Problem 148

We note that the only thing that affects the parity of the factor are the powers of 2. There are $10+5+2+1 = 18$ factors of 2 in the number. Thus, there are $18$ cases in which a factor of $21!$ would be even (have a factor of $2$ in its prime factorization), and $1$ case in which a factor of $21!$ would be odd. Therefore, the answer is $\boxed{\textbf{(B)} \frac 1{19}}$.
Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21! into 2 groups, one with powers of 2 and the others odd factors. For each power of 2, the factors combine a certain number of 2's from the first group and numbers from the odd group. That is why symmetry occurs here.

We can consider a factor of $21!$ to be odd if it does not contain a $2$; hence, finding the exponent of $2$ in the prime factorization of $21!$ will help us find our answer. We can start off with all multiples of $2$ up to $21$, which is $10$. Then, we find multiples of $4$, which is $5$. Next, we look at multiples of $8$, of which there are $2$. Finally, we know there is only one multiple of $16$ in the set of positive integers up to $21$. Now, we can add all of these to get $10+5+2+1=18$. We know that, in the prime factorization of $21!$, we have $2^{18}$, and the only way to have an odd number is if there is not a $2$ in that number's prime factorization. This only happens with $2^{0}$, which is only one of the 19 different exponents of 2 we could have (of which having each exponent is equally likely). Hence, we have $\boxed{\text{(B)} \dfrac{1}{19}}.$
Solution by: armang32324