Question

若式子(n^2+100)/(n+10)的值为正整数,则正整数n可取的最大值

Answer 1

(n^2+100)/(n+10)

=(n^2+20n+100-20n-200+200)/(n+10)

=[(n+10)^2-20*(n+10)+200]/(n+10)

=n+10-20+200/(n+10)

=n-10+200/(n+10)

要使n-10+200/(n+10)是正整数,那么200/(n+10)是正整数,n最大为190