Difference between revisions of "2020 AIME I Problems/Problem 12"

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Now, multiplying <math>n</math> by <math>4</math>, we see <cmath>v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})</cmath> and since <math>145^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4n}-2^{4n})=1</math> meaning that we have that by LTE, <math>4 \cdot 5^4</math> divides <math>n</math>.  
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Now, multiplying <math>n</math> by <math>4</math>, we see <cmath>v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})</cmath> and since <math>149^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4n}-2^{4n})=1</math> meaning that we have that by LTE, <math>4 \cdot 5^4</math> divides <math>n</math>.  
  
 
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.
 
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.

Revision as of 16:32, 12 March 2020

Problem

Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$

Solution

Lifting the Exponent shows that \[v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$. It also shows that \[v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$.


Now, multiplying $n$ by $4$, we see \[v_5(149^{4n}-2^{4n}) = v_5(149^{4n}-16^{n})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4n}-2^{4n})=1$ meaning that we have that by LTE, $4 \cdot 5^4$ divides $n$.

Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.

~kevinmathz

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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