Find numbers that fit each riddle look for more than one answer

Question

There are two $2$ digits numbers. The first number is greater than $50$ and ends in $0$. When you subtract one number from the other number the difference is $29$

Answer 1

Hint: Try something like $x-y = 29$ where $x = 10n, n \in \{6,7,8,9\}$

Answer 2

$|a0 - bc| = 29$. $a0 > 50.$

Case 1: $a0 > bc$

so $a0 - bc = 10a - (10b + c) = 10(a - b) - c = 29 \implies c = 1; a-b = 3$. Possibilities: $(a0, [a-3]1) =\{(60, 31)(70,41)(80,51)(90,61)\}$.

Case 2: $bc > a0$

so $bc - a0 = 10b + c - 10a = 10(b-a) + c = 29 \implies c = 9; b- a = 2$. Possibilities:$ ([a+2]9, a0) = \{(89,60)(99,70)\}$

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Or if you don't like algebra.

One number is a= 60, 70, 80, or 90 and the other is $a \pm 29$. There are 8 possibilities but two of them are impossible.