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? 
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.

"What is the largest 3-digit number...."
Let $N = abc$ have three digits.  $N = 100a + 10b + c$.  That's what the writing numbers of digits mean.  $593 = 500 + 90 +3$ and $abc = 100a + 10b + c$. 
" which has all 3 digits different"
So $a \ne b; a\ne c; b \ne c$.
" and is equal to 37 times the sum of it's digits?"
So $N = 37*(a+b+c) = 37a + 37b + 37c$
And $N = 100a + 10b + c$.
So $100a + 10b + c = 37a + 37b + 37c$
So $(100a - 37a)= (37 - 10)b + (37-1)c$
So $63a = 27b + 36c$
So $\frac {63}9 a = \frac {27}9 b + \frac {36}9 c$
So $7a = 3b + 4c$.
Obviously $a = 1, b= 1, c=1$ is a solution but 1) It's probably not the largest and 2) The digits aren't all different.
$b \ne c$ so $c = b \pm k$ for some $k \ne 0$.
So $7a = 3b + 4c = 3b + 4(b \pm k)  = 7b \pm 4k$
So $a = b \pm \frac 47k$ so $7$ must divide $k$.  But $b$ and $c$ are single digits and $c = b \pm k$ so $k = 7$ and $a = b \pm 4$
The possible answers are $b =0; c = 0+7=7; a = 0+4=4$
$b=1; c = 1+7= 8; a = 1+4 =5$
$b=2; c = 2+7 =9; a= 2+4 =6$
$b = 7; c = 7-7=0 ; a=7-4 = 3$
$b = 8; c = 8-7; a = 8-4 = 4$
$b = 9; c = 9 -7 =2; a = 9-4=5$.
Of those $a=6;b=2; c= 9$ and $N = 629 = 37*17 = 37(6+2+9)$ is the largest such answer.
