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M19-11

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M19-11 [#permalink]

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New post 16 Sep 2014, 00:05
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Expert Post
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M19-11 [#permalink]

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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.


Answer: C
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Re: M19-11 [#permalink]

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New post 07 Jul 2015, 16:26
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.


Answer: C



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
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Re: M19-11 [#permalink]

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New post 08 Jul 2015, 01:17
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Expert's post
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.


Answer: C



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
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: M19-11 [#permalink]

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New post 08 Jul 2015, 07:17
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.


Answer: C



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
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Re: M19-11 [#permalink]

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New post 07 Apr 2017, 03:36
What method are you using to derive this equation: (a−b)^2=(a^2+b^2)−2ab
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Re: M19-11 [#permalink]

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New post 07 Apr 2017, 04:11
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Re: M19-11 [#permalink]

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New post 07 Apr 2017, 05:21
Thank you, couldn't see what was staring me right in the face.
Re: M19-11   [#permalink] 07 Apr 2017, 05:21
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