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The 5774 Ulpaniada (part 2 of it, which was taken yesterday) includes the following question:

How many three digit numbers are there which are equal to $34$ times the sum of their digits?

It offers five choices: $10$, $8$, $6$, $4$, and $0$.

I don't want to check the sums of digits of $3\times34,4\times34,\ldots,29\times34$ (all the three-digit products). There must be a better method: any suggestions?

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If the product is written $abc$ in decimal, then we have$$100a+10b+c=34(a+b+c)$$($a,b,c<10,a\ne0$). Thus$$66a-33c=24b$$and, since $11$ divides the left-hand side, $11|b$. Since $b<10$, $b=0$. We're left with two requirements: $2a=c$ and $34|a0c_{10}$. Testing the possibilities afforded by $2a=c$, we see $102$ is divisible by $34$, so all its multiples are answers also, and the answer is $4$.

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