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605-655 (Medium)|   Algebra|   Exponents|      
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Romannepal
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 Updated on: 05 Dec 2025, 05:10
Originally posted by Romannepal on 08 Jun 2017, 04:44.
Last edited by Bunuel on 05 Dec 2025, 05:10, edited 2 times in total.
Renamed the topic and edited the question.
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Difficulty:

45% (medium)

Question Stats:

78% (02:55) correct 22% (02:51) wrong based on 452 sessions

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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
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C
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chetan2u
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 08 Jun 2017, 05:14
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Romannepal
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
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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
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Romannepal
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 08 Jun 2017, 10:03
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chetan2u
Romannepal
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
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Thanks for your solution.Really comprehensive. Cheers.

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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 18 Jul 2017, 07:18
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Bunuel
Any quick approach to solve this question the one explained above is lengthy I suppose
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Romannepal
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 18 Jul 2017, 08:08
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Bunuel
Any quick approach to solve this question the one explained above is lengthy I suppose
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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.

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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 26 Jul 2017, 07:44
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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.
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 19 Sep 2018, 11:46
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Romannepal
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
Show more



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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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 01 Apr 2026, 15:37
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