number Theory: Euler Totient (Variant C) | 10 | Solve (and briefly cross-validate): Compute Euler's totient function $\varphi(n)$.
Here $n=2625315593=17^5\cdot 43^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts... | [
{
"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": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{2413422816}$.\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 2413422816.",
"robustness_analysis": "If ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2413422816}$.) |
math-019349 | Number Theory: Euler Totient (Variant C) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=98=7^2\cdot 2^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit... | [
{
"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 $q=2$ are distinct primes, $\\gcd(p^2,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{42}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=2$ yields the same integer 42.",
"robustness_analysis": "If the problem were perturbed: Both methods ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{42}$.) |
math-019350 | Number Theory: Euler Totient (Variant A) | 10 | Proceed methodically: Compute Euler's totient function $\varphi(n)$.
Here $n=10125=3^4\cdot 5^3$.
(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 explicit about ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=3$ and $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.",... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{5400}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=5$ yields the same integer 5400.",
"robustness_analysis": "Sensitivity analy... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019351 | Number Theory: Euler Totient (Variant B) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=33=11^1\cdot 3^1$.
(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 $\varphi(n)$ 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 $q=3$ are distinct primes, $\\gcd(p^1,q^1)=1$.",
"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: $\\boxed{20}$.\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 20.",
"robustness_analysis": "If the problem were perturbed: Both m... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{20}$.) |
math-019352 | Number Theory: Euler Totient (Variant A) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=112999=17^3\cdot 23^1$.
(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 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 $q=23$ 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: $\\boxed{101728}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=23$ yields the same integer 101728.",
"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... | 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 $\boxed{101728}$.) |
math-019353 | Number Theory: Euler Totient (Variant A) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=176=2^4\cdot 11^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expl... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=2$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{80}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=11$ yields the same integer 80.",
"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... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{80}$.) |
math-019354 | Number Theory: Euler Totient (Variant A) | 10 | Checkpoint: Compute Euler's totient function $\varphi(n)$.
Here $n=1159171=13^2\cdot 19^3$.
(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 about why mu... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=13$ and $q=19$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1013688}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=19$ yields the same integer 1013688.",
"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 $\boxed{1013688}$.) |
math-019355 | Number Theory: Euler Totient (Core) | 10 | Exercise: Compute Euler's totient function $\varphi(n)$.
Here $n=756059=29^3\cdot 31^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multi... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=31$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{706440}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=31$ yields the same integer 706440.",
"robustness_analysis": "If the problem were perturbed: 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... | 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-019356 | Number Theory: Euler Totient (Core) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=81947069=29^1\cdot 41^4$.
(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 exp... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=29$ and $q=41$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{77191520}$.\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 77191520.",
"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... | 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 $\boxed{77191520}$.) |
math-019357 | Number Theory: Euler Totient (Variant A) | 10 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=281219=41^1\cdot 19^3$.
(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 e... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=41$ and $q=19$.",
"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: $\\boxed{259920}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=19$ yields the same integer 259920.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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-019358 | Number Theory: Euler Totient (Variant B) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=29282=11^4\cdot 2^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=11$ and $q=2$ are distinct primes, $\\gcd(p^4,q^1)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{13310}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=2$ yields the same integer 13310.",
"robustness_analysis": "Generality note: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{13310}$.) |
math-019359 | Number Theory: Euler Totient (Variant C) | 10 | Do not skip justification steps: Compute Euler's totient function $\varphi(n)$.
Here $n=810448=37^3\cdot 2^4$.
