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 | + | 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> |
== See also == | == See also == |
Revision as of 22:37, 22 November 2013
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 12, we see that there are factors of 2, factors of 3, and factors of 5. 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
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 |
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