Difference between revisions of "2013 AMC 12B Problems/Problem 9"
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==Solution== | ==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> | + | 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>. 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> |
==Video Solution== | ==Video Solution== | ||
Revision as of 16:55, 21 December 2020
Contents
Problem
What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides 
?
Solution
Looking at the prime numbers under 
, we see that there are 
factors of 
, 
factors of 
, and 
factors of 
. All greater primes are represented once or not at all in , 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 
of the factors of . The prime factorization of the square is therefore 
. To find the square root of this, we halve the exponents, leaving 
. The sum of the exponents is 
Video Solution
~someone
See also
| 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. 
