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

Revision as of 07:18, 13 August 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 $3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}$ .

Video Solutions

https://youtu.be/ZiZVIMmo260

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.

@@ Line 12: / Line 12: @@
 </math>
-==Solution 1==
+==Solution 1 (Algebra)==
-The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum integer value for <math>n</math> is <math>3</math>, and the minimum integer value for <math>n</math> is <math>-5</math>. Then we must find the range from <math>-5</math> to <math>3</math>, which is <math>3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}</math>.
-==Solution 2==
 Note that <cmath>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}.</cmath>
 Since this expression is an integer, we need
@@ Line 22: / Line 19: @@
    <li><math>3-n\geq0,</math> from which <math>n\leq3.</math></li><p>
 </ol>
-Taking the intersection gives <math>-5\leq n\leq3.</math> So, there are <math>\boxed{\textbf{(E) }9}</math> integer values of <math>n.</math>
+Taking the intersection gives <math>-5\leq n\leq3.</math> So, there are <math>3-(-5)+1=\boxed{\textbf{(E) }9}</math> integer values of <math>n.</math>
 ~MRENTHUSIASM
+==Solution 2 (Arithmetic)==
+The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum integer value for <math>n</math> is <math>3</math>, and the minimum integer value for <math>n</math> is <math>-5</math>. Then we must find the range from <math>-5</math> to <math>3</math>, which is <math>3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}</math>.
 ==Video Solutions==

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