Question 213493
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Let *[tex \Large x] represent the hundreds digit.  The ones digit is three times that, so the ones digit must be *[tex \Large 3x].  The tens digit is 1 less than the ones digit, so the tens digit must be *[tex \Large 3x - 1].


Next, the product of the hundreds digit and the tens digit is:  *[tex \Large x(3x -1) = 3x^2 - x], and the ones digit is 15 less than that product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 - x - 15\ =\ 3x]


Put the quadratic into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x^2 - 4x - 15\ =\ 0]


Solve the quadratic by any means (this one factors, by the way) and exclude any negative or fractional roots.  The positive integer root will be the first (hundreds) digit of the answer, and you can easily calculate the other two roots from the given relationships. 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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