Difference between revisions of "2013 AMC 12B Problems/Problem 9"

Revision as of 20:20, 27 January 2019

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$ ?

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AMC 12 Problems and Solutions

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 ==Solution==
-Looking at the prime numbers under 12, we see that there are <math>\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10</math> factors of 2, <math>\lfloor\frac{12}{3}\rfloor+\lfloor\frac{12}{3^2}\rfloor=4+1=5</math> factors of 3, and <math>\lfloor\frac{12}{5}\rfloor=2</math> factors of 5. All greater primes are represented once or not at all in <math>12!</math>, so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use <math>4</math> of the <math>5</math> factors of <math>3</math>. The prime factorization of the square is therefore <math>2^{10}*3^4*5^2</math>. To find the square root of this, we halve the exponents, leaving <math>2^5*3^2*5</math>. The sum of the exponents is <math>\boxed{\textbf{(C) }8}</math>
 == See also ==