Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 4"
(→Solution) |
|||
Line 3: | Line 3: | ||
== Solution == | == 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. | ||
− | <math> | + | 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=\boxed149</math> positive integer factors that are not perfect squares. | ||
+ | |||
+ | ~Ultraman | ||
== See also == | == See also == |
Revision as of 14:02, 2 January 2020
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
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 |