Question 758948
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The number itself, is, as you state: *[tex \LARGE 10x\ +\ y] IF you let *[tex \LARGE x] represent the 10s digit and *[tex \LARGE y] represent the 1s digit.


Then the product of the digits is *[tex \LARGE xy] and "less than 4 times the product of its digits by 5" is *[tex \LARGE 4xy\ -\ 5], leaving us:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ =\ 4xy\ =\ 5]


But then we are also given that the 10s digit is 1 more than the 1s digit, in other words:  *[tex \LARGE x\ =\ y\ +\ 1].  So then, by substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\(y\ +\ 1)\ +\ y\ =\ 4(y\ +\ 1)y\ =\ 5]


Solve for *[tex \LARGE y], then calculate *[tex \LARGE y\ +\ 1]


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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