Question 279490
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Let *[tex \Large x] represent the 10s digit.  Let *[tex \Large y] represent the 1s digit.


The sum of the digits is 12 so:  *[tex \Large x\ +\ y\ =\ 12]


The value of the original number must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y]


Adding 36


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ +\ 36]


The value of the number with the digits reversed must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10y\ +\ x]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ +\ 36\ =\ 10y\ +\ x]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ 9y\ =\ -36]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ -4]


Add that last equation to the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ x\ y\ -\ y\ =\ 12\ -\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4]


That means *[tex \Large y] must be 12 - 4 = 8 and that makes the original number be 48.


Check 48 plus 36 is 84 which is 48 with the digits reversed.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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