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# If a and b are positive integers, is a^2 + b^2 divisible by

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Retired Moderator
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If a and b are positive integers, is a^2 + b^2 divisible by [#permalink]

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11 May 2009, 14:45
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If a and b are positive integers, is a^2 + b^2 divisible by 5 ?

a) 2ab is divisible by 5
b) a-b is divisible by 5

I'm interested in how to merge the information in a) and b)
SVP
Joined: 29 Aug 2007
Posts: 2453

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11 May 2009, 19:06
1
KUDOS
If a and b are positive integers, is a^2 + b^2 divisible by 5 ?

a) 2ab is divisible by 5
b) a-b is divisible by 5

I'm interested in how to merge the information in a) and b)

Squarinng B: (a-b)^2 = a^2 -2ab + b^2
If a-b is divisible by 5, its square is also divisible by 5. If 2ab is divisible by 5, (a^2 + b^2) should also be divisible by 5.

C.
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GMAT Tutor
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11 May 2009, 21:34
If a and b are positive integers, is a^2 + b^2 divisible by 5 ?

a) 2ab is divisible by 5
b) a-b is divisible by 5

I'm interested in how to merge the information in a) and b)

GMATTiger has a great solution above. Alternatively you could look at both statements together as follows:

If 2ab is divisible by 5, then at least one of a or b must be divisible by 5. If, say, a is divisible by 5, then a^2 will certainly be divisible by 5 (in fact, by 25), so a^2 + b^2 will only be divisible by 5 if b^2 is also divisible by 5. That is, S1 tells us that at least one of a or b is divisible by 5; we then need to know if both a and b are divisible by 5.

Now using S2, if a-b is divisible by 5, and one of a or b is divisible by 5, the other must be as well. So together we know a and b are both multiples of 5, and a^2 + b^2 must be as well.
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Retired Moderator
Joined: 18 Jul 2008
Posts: 920

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12 May 2009, 07:20
Good explanations. OA is C.
Manager
Joined: 14 May 2009
Posts: 189

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14 May 2009, 15:17
Ok, here is my explanation.

Question:( $$(a^2 + b^2)/5$$ an integer? )

(1) 2ab is a multiple of 5

This implies that either a or b (or both) is a multiple of 5. If one is a multiple of 5, and the other one isn't, the answer is NO, and if they both are, the answer is YES.

YES&NO ==> Insufficient.

(2) a-b is a multiple of 5

Quick case a=5,b=0, YES
case a=6,b=1, NO

YES&NO ==> Insufficient.

(1&2) From (1) we know either a or b is a multiple of 5. 2 tells us that their difference is a multiple of 5. If we look at say variable a, and then add/subtract a bunch of 5's to get variable b, then variable b is still going to be a multiple of 5. You can pick a few cases to find this out via Finding the Pattern. This tells us both numbers are multiples of 5.

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Re: Divisibility   [#permalink] 14 May 2009, 15:17
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