How to solve $7200a+720b+72c=1000x+340+y<10000$? What is the easiest way to solve $7200a+720b+72c=1000x+340+y<10000$ where all variables are one digit natural numbers?
Trial and error method seems to be tedious.
 A: There is no solution. Since $7200a<10000$, $a=1$. Equations change to $7200+720b+72c=1000x+340+y<10000$. That means $1000x>7200$ and $1000x<10000$ so $x=8$ or $x=9$. Since left-hand-part is a multiple of $72$, so is the middle-part. If $x=8$, there is no multiple of $72$ between $8340$ and $8350$. If $x=9$, there is no multiple of $72$ between $9340$ and $9350$.
A: It’s immediately clear that $a$ must be $1$, which implies that $x$ must be at least $7$.
$7200a+720b+72c$ is clearly divisible by $9$. The digits of $1000x+340+y$ are $x,3,4$, and $y$, so $x+3+4+y$ must be a multiple of $9$, so either $x=7$ and $y=4$, $x=8$ and $y=3$, or $x=9$ and $y=2$. Then $1000x+340+y$ is $7344$, $8343$, or $9342$. Only the first of these is divisible by $72$: it’s $72\cdot102$.
It follows that $100a+10b+c=102$; if $a,b$, and $c$ must be single digits, this implies that $a=1$, $b=0$, and $c=2$. Thus, it’s impossible to meet the conditions: if $a,b,c,x$, and $y$ are all single digits, one of them must be $0$.
