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

Revision as of 20:48, 17 December 2020

Problem

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

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

Solution

Looking at the prime numbers under 12, we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of 2, $\lfloor\frac{12}{3}\rfloor+\lfloor\frac{12}{3^2}\rfloor=4+1=5$ factors of 3, and $\lfloor\frac{12}{5}\rfloor=2$ factors of 5. All greater primes are represented once or not at all in $12!$ , 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 $4$ of the $5$ factors of $3$ . The prime factorization of the square is therefore $2^{10}*3^4*5^2$ . To find the square root of this, we halve the exponents, leaving $2^5*3^2*5$ . The sum of the exponents is $\boxed{\textbf{(C) }8}$

Video Solution

https://youtu.be/a-3CAo4CoWc

~someone

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

Revision as of 20:48, 17 December 2020

Contents

Problem

Solution

Video Solution

See also

@@ Line 11: / Line 11: @@
 https://youtu.be/a-3CAo4CoWc
-~no one
+~someone
 == See also ==