Question 187305
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10s digit: <i>x</i>


1s digit: <i>y</i>


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x + y = 11 \ \ \Rightarrow\ \ x = 11 - y]


Also the number is then:


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


And the number with the digits reversed is:


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


And we are given that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  10x + y + 45 = 10y + x \ \ \Rightarrow\ \ 9x - 9y = -45 \ \ \Rightarrow\ \ x - y = -5]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  (11 - y) - y = 5 \ \ \Rightarrow\ \ -2y = -16 \ \ \Rightarrow\ \ y = 8 \ \ \Rightarrow\ \ x = 3]


And the number is 38.


Note that 38 + 45 = 83 and 3 + 8 = 11, so the answer checks.


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