Question 1023919
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If you reverse the digits of a two-digit number and get a number that differs from the original by 9, then the difference between the two digits must be 1.  (If the two numbers differ by 18, then the difference between the two digits is 2, and so on).


So you need to find two single digits that differ by 1 where the product of the digits is 12.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ -\ x\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  xy\ =\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x(x\ +\ 1)\ =\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ x\ -\ 12\ =\ 0]


Solve the quadratic.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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