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 TRUE?
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