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

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== Solution ==
 
== Solution ==
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[url=https://brilliant.org/wiki/lifting-the-exponent/]Lifting the Exponent[/url] shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
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[url=https://brilliant.org/wiki/lifting-the-exponent]Lifting the Exponent[/url] shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>.  
  
  

Revision as of 16:16, 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

[url=https://brilliant.org/wiki/lifting-the-exponent]Lifting the Exponent[/url] 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 $145^{4} \equiv 1 \pmod{25}$ 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 (Problems • Answer Key • Resources)
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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