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# M26-29

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Math Expert
Joined: 02 Sep 2009
Posts: 53066

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16 Sep 2014, 00:26
13
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Difficulty:

45% (medium)

Question Stats:

67% (01:19) correct 33% (01:26) wrong based on 265 sessions

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If $$357^x*117^y=a$$, where $$x$$ and $$y$$ are positive integers, what is the units digit of $$a$$?

(1) $$100 \lt y^2 \lt x^2 \lt 169$$

(2) $$x^2-y^2=23$$

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16 Sep 2014, 00:26
3
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Official Solution:

(1) $$100 \lt y^2 \lt x^2 \lt 169$$. Since both $$x$$ and $$y$$ are positive integers then $$x^2$$ and $$y^2$$ are perfect squares. Now, there are only two perfect squares in the given range $$121=11^2$$ and $$144=12^2$$, so $$y=11$$ and $$x=12$$. Sufficient. (As cyclisity of units digit of 7 in integer power is 4, therefore the units digit of $$7^{23}$$ is the same as the units digit of $$7^3$$, so 3).

(2) $$x^2-y^2=23$$. From that: $$(x-y)(x+y)=23=prime$$ and since both $$x$$ and $$y$$ are positive integers then: $$x-y=1$$ and $$x+y=23$$. Solving for $$x$$ and $$y$$ gives: $$y=11$$ and $$x=12$$. Sufficient.

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Joined: 06 Nov 2014
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21 Feb 2015, 09:18
Hi,

please explain why statement 2 is correct.?
Math Expert
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22 Feb 2015, 05:34
Richa07 wrote:
Hi,

please explain why statement 2 is correct.?

You should specify what is unclear in the solution above.
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12 Mar 2016, 13:02
Probably very basic question, anyway..

why is x-y in statement 2 equal 1?
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Joined: 02 Sep 2009
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13 Mar 2016, 07:11
3
Avigano wrote:
Probably very basic question, anyway..

why is x-y in statement 2 equal 1?

We are told that x and y are positive integers. Thus x+y is positive too. (2) says that $$(x-y)(x+y)=23=prime$$. Prime numbers can only be positive and they have only two factors: 1 and itself. Therefore, $$x-y=1$$ and $$x+y=23$$.

Hope it's clear.
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Schools: Goizueta '19 (A)
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GMAT 2: 620 Q45 V31
GMAT 3: 640 Q46 V32
GPA: 3.97

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13 Mar 2016, 09:56
Bunuel wrote:
Avigano wrote:
Probably very basic question, anyway..

why is x-y in statement 2 equal 1?

We are told that x and y are positive integers. Thus x+y is positive too. (2) says that $$(x-y)(x+y)=23=prime$$. Prime numbers can only be positive and they have only two factors: 1 and itself. Therefore, $$x-y=1$$ and $$x+y=23$$.

Hope it's clear.

Thank you, very clear!
clear as the greek waters I would say!
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Joined: 19 Apr 2018
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27 Aug 2018, 06:38
Hi Bunuel
In explanation of statement 2.
Why the possible answer of x+y only y=11 and x=12 ?
There are many possibilities, why not x=11 and y=12, or other combinations.
Thank you Bunuel

Posted from my mobile device
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27 Aug 2018, 06:50
bonion wrote:
Hi Bunuel
In explanation of statement 2.
Why the possible answer of x+y only y=11 and x=12 ?
There are many possibilities, why not x=11 and y=12, or other combinations.
Thank you Bunuel

Posted from my mobile device

Explained here: https://gmatclub.com/forum/m26-184468.html#p1658546 Hope it helps.
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27 Aug 2018, 07:10
Bunuel wrote:
bonion wrote:
Hi Bunuel
In explanation of statement 2.
Why the possible answer of x+y only y=11 and x=12 ?
There are many possibilities, why not x=11 and y=12, or other combinations.
Thank you Bunuel

Posted from my mobile device

Explained here: https://gmatclub.com/forum/m26-184468.html#p1658546 Hope it helps.

I understand now. The x-y=1 part limits the only possible answer.
Thank you Bunuel.
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27 Aug 2018, 07:24
1) two values ofx & y are possible, y=11,12
as x is greater than y so
y=11
x=12
SUFFICIENT

2) in this we have to choose two positive integers x and y whose square's difference is equal to 23. so only one pair is possible and that is x=12, y=11
SUFFICIENT

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Hasnain Afzal

"When you wanna succeed as bad as you wanna breathe, then you will be successful." -Eric Thomas

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Joined: 23 May 2018
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27 Aug 2018, 08:16
if we know the unit digit of the power the number is raised to, we can answer for the unit digit of a.

For statement 1

$$100<y^2<x^2<169$$ can be read as $$10^2<y^2<x^2<13^2$$

Which means y has to be 11 and x has to be 12. Thus, we can figure out the unit digit.

For statement 2

$$x^2-y^2=23$$

We can either do it by adding perfect squares to 23 and see when we reach another perfect square (it's going to take precious time, of course) or we could solve it with $$(x+y)(x-y)=23$$ and get the values for x and y. Thus, figuring out the unit digit of a.

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Joined: 26 May 2015
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27 Aug 2018, 10:18
Statement 2 : x^2-y^2=23=23 * 1,
so avg = (23+1)/2=12, so 23*1 can be further divided as = (12+11)(12-11)=12^2-11^2
so x=12 and y=11, these are the only solutions as 23 is prime.
So, statement 2 is sufficient
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18 Sep 2018, 06:05
Great question! Totally was stuck when I got to (x-y)(x+y) = 23 when doing the problem... felt the answer was probably D but didn't know why!
Re: M26-29   [#permalink] 18 Sep 2018, 06:05
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# M26-29

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