(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 ex... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=37$ and $q=2$.",
"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: $\\boxed{394272}$.\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 394272.",
"robustness_analysis": "Generality n... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{394272}$.) |
math-019360 | Number Theory: Euler Totient (Variant A) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=589=31^1\cdot 19^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=31$ and $q=19$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{540}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=19$ yields the same integer 540.",
"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 $n(1-1/p)(1-1/q)$. |
math-019361 | Number Theory: Euler Totient (Variant B) | 10 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=78608=2^4\cdot 17^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit ab... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=17$ 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: $\\boxed{36992}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=2,q=17$ yields the same integer 36992.",
"robustness_analysis": "If th... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019362 | Number Theory: Euler Totient (Variant B) | 10 | Explain why your operations are valid: Compute Euler's totient function $\varphi(n)$.
Here $n=248897=17^1\cdot 11^4$.
(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.... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=17$ and $q=11$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"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: $\\boxed{212960}$.\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 212960.",
"robustness_analysis": "Robustness ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{212960}$.) |
math-019363 | Number Theory: Euler Totient (Core) | 10 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=14115049=17^4\cdot 13^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence wha... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=17$ and $q=13$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{12262848}$.\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 12262848.",
"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 $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{12262848}$.) |
math-019364 | Number Theory: Euler Totient (Core) | 10 | Problem: Compute Euler's totient function $\varphi(n)$.
Here $n=3150464641=37^4\cdot 41^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why mu... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=37$ and $q=41$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{2990553120}$.\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 2990553120.",
"robustness_analysis": "Generality note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019365 | Number Theory: Euler Totient (Variant A) | 10 | Solve and justify each step: Compute Euler's totient function $\varphi(n)$.
Here $n=41323=31^2\cdot 43^1$.
(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 explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=31$ 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": "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 $p=31,q=43$ 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... | 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-019366 | Number Theory: Euler Totient (Core) | 10 | Answer with a short justification: Compute Euler's totient function $\varphi(n)$.
Here $n=3424361=41^1\cdot 17^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $p=41$ and $q=17$ are distinct primes, $\\gcd(p^1,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{3144320}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=17$ yields the same integer 3144320.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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-019367 | Number Theory: Euler Totient (Variant B) | 10 | Provide a rigorous solution: Compute Euler's totient function $\varphi(n)$.
Here $n=2825616886189=41^5\cdot 29^3$.
(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.
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 $p=41$ and $q=29$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2661640801120}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=29$ yields the same integer 2661640801120.",
"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... | 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-019368 | Number Theory: Euler Totient (Variant A) | 10 | Work carefully and justify each inference: Compute Euler's totient function $\varphi(n)$.
Here $n=621=3^3\cdot 23^1$.
(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.... | [
{
"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 $q=23$ 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: $\\boxed{396}$.\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 396.",
"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-019369 | Number Theory: Euler Totient (Core) | 10 | Warm-up: Compute Euler's totient function $\varphi(n)$.
Here $n=1125=3^2\cdot 5^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about why multiplica... | [
{
"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 $q=5$ are distinct primes, $\\gcd(p^2,q^3)=1$.",
"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: $\\boxed{600}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=5$ yields the same integer 600.",
"robustness_analysis": "Generality note: 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 $n(1-1/p)(1-1/q)$. |
math-019370 | Number Theory: Euler Totient (Core) | 10 | Keep the final answer in boxed form: Compute Euler's totient function $\varphi(n)$.
Here $n=925=5^2\cdot 37^1$.
(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 $p=5$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{720}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=37$ yields the same integer 720.",
"robustness_analysis": "If the problem were perturbed: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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 $\boxed{720}$.) |
math-019371 | Number Theory: Euler Totient (Variant B) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=594473=11^2\cdot 17^3$.
(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 $\varphi(n)$ 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 $p=11$ and $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{508640}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=17$ yields the same integer 508640.",
"robustness_analysis": "If the prob... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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 $\boxed{508640}$.) |
math-019372 | Number Theory: Euler Totient (Variant A) | 10 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=138462289=41^4\cdot 7^2$.
(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 e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=7$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{115787280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=41,q=7$ yields the same integer 115787280.",
"robustness_analysis": "Genera... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{115787280}$.) |
math-019373 | Number Theory: Euler Totient (Variant C) | 10 | Track quantifiers carefully: Compute Euler's totient function $\varphi(n)$.
