20. In a school there are $n$ students, each with a different student number. Each student number is a positive factor of $60^{60}$, and the H.C.F. two student numbers is not a student number in the school. Find the greatest possible value of $n$.
(2 marks)
某學校有 $n$ 名學生, 每人均有一個不同的編號。每名學生的編號均是 $60^{60}$ 的一個正因數，而且任意兩名學生的編號的最大公因數均不是該校的學生編號。求 $n$ 的最大可能值。
（２分）
20. 3721
20. Firstly, we show that $n$ can be 3721 . Since $60^{60}=2^{120} \times 3^{60} \times 5^{60}$, the factors of $60^{60}$ are in the form $2^{a} \times 3^{b} \times 5^{c}$ where $0 \leq a \leq 120$ and $0 \leq b, c \leq 60$. Suppose all the student numbers take this form and satisfy the condition $a+b+c=120$. Then each student number is uniquely
determined by $b$ and $c$, so that there are altogether $61^{2}=3721$ students. To show that the condition of the question is satisfied in this case, it suffices to show that no number of this form is a proper factor of the other (since the H.C.F. of two distinct numbers must be a proper factor of at least one of these numbers). Indeed, if $2^{a} \times 3^{b} \times 5^{c}$ is a proper factor of $2^{x} \times 3^{y} \times 5^{z}$, then $a \leq x, b \leq y$ and $c \leq z$, with at least one strict inequality. But this is a contradiction, since $120=a+b+c<x+y+z=120$. Hence 3721 is a possible value of $n$.
Now suppose there are at least 3722 student numbers. Since there are only $61^{2}=3721$ choices for the exponents $b$ and $c$, there are two student numbers having the same pairs of exponents $(b, c)$ by the pigeonhole principle. In that case, the smaller one is a proper factor of the other and hence is the H.C.F. of these two numbers. It follows that the answer is 3721 .