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M19-11

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M19-11  [#permalink]

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New post 16 Sep 2014, 01:05
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A
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C
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E

Difficulty:

  55% (hard)

Question Stats:

58% (01:55) correct 42% (01:17) wrong based on 43 sessions

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M19-11  [#permalink]

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New post 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.


Answer: C
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Re: M19-11  [#permalink]

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New post 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.


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, 02:17
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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
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Re: M19-11  [#permalink]

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New post 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.


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
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Re: M19-11  [#permalink]

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New post 07 Apr 2017, 04:36
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, 05:11
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Re: M19-11  [#permalink]

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New post 07 Apr 2017, 06:21
Thank you, couldn't see what was staring me right in the face.
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M19-11  [#permalink]

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New post 05 Feb 2019, 08:40
Bunuel wrote:
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


I could be wrong and overlooking something but I think 2 alone is sufficient. If a-b is divisible by 5 shouldn't a^2+b^2 be divisible by 5.

By factoring: a^2 + b^2 = (a+b)(a-b) and if a-b is divisible by 5 then (a+b)(a-b) should be too.

Could someone help me see where I am going wrong?
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Re: M19-11  [#permalink]

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New post 05 Feb 2019, 10:04
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bcal95 wrote:
Bunuel wrote:
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


I could be wrong and overlooking something but I think 2 alone is sufficient. If a-b is divisible by 5 shouldn't a^2+b^2 be divisible by 5.

By factoring: a^2 + b^2 = (a+b)(a-b) and if a-b is divisible by 5 then (a+b)(a-b) should be too.

Could someone help me see where I am going wrong?


\((a+b)(a-b)=a^2-b^2\) not \(a^2 + b^2\).
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Re: M19-11  [#permalink]

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New post 05 Feb 2019, 21:27
Bunuel wrote:
bcal95 wrote:
Bunuel wrote:
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


I could be wrong and overlooking something but I think 2 alone is sufficient. If a-b is divisible by 5 shouldn't a^2+b^2 be divisible by 5.

By factoring: a^2 + b^2 = (a+b)(a-b) and if a-b is divisible by 5 then (a+b)(a-b) should be too.

Could someone help me see where I am going wrong?


\((a+b)(a-b)=a^2-b^2\) not \(a^2 + b^2\).


Thanks! I feel a little silly now.

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Re: M19-11  [#permalink]

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New post 30 Jul 2019, 21:43
I got this wrong, but I'm trying to figure out how close I got to the right answer.

My thought process:

(1) 2ab is divisible by 5: "something" in the product of a*b is a factor of 5 - could be "a", "b", or both; since we don't know for sure, it's not sufficient

(2) a-b is divisible by 5: implies "a" and "b" both have a factor of 5, so a^2 + b^2 is also divisible by 5.

I realized one of my mistakes wasn't considering the "1" and "0" cases for (2).

Right but insufficient thinking or just completely wrong?
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Re: M19-11  [#permalink]

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New post 30 Jul 2019, 21:49
rwx5861 wrote:
I got this wrong, but I'm trying to figure out how close I got to the right answer.

My thought process:

(1) 2ab is divisible by 5: "something" in the product of a*b is a factor of 5 - could be "a", "b", or both; since we don't know for sure, it's not sufficient

(2) a-b is divisible by 5: implies "a" and "b" both have a factor of 5, so a^2 + b^2 is also divisible by 5.

I realized one of my mistakes wasn't considering the "1" and "0" cases for (2).

Right but insufficient thinking or just completely wrong?


a - b will be divisible by 5 not only when both a and b are multiples of 5. For example, a = 7 and b = 2.
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Re: M19-11   [#permalink] 30 Jul 2019, 21:49
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