Difference between revisions of "2013 AMC 12B Problems/Problem 9"
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==Solution== | ==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 | + | 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== | == Video Solution by OmegaLearn== |
Latest revision as of 20:25, 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 ?
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 none 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
. Therefore, the prime factorization of the square is
. To find the square root of this, we halve the exponents, leaving
. The sum of the exponents is
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=2694
~ pi_is_3.14
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.