15. Let $L$ denote the L.C.M. of $1,2, \ldots, 20$. How many positive factors of $L$ are divisible by exactly 18 of the 20 numbers $1,2, \ldots, 20$ ?
(2 marks)
設 $L$ 爲 $1 、 2 、 \cdots 、 20$ 的最小公倍數。那麼 $L$ 有多少個正因數可被 1 、 2、‥、20 這 20 個數當中的剛好 18 個整除？
15. 13
15. Note that $L=2^{4} \times 3^{2} \times 5 \times 7 \times 11 \times 13 \times 17 \times 19$. Suppose $L \div n$ is a factor of $L$ and is divisible by exactly 18 of the numbers $1,2, \ldots, 20$. We say that a number $m(1 \leq m \leq 20)$ is 'killed' if $L \div n$ is not divisible by $m$. Hence we are looking for choices of $n$ which 'kill' exactly two of 1 , $2, \ldots, 20$.
Note that if $n$ is even, then 16 is 'killed'; if $n$ is divisible by 4 , then 8 and 16 are 'killed'; if $n$ is divisible by 8 , then more than two numbers are killed. Likewise, if $n$ is divisible by 3 , then 9 and 18 are 'killed'; if $n$ is divisible by 9 , then more than two numbers are 'killed'. On the other hand, the inclusion of the prime factor 5 in $n$ 'kills' four numbers $(5,10,15,20)$, the inclusion of 7 'kills' two (7 and 14), while the inclusion of $11,13,17$ or 19 'kills' one.
To 'kill' exactly two numbers, the possible values of $n$ are therefore $4,3,7$ or the product of exactly two of $2,11,13,17$ and 19. It follows that the answer is $3+C_{2}^{5}=13$.