9.6. Find all three-digit numbers that are five times the product of their digits. Answer: 175.
Solution. Let $\overline{a b c}$ be the desired three-digit number. Then, by the condition, $100 a+10 b+c=5 a b c$. From this, we get $c=5(a b c-2 b-20 a)$, so $c$ is divisible by 5. But $c$ cannot be zero, since otherwise the product of the digits is also zero. Therefore, $c=5$. Thus, we have $100 a+10 b+5=25 a b \Leftrightarrow 20 a+2 b+1=5 a b \Leftrightarrow 2(10 a+b)=5 a b-1$. The number $5 a b-1$ gives a remainder of 4 when divided by 5, so the number $10 a+b$ gives a remainder of 2 when divided by 5. This is only possible if $b=2$ or $b=7$. The case $b=2$ does not work, because otherwise the number $5 a b=2(10 a+b)+1$ would have to be even, which is not true. Therefore, $b=7$, and for $a$ we get the equation $20 a+15=35 a \Leftrightarrow a=1$.