2020 AIME I Problems/Problem 12

Revision as of 16:14, 12 March 2020 by Kevinmathz (talk | contribs) (Solution)

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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. ! Missing $ inserted.)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$(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.

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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