Here $n=3969=3^4\cdot 7^2$.
(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 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 $q=7$ are distinct primes, $\\gcd(p^4,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})$... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{2268}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=3,q=7$ yields the same integer 2268.",
"robustness_analysis": "If the p... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{2268}$.) |
math-019374 | Number Theory: Euler Totient (Core) | 10 | State any required conditions first: Compute Euler's totient function $\varphi(n)$.
Here $n=204336469=19^3\cdot 31^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ 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 $p=19$ 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: $\\boxed{187337340}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=31$ yields the same integer 187337340.",
"robustness_analysis": "If th... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{187337340}$.) |
math-019375 | Number Theory: Euler Totient (Variant C) | 10 | Solve and then verify: Compute Euler's totient function $\varphi(n)$.
Here $n=1229206451=31^4\cdot 11^3$.
(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 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 $p=31$ and $q=11$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{1081413300}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=11$ yields the same integer 1081413300.",
"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-019376 | Number Theory: Euler Totient (Variant B) | 10 | Task: Compute Euler's totient function $\varphi(n)$.
Here $n=262608484333=13^5\cdot 29^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{234048940944}$.\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 234048940944.",
"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... | 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 $\boxed{234048940944}$.) |
math-019377 | Number Theory: Euler Totient (Variant C) | 10 | Use two approaches if possible: Compute Euler's totient function $\varphi(n)$.
Here $n=413674921=11^2\cdot 43^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
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 $q=43$ are distinct primes, $\\gcd(p^2,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{367322340}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=43$ yields the same integer 367322340.",
"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... | 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 $\boxed{367322340}$.) |
math-019378 | Number Theory: Euler Totient (Variant C) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=38021875=5^5\cdot 23^3$.
(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 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 $p=5$ 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: $\\boxed{29095000}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=5,q=23$ yields the same integer 29095000.",
"robustness_analysis": "Sensitivity analysis: 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 $\boxed{29095000}$.) |
math-019379 | Number Theory: Euler Totient (Variant C) | 10 | Provide both a computational and a conceptual explanation: Compute Euler's totient function $\varphi(n)$.
Here $n=978121=43^2\cdot 23^2$.
(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 ... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=43$ and $q=23$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{913836}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=23$ yields the same integer 913836.",
"robustness_analysis": "Se... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Core principle: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
math-019380 | Number Theory: Euler Totient (Variant B) | 10 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=1351619=17^1\cdot 43^3$.
(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 explicit... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=17$ and $q=43$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{1242528}$.\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 1242528.",
"robustness_analysis": "Robustness note: Both meth... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019381 | Number Theory: Euler Totient (Variant C) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=184=2^3\cdot 23^1$.
(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 explicit ... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=2$ and $q=23$ 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: $\\boxed{88}$.\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 88.",
"robustness_analysis": "Generality note: Both methods rely on... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{88}$.) |
math-019382 | Number Theory: Euler Totient (Variant A) | 10 | Give reasoning, not just computation: Compute Euler's totient function $\varphi(n)$.
Here $n=126495637=37^1\cdot 43^4$.
(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 $\varphi(n)$ 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 $p=37$ 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: $\\boxed{120214584}$.\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 120214584.",
"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... | 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-019383 | Number Theory: Euler Totient (Core) | 10 | Indicate where a theorem is used: Compute Euler's totient function $\varphi(n)$.
Here $n=134036773=13^5\cdot 19^2$.
(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.
... | [
{
"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 $q=19$ 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: $\\boxed{117214344}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=19$ yields the same integer 117214344.",
"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... | 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 $\boxed{117214344}$.) |
math-019384 | Number Theory: Euler Totient (Core) | 10 | Give a fully justified solution: Compute Euler's totient function $\varphi(n)$.
