View source for 2018 UNCO Math Contest II Problems/Problem 4

== Problem ==
How many positive integer factors of <math>36,000,000</math> are not perfect squares?

== Solution ==
We can use [[complementary counting]]. Taking the prime factorization of <math>36,000,000</math>, we get <math>2^8\cdot3^2\cdot5^6</math>.So the total number of factors of <math>36,000,000</math> is <math>(8+1)(2+1)(6+1) = 189</math> factors. Now we need to find the number of factors that are perfect squares. So back to the prime factorization, <math>2^8\cdot3^2\cdot5^6 = 4^4\cdot9^1\cdot5^3</math>. Now we get <math>(4+1)(1+1)(3+1)=40</math> factors that are perfect squares. So there are <math>189-40=\boxed{149}</math> positive integer factors that are not perfect squares.

~Ultraman

== Solution 2 ==
There is a similar way to the previous solution. The prime factorization of <math>36,000,000</math> is <math>2^8\cdot3^2\cdot5^6</math>. We need to find the amount of factors of that, so add <math>1 to each exponent and multiply to get </math>(8+1)(2+1)(6+1) = 189<math> factors. Now we need to find the number of factors that are perfect squares. Perfect squares are numbers in the prime factorization with exponents of </math>0, 2, 4, 6, etc. You find the max amount of the exponent that is less than the exponent in the prime factorization. There is a trick to that. You take the exponent in the prime factorization, for example 4. You divide by 2 and add 1 to the result to find the perfect squares in the number and exponent in the prime factorization. You also round up and do not add 1 if your answer is a decimal. You do <math>8/2 + 1</math> to get 5, <math>2/2 + 1 to get 2, and 7/2 and round up to get 4. Multiply those answers to get 5^2^4 to get 40 perfect squares. Subtract it from </math>189 to get <math>149. So there are </math>\boxed{149}$ positive integer factors that are not perfect squares.

~Aarushgoradia18

== See also ==
{{UNCO Math Contest box|year=2018|n=II|num-b=3|num-a=5}}

[[Category:Intermediate Number Theory Problems]]