take any 3-digit integer, reverse its digits, and subtract.
For example 742-247 = 495
This difference is divisible by 9.
Prove that, this must happen for all 3-digit numbers
Thank you in advance
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Any 3 digit number can be represented as
100x + 10y + z and its reverse would be
100z + 10y - x
Their difference would be
100(x-z) + (z-x) which can be written as 99(x-z).
99(x-z) is always divisable by 9
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