2018 AMC 10A Problems/Problem 7

The following problem is from both the 2018 AMC 10A #7 and 2018 AMC 12A #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 (Observations)

Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$. We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$. For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$. Also keep in mind that the expression will be an integer for $n=0$, which gives us a total of $\boxed{\textbf{(E) }9}$ for $n.$

Solution 3 (Brute Force)

The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$

The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.

In total, there are $\boxed{\textbf{(E) }9}$ values for $n.$

~Little ~MRENTHUSIASM

Video Solution

https://youtu.be/ZiZVIMmo260

~IceMatrix

Video Solution

https://youtu.be/2vz_CnxsGMA

~savannahsolver

Video Solution

https://youtu.be/vzyRAnpnJes

Education, the Study of Everything

Video Solution

https://youtu.be/ZhAZ1oPe5Ds?t=1763

~ pi_is_3.14

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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