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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0

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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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Updated on: 08 Jun 2017, 04:48
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Difficulty:

35% (medium)

Question Stats:

74% (03:06) correct 26% (02:52) wrong based on 177 sessions

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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

Originally posted by Romannepal on 08 Jun 2017, 04:44.
Last edited by Bunuel on 08 Jun 2017, 04:48, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Aug 2009
Posts: 6975
If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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08 Jun 2017, 05:14
3
4
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...
$$(y^3-49y)(y^2-7y-18)=0........ y (y-7)(y+7)(y-9)(y+2)=0$$..
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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Re: If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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08 Jun 2017, 10:03
1
chetan2u wrote:
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.

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Re: If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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

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18 Jul 2017, 08:08
1
StockAnalyst wrote:
Bunuel
Any quick approach to solve this question the one explained above is lengthy I suppose

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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Re: If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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26 Jul 2017, 07:44
2
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)
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If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0  [#permalink]

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19 Sep 2018, 11:46
1
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

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 ?
If 2^(5x) is an integer, x + 2 = y - 3 and (y^3 - 49y)(y^2 - 7y -18)=0 &nbs [#permalink] 19 Sep 2018, 11:46
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