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If b is an integer, is sq(a^2+b^2) an integer?

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If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 26 Feb 2011, 15:45
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If b is an integer, is \(\sqrt {a^2+b^2}\) an integer?

(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 26 Feb 2011, 15:55
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banksy wrote:
If b is an integer, is sq(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a^2 – 3b^2 = 0


If b is an integer, is sqrt(a^2+b^2) an integer?

(1) a^2 + b^2 is an integer. Clearly insufficient: if a^2 + b^2 is a perfect square, say 4, then the answer is YES but if a^2 + b^2 is NOT a perfect square, say 5, then the answer is NO. Not sufficient.

(2) a^2 – 3b^2 = 0 --> \(a^2=3b^2\) --> \(\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2|b|=integer\). Sufficient.

Answer: B.
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 17 Mar 2016, 16:40
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Request to please format the question properly. "sq" does not imply sqrt and is misleading.

Thanks.
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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If b is an integer, is sqrt {a^2+b^2} an integer?  [#permalink]

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New post 24 Nov 2018, 08:05
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Project DS Butler: Day 19: Data Sufficiency (DS38)


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If b is an integer, is \(\sqrt {a^2+b^2}\) an integer?

(1) \(a^2 + b^2\) is an integer.
(2) \(a^2 – 3b^2\) = 0
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Re: If b is an integer, is sqrt {a^2+b^2} an integer?  [#permalink]

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New post 24 Nov 2018, 08:15
(1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer.
At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer.

Since we can get a Yes or No for the question, statement is Insufficient.

(2) \(a^2 = 3b^2\)
Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient.

IMO, Option B
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 24 Nov 2018, 11:48
nkin wrote:
(1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer.
At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer.

Since we can get a Yes or No for the question, statement is Insufficient.

(2) \(a^2 = 3b^2\)
Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient.

IMO, Option B



Hi nkin, I think didn't get when you Let a be \(\sqrt{3}\)


just double check if I understand your reasoning correctly, so when we plug in these values we get the following ?

\(\sqrt{\sqrt({3})^2+2^2}\) = \(\sqrt{3+4}\) = \(\sqrt{7}\)

Do i understand you correctly ? :?

thank you :)
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 24 Nov 2018, 11:52
Bunuel wrote:
banksy wrote:
If b is an integer, is sq(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a^2 – 3b^2 = 0


If b is an integer, is sqrt(a^2+b^2) an integer?

(1) a^2 + b^2 is an integer. Clearly insufficient: if a^2 + b^2 is a perfect square, say 4, then the answer is YES but if a^2 + b^2 is NOT a perfect square, say 5, then the answer is NO. Not sufficient.
(2) a^2 – 3b^2 = 0 --> \(a^2=3b^2\) --> \(\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2|b|=integer\). Sufficient.

Answer: B.


Bunuel can you pls explain how would you get 5 ?
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 24 Nov 2018, 11:58
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dave13 wrote:
nkin wrote:
(1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer.
At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer.

Since we can get a Yes or No for the question, statement is Insufficient.

(2) \(a^2 = 3b^2\)
Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient.

IMO, Option B



Hi nkin, I think didn't get when you Let a be \(\sqrt{3}\)


just double check if I understand your reasoning correctly, so when we plug in these values we get the following ?

\(\sqrt{\sqrt({3})^2+2^2}\) = \(\sqrt{3+4}\) = \(\sqrt{7}\)

Do i understand you correctly ? :?

thank you :)


dave13, yes that is correct. Statement 1 says \(a^2 + b^2\) is an integer, so here \(a^2 + b^2\) is 7, which is an integer but sqrt(a^2 + b^2) or sqrt(7) is not an integer as asked in the question stem. This is just to give an example where using statement (1), we are not able to answer the question stem definitively.
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Re: If b is an integer, is sq(a^2+b^2) an integer?  [#permalink]

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New post 24 Nov 2018, 12:01
1
dave13 wrote:
Bunuel wrote:
banksy wrote:
If b is an integer, is sq(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a^2 – 3b^2 = 0


If b is an integer, is sqrt(a^2+b^2) an integer?

(1) a^2 + b^2 is an integer. Clearly insufficient: if a^2 + b^2 is a perfect square, say 4, then the answer is YES but if a^2 + b^2 is NOT a perfect square, say 5, then the answer is NO. Not sufficient.
(2) a^2 – 3b^2 = 0 --> \(a^2=3b^2\) --> \(\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2|b|=integer\). Sufficient.

Answer: B.


Bunuel can you pls explain how would you get 5 ?


dave13, Here's what Bunuel is saying: If you would get 5, i.e. if a = 1 and b = 2, that a^2 + b^2 = 1 + 4 = 5, then 5 is an integer as per statement 1, but then again, sqrt (a^2 + b^2) = sqrt(5) is not an integer, as per the question stem's ask.
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Re: If b is an integer, is sq(a^2+b^2) an integer? &nbs [#permalink] 24 Nov 2018, 12:01
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