2018 UNCO Math Contest II Problems/Problem 4
Contents
Problem
How many positive integer factors of 
are not perfect squares?
Solution
We can use complementary counting. Taking the prime factorization of , we get 
.So the total number of factors of is 
factors. Now we need to find the number of factors that are perfect squares. So back to the prime factorization, 
. Now we get 
factors that are perfect squares. So there are 
positive integer factors that are not perfect squares.
~Ultraman
Solution 2
There is a similar way to the previous solution. The prime factorization of is . We need to find the amount of factors of that, so add 
(8+1)(2+1)(6+1) = 189
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 
to get 5, $2/2 + 1 to get 2, and 7/2 and round up to get 4. Multiply those answers to get 5^2^4 to get 40 perfect squares. Subtract it from$ (Error compiling LaTeX. Unknown error_msg)189 to get 
\boxed{149}$ positive integer factors that are not perfect squares.
~Aarushgoradia18
See also
| 2018 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 | ||
