Mongolian Mathematical Olympiad

Let the three-digit number be $N=100a+10b+c$, where $a,b,c$ are digits and $a\ne 0$.

We are told that $N=7S+P$, where $S=a+b+c$ is the sum of the digits and $P=abc$ is the product of the digits.

So: $$100a+10b+c=7(a+b+c)+abc$$ $$100a+10b+c-7a-7b-7c=abc$$ $$(100a-7a)+(10b-7b)+(c-7c)=abc$$ $$93a+3b-6c=abc$$

So: $$abc=93a+3b-6c$$

Since $a,b,c$ are digits ($a=1,\dots ,9$, $b,c=0,\dots ,9$), and $abc$ is the product of the digits, we can try small values for $a$.

Let us try $a=9$:

Then: $$abc=93\times 9+3b-6c=837+3b-6c$$ But $abc\le 9\times 9\times 9=729<837$, so $a$ cannot be $9$. Try $a = 8$: abc = 93 \times 8 + 3b - 6c = 744 + 3b - 6c But $abc \leq 8 \times 9 \times 9 = 648 < 744$, so $a$ cannot be $8$.

Try $a=7$: $$abc=93\times 7+3b-6c=651+3b-6c$$ $abc\le 7\times 9\times 9=567<651$, so $a$ cannot be $7$. Try $a = 6$: abc = 93 \times 6 + 3b - 6c = 558 + 3b - 6c $abc \leq 6 \times 9 \times 9 = 486 < 558$, so $a$ cannot be $6$.

Try $a=5$: $$abc=93\times 5+3b-6c=465+3b-6c$$ $abc\le 5\times 9\times 9=405<465$, so $a$ cannot be $5$. Try $a = 4$: abc = 93 \times 4 + 3b - 6c = 372 + 3b - 6c $abc \leq 4 \times 9 \times 9 = 324 < 372$, so $a$ cannot be $4$.

Try $a=3$: $$abc=93\times 3+3b-6c=279+3b-6c$$ $abc\le 3\times 9\times 9=243<279$, so $a$ cannot be $3$. Try $a = 2$: abc = 93 \times 2 + 3b - 6c = 186 + 3b - 6c $abc \leq 2 \times 9 \times 9 = 162 < 186$, so $a$ cannot be $2$.

Try $a=1$: $$abc=93\times 1+3b-6c=93+3b-6c$$ $abc\le 1\times 9\times 9=81<93$, so $a$ cannot be $1$. $$

Therefore, there are no three-digit numbers that can be obtained in this way.

Answer: There are no such three-digit numbers.