Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == Solution == | ||
− | We first factorize the product as <math>2^7\cdot3^2\cdot5\cdot7</math>. | + | We first factorize the product as <math>2^7\cdot3^2\cdot5\cdot7</math>. Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are four even powers of two that are less than or equal to <math>2^7</math>, namely <math>2^0</math>, <math>2^2</math>, <math>2^4</math>, <math>2^6</math>, as <math>0</math> is even. Repeating this process with three, five, and seven, we find that three has two even power, <math>3^0</math>, and <math>3^2</math>, and that five and seven have only one even power, <math>5^0</math> and <math>7^0</math> respectively. Multiplying this we have <math>4\cdot2\cdot1\cdot1 = 8</math>. Therefore, our answer is <math>8</math>. |
== See also == | == See also == |
Revision as of 22:54, 25 November 2016
Problem
How many perfect squares are divisors of the product ? (Here, for example, means )
Solution
We first factorize the product as . Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are four even powers of two that are less than or equal to , namely , , , , as is even. Repeating this process with three, five, and seven, we find that three has two even power, , and , and that five and seven have only one even power, and respectively. Multiplying this we have . Therefore, our answer is .
See also
2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |