Difference between revisions of "2018 AMC 10A Problems/Problem 7"

Revision as of 01:41, 30 October 2021

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:

$5+n\geq0,$ from which $n\geq-5.$

$3-n\geq0,$ 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 $3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}.$

Solution 3 (Brute Force)

The possible 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 $\boxed{\textbf{(E) }9}$ possible values for $n.$

~Little

Video Solutions

https://youtu.be/ZiZVIMmo260

~IceMatrix

https://youtu.be/2vz_CnxsGMA

~savannahsolver

https://youtu.be/vzyRAnpnJes

Education, the Study of Everything

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

~ pi_is_3.14

2018 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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