Difference between revisions of "2013 AMC 12B Problems/Problem 9"
Sugar rush (talk | contribs) |
Megaboy6679 (talk | contribs) m (→Solution) |
||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides <math>12!</math> ? | + | What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides <math>12!</math>? |
<math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12 </math> | <math>\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12 </math> | ||
| Line 6: | Line 6: | ||
==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 | + | 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== | ||
| + | https://youtu.be/ZhAZ1oPe5Ds?t=2694 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/a-3CAo4CoWc | https://youtu.be/a-3CAo4CoWc | ||
| − | ~ | + | ~someone |
== See also == | == See also == | ||
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 
?
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. 
