Question 524554
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Let *[tex \Large x] represent the larger of the two digits.  Let *[tex \Large y] represent the smaller.  We know that they have to be different, and therefore there exists a smaller and a larger because the difference is non-zero.


The difference of the digits is 5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ 5]


The difference of the squares is 45:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ y^2\ =\ 45]


Add *[tex \Large y] to both sides of the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ y\ +\ 5]


Substitute *[tex \Large y\ +\ 5] for *[tex \Large x] in the second equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\ +\ 5)^2\ -\ y^2\ =\ 45]


Expand the binomial and simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 10y\ +\ 25\ -\ y^2\ =\ 45]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10y\ +\ 25\ =\ 45]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2]


Now you know that one of the digits of your two digit number is 2.


Go back to the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 2\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 7]


Now you know that your two digit number is either 72 or 27.  Only one of these is odd.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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