Question 719734
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Let *[tex \LARGE x] represent the 10s digit of the desired number.  Let *[tex \LARGE y] represent the 1s digit of the desired number.  Then the desired number can be represented by:


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


and the sum of the digits can be represented by 


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


So if the number is twice as large as the sum of its digits:


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


Which simplifies to:


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


The new number when the digits reversed would be represented by:


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


Five times the original number, less 9:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5(10x\ +\ y)\ -\ 9]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10y\ +\ x\ =\ 5(10x\ +\ y)\ -\ 9]


Which simplifies to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5y\ -\ 49x\ =\ -\ 9]


Since we know that *[tex \LARGE 8x\ =\ y], substitute *[tex \LARGE 8x] for *[tex \LARGE y] in *[tex \LARGE 5y\ -\ 49x\ =\ -\ 9] and solve for *[tex \LARGE x].  Then substitute the computed value of *[tex \LARGE x] into *[tex \LARGE 8x\ =\ y] to find the value of *[tex \LARGE y].


Actually, you could have stopped right after you had established that *[tex \LARGE 8x\ =\ y].  There is only one pair of single-digit positive integers that satisfies this equation.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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