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The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. (A) 39 (B) 57 (C) 66 (D) 93
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Suppose the number is \(10a+b\), where \(1\le a\le9\) and \(0\le b\le9\). Then \(a+b=12\). The number with its digits reversed would be \(10b+a\). You're told that this new number is larger than the original number by \(54\), which means \(10b+a=10a+b+54\), or \(9b-9a=54\). Solve for \(a\) and \(b\), then find \(10a+b\).
Thanks a lot! @HolsterEmission
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