Find a two digit number 15 more than 4 times its reverse. I acknowledge the answer is 91 since the reverse of 91 is 19, 4 times 19 is 76, which is 15 less than 91. I used trial and error, but I believe there is a more algebraically method to model this problem. I have developed something that looks as follows:
$x = $ first digit
$y = $ second digit
$10x + y = 15 + 4 * (10y + x)$
$= 15 + 4(y * 10) + 4x$
$= 15 + 40y + 4x$
$10x = 15 + 39y + 4x$
$6x = 15 + 39y$
I'm not sure if i am heading in the correct direction. I am confused how i can change the last expression into anything else.
Best Answer
You are on the right track.
Note that $x,y\in\{1,2,3,4,5,6,7,8,9\}$
Now it has to hold $6x=15+39y$. The LHS is at most $6\cdot 9=54$. While the RHS gives way greater values, even for small $y$. So the only value which can be $y$ is $1$, since already for $y=2$ we would get $15+78=93$.
So $y=1$ and $x=9$. The only solution.