2009 AMC 12A Problems/Problem 12

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Problem

How many positive integers less than $1000$ are $6$ times the sum of their digits?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 12$

Solution

The sum of the digits is at most $9+9+9=27$. Therefore the number is at most $6\cdot 27 = 162$. Out of the numbers $1$ to $162$ the one with the largest sum of digits is $159$, and the sum is $1+5+9=15$. Hence the sum of digits will be at most $15$.

Also, each number with this property is divisible by $6$, therefore it is divisible by $3$, and thus also its sum of digits is divisible by $3$.

We only have five possibilities left for the sum of the digits: $3$, $6$, $9$, $12$, and $15$. These lead to the integers $18$, $36$, $54$, $72$, and $90$. But for $18$ the sum of digits is $1+8=9$, which is not $3$, therefore $18$ is not a solution. Similarly we can throw away $36$, $72$, and $90$, and we are left with just $\boxed{1}$ solution: the number $54$.

See Also

2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 AMC 12 Problems and Solutions