6. If $m$ is a positive integer not exceeding 1000 and the fraction $\frac{m+4}{m^{2}+7}$ is in its lowest term, how many possible values of $m$ are there?
(1 mark)
若 $m$ 是不超過 1000 的正整數, 且 $\frac{m+4}{m^{2}+7}$ 是最簡分數, 問 $m$ 有多少個可能值？
6. 957
6. Note that $\frac{m+4}{m^{2}+7}$ is in lowest term if and only of $\frac{m^{2}+7}{m+4}$ is in lowest term. Since
$$
\frac{m^{2}+7}{m+4}=\frac{m^{2}-16}{m+4}+\frac{23}{m+4}=m-4+\frac{23}{m+4},
$$
the fraction is in lowest term except when $m+4$ is a multiple of 23 . Since $m$ may be equal to $1,2, \ldots, 1000$, we shall count the number of multiples of 23 from 5 to 1004 . The first one is $23=23 \times 1$ and the last one is $989=23 \times 43$. Hence there are 43 such multiples. It follows that the answer is $1000-43=957$.