Question 910345
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Let *[tex \Large x] represent the ones digit of the original number.  Then *[tex \Large 2x] represents the tens digit of the original number, and the original number is formed by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10(2x)\ +\ x\ =\ 20x\ +\ x\ =\ 21x]


Reversing the digits, we get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10(x)\ +\ 2x\ =\ 10x\ +\ 2x\ =\ 12x]


Doubling the number with the reversed digits gives us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 24x]


And this is the same as the original number plus 9.  "The same as" lets us form an equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 24x\ =\ 21x\ +\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3]


So the original number is *[tex \Large 63].


Check the answer: Reverse the digits to get 36.  Double 36 to get 72, which is 9 more than the original 63.


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