Here $n=3751=31^1\cdot 11^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=31$ and $q=11$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{3300}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=31,q=11$ yields the same integer 3300.",
"robustness_analysis": "Generality note: Both methods re... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | 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-019385 | Number Theory: Euler Totient (Variant A) | 10 | Solve with verification: Compute Euler's totient function $\varphi(n)$.
Here $n=68=17^1\cdot 2^2$.
(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 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 $p=17$ and $q=2$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{32}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=17,q=2$ yields the same integer 32.",
"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-019386 | Number Theory: Euler Totient (Variant A) | 10 | Give an answer and a quick verification: Compute Euler's totient function $\varphi(n)$.
Here $n=206806264579=19^5\cdot 17^4$.
(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 $\varphi(n)$... | [
{
"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 $q=17$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{184396917024}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=19,q=17$ yields the same integer 184396917024.",
"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-019387 | Number Theory: Euler Totient (Core) | 10 | Make each step logically reversible (or explain if not): Compute Euler's totient function $\varphi(n)$.
Here $n=783=29^1\cdot 3^3$.
(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 $\varp... | [
{
"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, $\\gcd(p^1,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{504}$.\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 504.",
"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-019388 | Number Theory: Euler Totient (Variant C) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=847=7^1\cdot 11^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explic... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=7$ and $q=11$ are distinct primes, $\\gcd(p^1,q^2)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{660}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=7,q=11$ yields the same integer 660.",
"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-019389 | Number Theory: Euler Totient (Variant A) | 10 | Carefully track domains: Compute Euler's totient function $\varphi(n)$.
Here $n=6996025=23^4\cdot 5^2$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $q=5$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$."... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{5353480}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=5$ yields the same integer 5353480.",
"robustness_analysis": "Robustness note: Both metho... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{5353480}$.) |
math-019390 | Number Theory: Euler Totient (Variant B) | 10 | Challenge: Compute Euler's totient function $\varphi(n)$.
Here $n=30613=23^1\cdot 11^3$.
(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 explicit about why multi... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=23$ and $q=11$.",
"Step 2: Apply the formula: $\\varphi(n)=n(1-\\frac1p)(1-\\frac1q)$.... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{26620}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=23,q=11$ yields the same integer 26620.",
"robustness_analysis": "Sensitivity analysis: Both methods r... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Key idea: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{26620}$.) |
math-019391 | Number Theory: Euler Totient (Variant B) | 10 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=985527=3^4\cdot 23^3$.
(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 explicit abou... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=3$ and $q=23$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{628452}$.\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 628452.",
"robustness_analysis": "Generality note: Both methods rely ... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Takeaway: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. (Here the result is $\boxed{628452}$.) |
math-019392 | Number Theory: Euler Totient (Core) | 10 | Explain each transformation: Compute Euler's totient function $\varphi(n)$.
Here $n=23784977251=37^5\cdot 7^3$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=37$ and $q=7$ are distinct primes, $\\gcd(p^5,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{19836120024}$.\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 19836120024.",
"robustness_analysis": "Ge... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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 $\boxed{19836120024}$.) |
math-019393 | Number Theory: Euler Totient (Variant A) | 10 | Be explicit about assumptions: Compute Euler's totient function $\varphi(n)$.
Here $n=24167=13^3\cdot 11^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be expl... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=13$ and $q=11$.",
"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: $\\boxed{20280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=13,q=11$ yields the same integer 20280.",
"robustness_analysis": "Robu... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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-019394 | Number Theory: Euler Totient (Variant C) | 10 | Answer using clear logical steps: Compute Euler's totient function $\varphi(n)$.
Here $n=344605=41^3\cdot 5^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be e... | [
{
"method_name": "Prime-Power + Multiplicativity",
"approach": "Use $\\varphi(p^e)=p^e-p^{e-1}$ and multiplicativity for coprime factors.",
"steps": [
"Step 1: Since $p=41$ and $q=5$ 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: $\\boxed{268960}$.\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 268960.",
"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... | 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 $\boxed{268960}$.) |
math-019395 | Number Theory: Euler Totient (Variant C) | 10 | Solve and include a self-check: Compute Euler's totient function $\varphi(n)$.
