Romannepal wrote:
If \(2^5^x\) is an integer, \(x+2=y-3\) and \((y^3-49y)(y^2-7y-18)=0\), what is the value of \(xy\) ?
A. -6
B. 0
C. 14
D. 84
E. 126
Hi,
From the equation involving y, we will get possible values of y..
From these we will find value of x for which\(2^{5x}\) is INTEGER..
Let's see the possible value of y...
\((y^3-49y)(y^2-7y-18)=0........ y (y-7)(y+7)(y-9)(y+2)=0\)..
So y can be 0, x will be 0-5=-5.... 2^5x will be a fraction... No
If y =7, x=y-5=7-5=2.... 2^5x will be 2^10... Yes.. xy=2*7=14....C
If y =-7, x=-7-5=-7-5=-12.... 2^5x will not be INTEGER... NO
.If y =9, x=9-5=4.... 2^5x will be 2^20... Yes.. xy=4*9=36... Not in choices
If y =-2 x=y-5=-2-5=-7... 2^5x will be not be INTEGER...No
C
Thanks for your solution.Really comprehensive. Cheers.