Question 369029
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Let *[tex \Large x] represent the 10s digit of the original number.  Let *[tex \Large y] represent the units digit of the original number.  The original number is then *[tex \Large 10x\ +\ y], the number with the digits reversed is *[tex \Large 10y\ +\ x], and the digits are related by *[tex \Large y\ =\ x\ +\ 6].


If reversing the digits increases the number by 54, then we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ = 10y\ +\ x\ - 54]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ 9y\ =\ -54]


Since we know that *[tex \Large y\ =\ x\ +\ 6], we can substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ 9(x\ +\ 6)\ =\ -54]


Which results in the triviality:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -54\ =\ -54]


Ordinarily when solving a system of equations such a result would indicate that the system was consistent and dependent, which is to say that the solution set has an infinite number of elements.  However, in this problem our variables are restricted to the set of single digit positive integers, namely *[tex \Large \{1,\,2,\,3,\,4,\,5,\,6,\,7,\,8,\,9\}]


The smallest value *[tex \Large x] can assume is 1 which implies that *[tex \Large y] would be 7 and the original number would be 17.  *[tex \Large x] could also be 2, *[tex \Large y\ =] 8, and the original number 28, or *[tex \Large x] could be 3, *[tex \Large y\ =] 9, and the original number 39.  This exhausts the possibilities.


Verification that each of the three results above fit the parameters given in the problem is left as an exercise for the student.


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