, 2013

Let $n$ be the number of cookies in the cookie jar.

Let $r$ be the number of raisins in each of the $n-1$ smaller, identical cookies.

This means that there are $r+1$ raisins in the larger cookie.

If we removed one raisin from the larger cookie, it too would have $r$ raisins and so each of the $n$ cookies would have the same number of raisins ($r$), and the total number of raisins in the cookies would be $100-1=99$.

From this, we obtain $nr=99$.

(We could also obtain this equation by noting that there are $n-1$ cookies containing $r$ raisins and 1 cookie containing $r+1$ raisins and $100$ raisins in total, so $(n-1)r+(r+1) = 100$ or $nr-r+r+1=100$ or $nr=99$.)

Since $n$ and $r$ are positive integers whose product is $99$, then the possibilities are: $$99 = 99\times 1 = 33\times 3 = 11\times 9 = 9 \times 11 = 3 \times 33 = 1 \times 99$$ Since $n$ is between 5 and 10, then we must have $99 = 9\times 11$; that is, $n=9$ and $r=11$.

Since there are 11 raisins in each of the smaller cookies, then there are $11+1=12$ raisins in the larger cookie.