What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?
-Question 28, Junior Division, AMC 2016
I found the solution here, page 51, however I don't understand it from the very beginning. How does $100a+10b+c$ equal $37a+37b+37c$? Can somebody explain to me what's happened in this solution step by step please? Or can somebody give an alternate solution, suitable for Year 7 and 8? Thanks!
P.S. If anybody knows a more accurate tag for this question, please feel free to edit my question or comment so I can edit it if you can't.
Best Answer
You have $\overline{abc}=100a +10b+ c=37 (a+b+c)$ or
$$63a-27b-36c=0 \ \ \Longrightarrow 7a=3b+4c$$
Now $a \le \frac{3b+4c}{7}\le \frac{3\times 8+4 \times 9}{7}<9$.
Let $a=8$. Then $3b+4c=56$. We have $3b=4(14-c)$. Thus $b$ is a multiple of $4$. This forces that $b=4$ and gives $c=11$, which is impossible!
Let $a=7$. Then $3b+4c=49$ and $b=\frac{49-4c}{3}=16+\frac{1-4c}{3}$. This gives $c=4$ and leads to $b=11$, which is impossible again.
Let $a=6$. Then $3b+4c=42$ and $4c=3(14-b)$. This gives $c=6, 9$ since $c$ must be a multiple of $3$. Then possible pairs for $(b, c)$ are $(6, 6), (2, 9)$.
Thus largest is $629$.