Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 4"
(→Solution) |
(→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. | + | 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. |
| − | |||
| − | 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=\ | ||
~Ultraman | ~Ultraman | ||
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 | ||
