+178 We choose a positive divisor of 20^{20} at random (with all divisors equally likely to be chosen). What is the probability that we chose a multiple of 5?
+9 Firstly, let's prime factorize \({20}^{20}\):
\({20}^{20}={(4\times5)}^{20}={2}^{40}\times{5}^{20}\)
When we choose a factor of \({20}^{20}\), we are choosing a number of the form \({2}^{a}\times{5}^{b}\), where \(0 \le a \le 40\) and \(0 \le b \le 20\).
Since all divisors are equally likely to be chosen, a and b are randomized within their ranges.
If our factor is a multiple of 5, then we have \(b \ge 1\). Otherwise, \(b=0\), and our factor is not a multiple of 5.
Therefore, the probability that our factor is a multiple of 5 is the probability that \(b \ge 1\).
There are 21 numbers between 0 and 20 inclusive, and there are 20 numbers from 1 to 20 inclusive, our answer is \(\frac{20}{21}\).
