MS_Kekas wrote: Find all three-digit numbers that are \( 5 \) times greater than the product of their digits. $100a+10b+c=5abc$ with $a\in\{1,2,...,9\}$ and $b,c\in\{0,1,...,9\}$ So $c\equiv 0\pmod 5$ and so $c=0$ or $c=5$ $c=0$ implies $100a+10b=0$, impossible $c=5$ implies $20a+2b+1=5ab$, which is $(5a-2)(b-4)=9$ and so, since $5a-2\ge 3$ : Either $5a-2=3$ and $b-4=3$ and solution $\boxed{175}$ Either $5a-2=9$ and $b-4=1$, which does not fit

MS_Kekas wrote: Find all three-digit numbers that are \( 5 \) times greater than the product of their digits. All such numbers will be of the form $\overline{abc}$ where $100a+10b+c=5abc$. Note $abc\ne0$ since if it was, the LHS wouldn't be a three digit number. Then $5\mid c$, so $c=5$. We are left with $5ab-20a-2b-1=0$ which simplifies to $(5a-2)(b-4)=9$. Now $5a-2\ge5\cdot1-2=3$ so we check all $2$ possibilities: $(5a-2,b-4)=(3,3)\Rightarrow(a,b)=(1,7)$ producing $\boxed{175}$ as a solution, which we can verify works (all steps are reversible). $(5a-2,b-4)=(9,1)\Rightarrow a\notin\mathbb Z$, hence no other solutions.

The answer is $175$, which obviously works. Nowe show it's the only one. Note that the number is a multiple of $5$, so it ends in a $0$ or $5$. However, if it ends in a $0$ it's product of digits is $0$, so it's $0$, bad. Therefore, it ends in a $5$, so all digits are odd. But then it's a multiple of $5 \cdot 5 = 25$, so it must end in $75$. Now, it must be a multiple of $7$, and since $175$ is a multiple of $7$, it can't be $375, 575, 775,$ or $975$ so we're done.