2020 AIME I Problems/Problem 12

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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}) $and since$ 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$ (Error compiling LaTeX. Unknown error_msg)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}$.

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2020 AIME I Problems/Problem 12

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