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# M19 Q11

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Manager
Joined: 13 May 2010
Posts: 122

Kudos [?]: 24 [1], given: 4

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05 Aug 2012, 22:41
1
KUDOS
If a and b are positive integers, is a^2 + b^2 divisible by 5 ?

(1) 2ab is divisible by 5
(2) a - b is divisible by 5

I have a question regarding this question. I want to verify the property of multiples that is related to divisibility. So

a) Mutliple of N + Mutliple of N = Multiple of N
b) Mutliple of N + Non -Mutliple of N = Non-Multiple of N
c) Non-Mutliple of N +Non- Mutliple of N = Can be both (multiple or non-multiple)

Does this property work for all integers, are there any exceptions?

Kudos [?]: 24 [1], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129505 [0], given: 12201

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06 Aug 2012, 00:56
teal wrote:
If $$a$$ and $$b$$ are positive integers, is $$a^2 + b^2$$ divisible by 5 ?

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

I have a question regarding this question. I want to verify the property of multiples that is related to divisibility. So

a) Mutliple of N + Mutliple of N = Multiple of N
b) Mutliple of N + Non -Mutliple of N = Non-Multiple of N
c) Non-Mutliple of N +Non- Mutliple of N = Can be both (multiple or non-multiple)

Does this property work for all integers, are there any exceptions?

There are no exceptions.

If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

As for the question:
If $$a$$ and $$b$$ are positive integers, is $$a^2+b^2$$ divisible by 5 ?

(1) $$2ab$$ is divisible by 5 --> if $$a=b=5$$ then the answer is YES but if $$a=5$$ and $$b=1$$ then the answer is NO. Not sufficient.

(2) $$a-b$$ is divisible by 5 --> if $$a=b=5$$ then the answer is YES but if $$a=b=1$$ then the answer is NO. Not sufficient.

(1)+(2) From (2) $$a-b$$ is divisible by 5 so $$(a-b)^2=(a^2+b^2)-2ab$$ is also divisible by 5. Next, since from (1) $$2ab$$ is divisible by 5 then $$a^2+b^2$$ must also be divisible by 5 in order their sum to be divisible by 5. Sufficient.

_________________

Kudos [?]: 129505 [0], given: 12201

Intern
Joined: 18 Mar 2012
Posts: 47

Kudos [?]: 270 [0], given: 117

GPA: 3.7
If a and b are positive integers [#permalink]

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25 Jan 2013, 06:09
If a and b are positive integers, is a^2 + b2^2 divisible by 5?

1) 2ab is divisible by 5
2) a-b is divisble by 5

How would you solve this question? Would you pick numbers or would you try algebra? Anyone know the algebraic way of solving this question?

Kudos [?]: 270 [0], given: 117

Math Expert
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129505 [0], given: 12201

Re: If a and b are positive integers [#permalink]

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25 Jan 2013, 06:37
alexpavlos wrote:
If a and b are positive integers, is a^2 + b2^2 divisible by 5?

1) 2ab is divisible by 5
2) a-b is divisble by 5

How would you solve this question? Would you pick numbers or would you try algebra? Anyone know the algebraic way of solving this question?

_________________

Kudos [?]: 129505 [0], given: 12201

Re: If a and b are positive integers   [#permalink] 25 Jan 2013, 06:37
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# M19 Q11

Moderator: Bunuel

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