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Romannepal
Joined: 12 Nov 2016
Last visit: 25 Sep 2017
Posts: 22
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE:Account Management (Accounting)
Posts: 22
Updated on: 05 Dec 2025, 04:10
Originally posted by Romannepal on 08 Jun 2017, 03:44.
Last edited by Bunuel on 05 Dec 2025, 04:10, edited 2 times in total.
Last edited by Bunuel on 05 Dec 2025, 04:10, edited 2 times in total.
Renamed the topic and edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
77% (02:52) correct
23%
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wrong
based on 527
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If \(2^{5x}\) 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
A. -6
B. 0
C. 14
D. 84
E. 126
Romannepal
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...
\((y^3-49y)(y^2-7y-18)=0........ y (y-7)(y+7)(y-9)(y+2)=0\)..
Possible values of y ..... 0, 7, -7, 9, -2
Let's check values of x for these and if \(2^{5x}\) is an integer.
1) So y can be 0..... x will be 0-5=-5.... 2^5x will be a fraction... No
2) If y =7........ x=y-5=7-5=2.... 2^5x will be 2^10... Yes.. xy=2*7=14....C
3) If y =-7....... x=-7-5=-7-5=-12.... 2^5x will not be INTEGER... NO
4).If y =9........ x=9-5=4.... 2^5x will be 2^20... Yes.. xy=4*9=36... Not in choices
5) If y =-2........ x=y-5=-2-5=-7... 2^5x will be not be INTEGER...No
C
_________________
General Discussion
Romannepal
Joined: 12 Nov 2016
Last visit: 25 Sep 2017
Posts: 22
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE:Account Management (Accounting)
Posts: 22
chetan2u
Thanks for your solution.Really comprehensive. Cheers.
Posted from my mobile device
Posts: 7
Romannepal
Joined: 12 Nov 2016
Last visit: 25 Sep 2017
Posts: 22
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE:Account Management (Accounting)
Posts: 22
StockAnalyst
2^5x is an integer means x mustnot be negative.Also given y-3=x+2, so y=x+5. Hence y is greater than or equal to 5. Calculate values of y. They are 0, 7, -7, 9 and -2. Only 7 and 9 satisfies the condition. So, y=7 and x=2 or y=9 and x=4. xy=14 or 36. xy=14 is in the option and hence the answer.
Posted from my mobile device
Posts: 249
nickrocks Thank you for this answer - posting it as I felt it is algebraic but sort of more detailed :
we have
1) 2 ^ 5x is an integer
2) x +2 = y - 3
==> x = y - 5
simplifying equation:
-> (y cube - 49y)(y sq - 7y -18)
-> y(y sq - 49 {which is 7^2}y)(y+2)(y+9) = 0
-> y(y+7)(y-7)(y+2)(y+9) = 0
y thus has values:
- 0
-7
+7
-2
9
since x = y- 5 if we substitute each value of y we get x:
- 5
-12
2
-7
4
x*y = 7*2 = 14
9*4 = 36 which isn't there.
so 14 (C) is the answer. Kudos, please
if this explanation helped.
we have
1) 2 ^ 5x is an integer
2) x +2 = y - 3
==> x = y - 5
simplifying equation:
-> (y cube - 49y)(y sq - 7y -18)
-> y(y sq - 49 {which is 7^2}y)(y+2)(y+9) = 0
-> y(y+7)(y-7)(y+2)(y+9) = 0
y thus has values:
- 0
-7
+7
-2
9
since x = y- 5 if we substitute each value of y we get x:
- 5
-12
2
-7
4
x*y = 7*2 = 14
9*4 = 36 which isn't there.
so 14 (C) is the answer. Kudos, please
if this explanation helped. Romannepal
Equation One \(x+2=y-3\)
Equation Two \((y^3-49y)(y^2-7y-18)=0\)
Factorize ----- > \((y^3-49y)\) ----> \(y(y^2-49)\) considering the fact that equation two, equals 0, then y(y^2-49) y must be 7
\(7(7^2-49)\) ---> \(7(49-49) = 0\)
hence \(0 *(y^2-7y-18)=0\)
if y is 7 , plug in this value into first equation \(x+2=7-3\) --- > \(x+2=4\) ------ > \(x = 2\)
hence \(7*2 = 14\) isnt it surprising ?
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