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.

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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