Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 4"
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== Solution 2 == | == 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 < | + | 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</math> to get 2, and <math>7/2</math> and round up to get <math>4</math>. 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}</math> positive integer factors that are not perfect squares. |
~Aarushgoradia18 | ~Aarushgoradia18 | ||
Revision as of 18:54, 21 December 2021
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, 
to get 2, and 
and round up to get 
. Multiply those answers to get 5^2^4 to get 40 perfect squares. Subtract it from 
149. So there are 
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 | ||