Here $n=27300339319=29^5\cdot 11^3$.
(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 $p=29$ 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": "Both approaches agree after simplification. Final answer: $\\boxed{23962680280}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=29,q=11$ yields the same integer 23962680280.",
"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... | 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 $\boxed{23962680280}$.) |
math-019396 | Number Theory: Euler Totient (Core) | 10 | Find the exact value: Compute Euler's totient function $\varphi(n)$.
Here $n=992=2^5\cdot 31^1$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be explicit about w... | [
{
"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 $q=31$ are distinct primes, $\\gcd(p^5,q^1)=1$.",
"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: $\\boxed{480}$.\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 480.",
"robustness_analysis": "If the pr... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, 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 $\boxed{480}$.) |
math-019397 | Number Theory: Euler Totient (Variant B) | 10 | Write the solution set clearly: Compute Euler's totient function $\varphi(n)$.
Here $n=30074733=13^5\cdot 3^4$.
(a) Use multiplicativity and the prime-power formula.
(b) Give an independent check using the product formula $\varphi(n)=n\prod_{r\mid n}(1-1/r)$.
(c) Explain in one sentence what $\varphi(n)$ counts.
Be 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 $p=13$ and $q=3$ are distinct primes, $\\gcd(p^5,q^4)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2})... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{18507528}$.\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 18507528.",
"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 $\boxed{18507528}$.) |
math-019398 | Number Theory: Euler Totient (Variant C) | 10 | Start by stating any domain restrictions: Compute Euler's totient function $\varphi(n)$.
Here $n=1998607065841=29^4\cdot 41^4$.
(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... | [
{
"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": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{1882623981280}$.\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 1882623981280.",
"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... | 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-019399 | Number Theory: Euler Totient (Core) | 10 | Give a theorem-based solution: Compute Euler's totient function $\varphi(n)$.
Here $n=5203=11^2\cdot 43^1$.
(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 expli... | [
{
"method_name": "Totient Product Formula",
"approach": "Use $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$ over distinct prime divisors $r$ of $n$.",
"steps": [
"Step 1: The distinct prime divisors of $n$ are $p=11$ and $q=43$.",
"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: $\\boxed{4620}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=11,q=43$ yields the same integer 4620.",
"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 $n(1-1/p)(1-1/q)$. |
math-019400 | Number Theory: Euler Totient (Core) | 10 | Where appropriate, name the theorem you use: Compute Euler's totient function $\varphi(n)$.
Here $n=41596551767=43^4\cdot 23^3$.
(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(... | [
{
"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 $q=23$ are distinct primes, $\\gcd(p^4,q^3)=1$.",
"Step 2: Therefore $\\varphi(n)=\\varphi(p^{e_1})\\varphi(q^{e_2}... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{38862703572}$.\nBoth computations are equivalent forms of the same theorem: $\\varphi(n)=n\\prod_{r\\mid n}(1-1/r)$. Plugging $p=43,q=23$ yields the same integer 38862703572.",
"robustness_analysis": "Robustness note: Both me... | [
{
"error_description": "Used $\\varphi(p^e)=p^e-1$ for $e>1$.",
"why_plausible": "It is true for $e=1$ that $\\varphi(p)=p-1$, so it feels like it should generalize.",
"why_wrong": "For $p^e$, exactly the multiples of $p$ are not coprime, and there are $p^{e-1}$ of them, giving $p^e-p^{e-1}$.",
"whi... | Remember: $\varphi(n)$ counts integers $1\le k\le n$ coprime to $n$. For $n=p^{e_1}q^{e_2}$ it equals $n(1-1/p)(1-1/q)$. |
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