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# M19-11

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Math Expert
Joined: 02 Sep 2009
Posts: 46047

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16 Sep 2014, 01:05
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Difficulty:

55% (hard)

Question Stats:

53% (01:37) correct 47% (01:01) wrong based on 45 sessions

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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

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Joined: 02 Sep 2009
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16 Sep 2014, 01:05
Official Solution:

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.

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Joined: 24 Jun 2015
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07 Jul 2015, 17:26
Bunuel wrote:
Official Solution:

(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.

Hi Bunuel,

Could you explain statement (2)...... Why If a=b=5 then the answer is YES? (5) -(5) = 0..... Does cero is a divisor of 5?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 46047

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08 Jul 2015, 02:17
1
luisnavarro wrote:
Bunuel wrote:
Official Solution:

(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.

Hi Bunuel,

Could you explain statement (2)...... Why If a=b=5 then the answer is YES? (5) -(5) = 0..... Does cero is a divisor of 5?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 5.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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Joined: 24 Jun 2015
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08 Jul 2015, 08:17
Bunuel wrote:
luisnavarro wrote:
Bunuel wrote:
Official Solution:

(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.

Hi Bunuel,

Could you explain statement (2)...... Why If a=b=5 then the answer is YES? (5) -(5) = 0..... Does cero is a divisor of 5?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

0 is not a divisor of any integer, but a multiple of every integer. So, 0 is divisible by 5.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html

So Cero is multiple of every integer (That is why is divisible by any integer except cero) but not a factor or divisor of any integer.

Thanks a lot Bunuel...

Luis Navarro
Looking for 700
Intern
Joined: 20 Feb 2017
Posts: 2

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07 Apr 2017, 04:36
What method are you using to derive this equation: (a−b)^2=(a^2+b^2)−2ab
Math Expert
Joined: 02 Sep 2009
Posts: 46047

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07 Apr 2017, 05:11
jklock1992 wrote:
What method are you using to derive this equation: (a−b)^2=(a^2+b^2)−2ab

This is one of the most basic algebraic property: $$(a−b)^2=a^2-2ab+b^2$$.
_________________
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Joined: 20 Feb 2017
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07 Apr 2017, 06:21
Thank you, couldn't see what was staring me right in the face.
Re: M19-11   [#permalink] 07 Apr 2017, 06:21
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# M19-11

Moderators: chetan2u, Bunuel

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