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Math Expert V
Joined: 02 Sep 2009
Posts: 58421

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Difficulty:   55% (hard)

Question Stats: 58% (01:55) correct 42% (01:17) wrong based on 43 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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Math Expert V
Joined: 02 Sep 2009
Posts: 58421

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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.

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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 V
Joined: 02 Sep 2009
Posts: 58421

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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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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
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What method are you using to derive this equation: (a−b)^2=(a^2+b^2)−2ab
Math Expert V
Joined: 02 Sep 2009
Posts: 58421

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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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Thank you, couldn't see what was staring me right in the face.
Intern  B
Joined: 07 Oct 2018
Posts: 7

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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.

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58421

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1
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.

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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Joined: 07 Oct 2018
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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.

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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Joined: 18 Jul 2018
Posts: 16

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58421

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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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