# 2000 AMC 8 Problems/Problem 11

## Problem

The number $64$ has the property that it is divisible by its unit digit. How many whole numbers between 10 and 50 have this property? $\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

## Solution

Casework by the units digit $u$ will help organize the answer. $u=0$ gives no solutions, since no real numbers are divisible by $0$ $u=1$ has $4$ solutions, since all numbers are divisible by $1$. $u=2$ has $4$ solutions, since every number ending in $2$ is even (ie divisible by $2$). $u=3$ has $1$ solution: $33$. $\pm 10$ or $\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$. $\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$, but those numbers are $3$ and $63$, which are out of the range of the problem. $u=4$ has $2$ solutions: $24$ and $44$. Adding or subtracting $10$ will kill divisibility by $4$, since $10$ is not divisible by $4$. $u=5$ has $4$ solutions: every number ending in $5$ is divisible by $5$. $u=6$ has $1$ solution: $36$. $\pm 10$ or $\pm 20$ will kill divisibility by $3$, and thus kill divisibility by $6$. $u=7$ has no solutions. The first multiples of $7$ that end in $7$ are $7$ and $77$, but both are outside of the range of this problem. $u=8$ has $1$ solution: $48$. $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$. $u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$.

Totalling the solutions, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions, giving the answer $\boxed{C}$, which is 17.

## Video Solution

https://youtu.be/m5D5-2YB0tI Soo, DRMS, NM

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