# 2018 AMC 10A Problems/Problem 7

The following problem is from both the 2018 AMC 12A #7 and 2018 AMC 10A #7, so both problems redirect to this page.

## Problem

For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer? $\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

## Solution 1 (Algebra)

Note that $$4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.$$ Since this expression is an integer, we need:

1. $5+n\geq0,$ from which $n\geq-5.$
2. $3-n\geq0,$ from which $n\leq3.$

Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\boxed{\textbf{(E) }9}$ integer values of $n.$

~MRENTHUSIASM

## Solution 2 (Arithmetic)

The prime factorization of $4000$ is $2^{5}\cdot5^{3}$. Therefore, the maximum integer value for $n$ is $3$, and the minimum integer value for $n$ is $-5$. Then we must find the range from $-5$ to $3$, which is $3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}$.

## Video Solutions

~savannahsolver

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