Question 181116
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I'm pretty sure you meant that the product <i>of the digits</i> is 18 and the sum <i>of the digits</i> is 9.


Let <i>x</i> be the tens digit and <i>y</i> be the ones digit.


Then we know that *[tex \Large xy = 18] and *[tex \Large x + y = 9]


*[tex \Large \text{          }\math x + y = 9 \ \ \rightarrow \ \ y = 9 - x]


Substituting:


*[tex \Large \text{          }\math x(9 - x) = 18 \ \ \rightarrow \ \ -x^2 + 9x - 18 = 0 \ \ \rightarrow \ \ (x - 3)(x - 6) = 0 \ \ \rightarrow \ \ x = 3 \text { or }\math x = 6]


*[tex \Large x = 3 \ \ \rightarrow \ \ y = 9 - 3 = 6]


*[tex \Large x = 6 \ \ \rightarrow \ \ y = 9 - 6 = 3]


But since we are also given that *[tex \Large x < y] we know that *[tex \Large x = 3 \text{ and } y = 6]


Therefore the number is 36.
 

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