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M26-29

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M26-29  [#permalink]

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New post 16 Sep 2014, 00:26
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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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Re M26-29  [#permalink]

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New post 16 Sep 2014, 00:26
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1
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.


Answer: D
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Re: M26-29  [#permalink]

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New post 21 Feb 2015, 09:18
Hi,

please explain why statement 2 is correct.?
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Re: M26-29  [#permalink]

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Re: M26-29  [#permalink]

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New post 12 Mar 2016, 13:02
Probably very basic question, anyway..

why is x-y in statement 2 equal 1?
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Re: M26-29  [#permalink]

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New post 13 Mar 2016, 07:11
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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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Re: M26-29  [#permalink]

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New post 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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M26-29  [#permalink]

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New post 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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Re: M26-29  [#permalink]

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New post 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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Re: M26-29  [#permalink]

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New post 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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Re: M26-29  [#permalink]

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New post 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

Answer D
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M26-29  [#permalink]

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New post 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.

Answer: D
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Re: M26-29  [#permalink]

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New post 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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Re: M26-29  [#permalink]

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New post 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!
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Re: M26-29   [#permalink] 18 Sep 2018, 06:05
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