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If x and y are prime numbers such that x > y > 2, then x^2 −

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Joined: 10 Jul 2013
Posts: 315
If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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Updated on: 20 Aug 2013, 11:34
2
3
00:00

Difficulty:

5% (low)

Question Stats:

85% (00:44) correct 15% (00:48) wrong based on 294 sessions

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If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

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Asif vai.....

Originally posted by Asifpirlo on 20 Aug 2013, 11:29.
Last edited by Bunuel on 20 Aug 2013, 11:34, edited 1 time in total.
Edited the question.
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Posts: 51121
Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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20 Aug 2013, 11:36
4
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

If x=5 and y=3, then x^2-y^2=16 and 16 is divisible only by 4 from the options, thus it must be correct.

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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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22 Aug 2013, 19:38
y can take values like 3, 5, 7, 9
x can take values like 5, 7, 9
Square of their diff is divisible by 4
Ans: B

PS : Numbers have to be PRIME
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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23 Aug 2013, 01:20
This can be termed as a property of the squares of the prime numbers....
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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02 Sep 2013, 14:01
2
1
In addition to what has already been stated, this is a number properties question, without testing any numbers (not sure if that's faster or slower) but:

x>y>2 and noting that they are prime means that X and y are odd.

also

X^2 - y^2= (x+y)(x-y) so now you have (odd+odd)(odd-odd) which = even * even, which means that at minimum the numbers have 2 powers of 2 in them so it must be divisible by 4.

Just how I thought of it....
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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10 Sep 2013, 00:14
if we take x=7 and y=5 then x2-y2 is 49-25=24 and this is divisible by 3,4 and 12, something is wrong with this question?
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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10 Sep 2013, 00:17
jkher wrote:
if we take x=7 and y=5 then x2-y2 is 49-25=24 and this is divisible by 3,4 and 12, something is wrong with this question?

The question asks "x^2 − y^2 must be divisible by which one of the following numbers" not "could be divisible".
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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02 Jan 2015, 03:48
1
we have

(x-y)*(x+y)=x^2-y^2

3,5,7,11,13,17,19,23,27,31......

difference is only 2 or 4 and when it is 2 the sum is always divisible by 4.

It is B
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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29 Jun 2017, 03:40
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

X^2 - Y^2
= ( X + Y ) * ( X –Y )
= ( odd + odd ) * ( odd – odd )
= even * even
= 2x * 2x
= 4 x^2 which must be divisible by 4
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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29 Jun 2017, 04:13
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

Lets take some examples:

$$x = 5 & y = 3$$

$$= 5^2 - 3^2$$

$$= 25 - 9$$

$$= 16$$

Similarly you take any other prime numbers you will get a difference which is always divisible by $$4$$.

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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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07 Jul 2017, 05:13
Bunuel wrote:
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

If x=5 and y=3, then x^2-y^2=16 and 16 is divisible only by 4 from the options, thus it must be correct.

Hi, what if x = 7 and y = 5, then x^2-y^2 = 49 -25 = 24 which is divisible by 12 (answer no. E)
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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07 Jul 2017, 05:17
zahinsarwar wrote:
Bunuel wrote:
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

If x=5 and y=3, then x^2-y^2=16 and 16 is divisible only by 4 from the options, thus it must be correct.

Hi, what if x = 7 and y = 5, then x^2-y^2 = 49 -25 = 24 which is divisible by 12 (answer no. E)

The question ask x^2 − y^2 MUST be divisible by which one of the following numbers, not COULD be divisible by which one of the following numbers. It COULD be divisible by 12 but it MUST be divisible only by 4 (from the options). So, it will ALWAYS be divisible by 4, and will be divisible by 12 only in specific cases.
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Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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15 Jul 2017, 07:23
Choice:B
Time taken=1:26
In this case I took 2 choices
first: x=5,y=3
x^2-y^2
25-9
16 only choice B satisfies
Second: x=7,y=5
x^2-y^2
49-25
24
Now A,B and E satisfies
So B is correct choice
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If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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01 Oct 2018, 05:16
Bunuel wrote:
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

If x=5 and y=3, then x^2-y^2=16 and 16 is divisible only by 4 from the options, thus it must be correct.

x^2-y^2
since x and y are greater than 2, I have taken x=7 and y=5

x^2-y^2 = (x+y)(x-y)
Now, x+y = 12

Thus [(x+y)(x-y)] / 12

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Joined: 02 Sep 2009
Posts: 51121
Re: If x and y are prime numbers such that x > y > 2, then x^2 −  [#permalink]

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01 Oct 2018, 05:59
780gmatpossible wrote:
Bunuel wrote:
Asifpirlo wrote:
If x and y are prime numbers such that x > y > 2, then x^2 − y^2 must be divisible by which one of the following numbers?

(A) 3
(B) 4
(C) 5
(D) 9
(E) 12

If x=5 and y=3, then x^2-y^2=16 and 16 is divisible only by 4 from the options, thus it must be correct.

x^2-y^2
since x and y are greater than 2, I have taken x=7 and y=5

x^2-y^2 = (x+y)(x-y)
Now, x+y = 12

Thus [(x+y)(x-y)] / 12

Hope it helps.
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Re: If x and y are prime numbers such that x > y > 2, then x^2 − &nbs [#permalink] 01 Oct 2018, 05:59
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