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Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 4"

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== Solution ==   
 
== Solution ==   
  
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We first factorize the product as <math>2^7\cdot3^2\cdot5\cdot7</math>. Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are four even powers of two that are less than or equal to <math>2^7</math>, namely <math>2^0</math>, <math>2^2</math>, <math>2^4</math>, <math>2^6</math>, as <math>0</math> is even. Repeating this process with three, five, and seven, we find that three has two even powers, <math>3^0</math>, and <math>3^2</math>, and that five and seven have only one even power, <math>5^0</math> and <math>7^0</math> respectively. Multiplying this we have <math>4\cdot2\cdot1\cdot1 = 8</math>. Therefore, our answer is <math>\boxed8</math>.
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We first factorize the product as <math>2^{7}\cdot3^{6}\cdot4^{5}\cdot5^{4}\cdot6^{3}\cdot7^{2}\cdot8 = 2^{23}\cdot3^{9}\cdot5^{4}\cdot7^{2}</math>. Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are twelve even powers of two that are less than or equal to <math>2^{23}</math>, namely <math>2^0</math>, <math>2^2</math>, <math>2^4</math>, <math>2^6</math>,<math>2^8</math>,<math>2^{10}</math>,<math>2^{12}</math>,<math>2^{14}</math>,<math>2^{16}</math>,<math>2^{18}</math>,<math>2^{20}</math>,<math>2^{22}</math>, as <math>0</math> is even. Repeating this process with three, five, and seven, we find that three has five even powers, <math>3^0</math>,<math>3^2</math>, <math>3^4</math>,<math>3^6</math> and <math>3^8</math>, and that five has 3 and and seven has two even powers. Multiplying this we have <math>12\cdot{5}\cdot{3}\cdot{2} = 360</math>. Therefore, our answer is <math>\boxed{360}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 00:50, 13 January 2019

Problem

How many perfect squares are divisors of the product $1!\cdot 2!\cdot 3!\cdot 4!\cdot 5!\cdot 6!\cdot 7!\cdot 8!$ ? (Here, for example, $4!$ means $4\cdot 3\cdot 2\cdot 1$)

Solution

We first factorize the product as $2^{7}\cdot3^{6}\cdot4^{5}\cdot5^{4}\cdot6^{3}\cdot7^{2}\cdot8 = 2^{23}\cdot3^{9}\cdot5^{4}\cdot7^{2}$. Since we want only perfect squares, we are looking for even powers in the prime factorization of the divisors. Working with each term in the prime factorization, we find that there are twelve even powers of two that are less than or equal to $2^{23}$, namely $2^0$, $2^2$, $2^4$, $2^6$,$2^8$,$2^{10}$,$2^{12}$,$2^{14}$,$2^{16}$,$2^{18}$,$2^{20}$,$2^{22}$, as $0$ is even. Repeating this process with three, five, and seven, we find that three has five even powers, $3^0$,$3^2$, $3^4$,$3^6$ and $3^8$, and that five has 3 and and seven has two even powers. Multiplying this we have $12\cdot{5}\cdot{3}\cdot{2} = 360$. Therefore, our answer is $\boxed{360}$.

See also

2009 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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