Question 777589
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The quotient of 2 and 3:  *[tex \LARGE \frac{2}{3}]


4 more than ... : *[tex \LARGE \frac{2}{3}\ +\ 4]


...times a number *[tex \LARGE k] :  *[tex \LARGE \left(\frac{2}{3}\ +\ 4\right)k]


...is 20:  *[tex \LARGE \left(\frac{2}{3}\ +\ 4\right)k\ =\ 20]


Then substitute 24 in place of *[tex \LARGE k], do the arithmetic, and see if you have a true statement or not.  If true, then 24 is a solution, otherwise not.


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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