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

Latest revision as of 19:25, 7 August 2023

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 $2$ , $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$ , and $\left\lfloor\frac{12}{5}\right\rfloor=2$ factors of $5$ . All greater primes are represented once or none 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 $3$ . Therefore, the prime factorization of the square is $2^{10}\cdot3^4\cdot5^2$ . To find the square root of this, we halve the exponents, leaving $2^5\cdot3^2\cdot5$ . The sum of the exponents is $\boxed{\textbf{(C) }8}$

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=2694

~ pi_is_3.14

Video Solution

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

~someone

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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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 ==Solution==
-Looking at the prime numbers under <math>12</math>, 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 <math>2</math>, <math>\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5</math> factors of <math>3</math>, and <math>\left\lfloor\frac{12}{5}\right\rfloor=2</math> factors of <math>5</math>. 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>. Therefore, the prime factorization of the square is <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>
+Looking at the prime numbers under <math>12</math>, 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 <math>2</math>, <math>\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5</math> factors of <math>3</math>, and <math>\left\lfloor\frac{12}{5}\right\rfloor=2</math> factors of <math>5</math>. All greater primes are represented once or none 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>. Therefore, the prime factorization of the square is <math>2^{10}\cdot3^4\cdot5^2</math>. To find the square root of this, we halve the exponents, leaving <math>2^5\cdot3^2\cdot5</math>. The sum of the exponents is <math>\boxed{\textbf{(C) }8}</math>
 == Video Solution by OmegaLearn==

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