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Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 4"

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== Solution 2 ==
 
== Solution 2 ==
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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</math> 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</math>, 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.
+
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</math> 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</math>, 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</math> to get <math>149</math>. So there are <math>\boxed{149}</math> positive integer factors that are not perfect squares.
  
 
~Aarushgoradia18
 
~Aarushgoradia18

Revision as of 18:55, 21 December 2021

Problem

How many positive integer factors of $36,000,000$ are not perfect squares?

Solution

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

~Ultraman

Solution 2

There is a similar way to the previous solution. The prime factorization of $36,000,000$ is $2^8\cdot3^2\cdot5^6$. We need to find the amount of factors of that, so add $1$ to each exponent and multiply to get $(8+1)(2+1)(6+1) = 189$ 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 $0, 2, 4, 6$, 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 $8/2 + 1$ 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 $189$ to get $149$. So there are $\boxed{149}$ 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
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