MathSolver94 wrote: A two-digit integer is divided by the sum of its digits. Find the largest remainder that can occur. Let $\overline {ab}$ be the number. The remainder is $<a+b\le 18$ So max remainder possible is $17$ and then $a+b=18$ and $\overline {ab}=99$ and remainder is in fact $9$ Next max remainder possible is $16$ and then $a+b=17,18$ $a+b=18$ $\implies$ $\overline {ab}=99$ and remainder is in fact $9$ $a+b=17$ $\implies$ $\overline {ab}=98$ or $89$ and remainder is in fact $13$ or $4$ Next max remainder possible is $15$ and then $a+b=16,17,18$ $a+b=17,18$ have been seen above and remainder is then $13,9$ or $4$ $a+b=16$ gives three cases : $\overline {ab}=97$ and remainder is then $1$ $\overline {ab}=88$ and remainder is then $8$ $\overline {ab}=79$ and remainder is then $15$ Hence the answer : $\boxed{15}$