+59 1. For how many positive integers n does a regular polygon with n sides have interior angles with measures equal to an integer number of degrees?
2. How many perfect squares divide 60^4?
+177 First, perform the factorization procedure on 60. I am not sure how to show this properly, so I will do my best. I use the red color to distinguish a prime number factor.
\(60 = \textcolor{red} 2 * 30 \\ 30 = \textcolor{red} 2 * 15 \\ 15 = \textcolor{red} 3 * \textcolor{red} 5\)
Now, that we have found the prime factorization, we know that \(60 = 2^2 \times 3 \times 5\). However, we need the factorization of \(60^4\). We can accomplish this by doing the following:
\(60 = 2^2 \times 3 \times 5 \\ 60^4 = \left(2^2 \times 3 \times 5\right)^4 \\ 60^4 = 2^8 \times 3^4 \times 5^4\)
Now, we have the prime factorization of 60^4. In order for a number to be a perfect square factor, the factor would need an even-numbered power of all the factors 2, 3, and 5.
For the factor 2, the even-numbered powers possible are 2, 4, 6, and 8 for a total of 4 choices.
For the factor 3, the even-numbered powers possible are 2 and 4 for a total of 2 choices.
For the factor 5, the even-numbered powers possible are 2 and 4 for a total of 2 choices.
These choices are completely independent choices, so we can use the multiplication rule to find the number of perfect square factors of \(60^4 \). The number of perfect square factors is \(4 * 2 * 2 = 16\) perfect square factors.
+177 First, perform the factorization procedure on 60. I am not sure how to show this properly, so I will do my best. I use the red color to distinguish a prime number factor.
\(60 = \textcolor{red} 2 * 30 \\ 30 = \textcolor{red} 2 * 15 \\ 15 = \textcolor{red} 3 * \textcolor{red} 5\)
Now, that we have found the prime factorization, we know that \(60 = 2^2 \times 3 \times 5\). However, we need the factorization of \(60^4\). We can accomplish this by doing the following:
\(60 = 2^2 \times 3 \times 5 \\ 60^4 = \left(2^2 \times 3 \times 5\right)^4 \\ 60^4 = 2^8 \times 3^4 \times 5^4\)
Now, we have the prime factorization of 60^4. In order for a number to be a perfect square factor, the factor would need an even-numbered power of all the factors 2, 3, and 5.
For the factor 2, the even-numbered powers possible are 2, 4, 6, and 8 for a total of 4 choices.
For the factor 3, the even-numbered powers possible are 2 and 4 for a total of 2 choices.
For the factor 5, the even-numbered powers possible are 2 and 4 for a total of 2 choices.
These choices are completely independent choices, so we can use the multiplication rule to find the number of perfect square factors of \(60^4 \). The number of perfect square factors is \(4 * 2 * 2 = 16\) perfect square factors.
+177 I have to confess that I committed a serious error in my calculations because I forgot to consider that a valid even-numbered exponent is 0. This means that there is one additional choice for each prime factor 2, 3, and 5. \(5 * 3 * 3 = 45 \) perfect square factors! I apologize in advance for this mistake!
