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2^x + 2^y = x^2 + y^2, where x and y are non-negative integers.

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KarunMendiratta
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2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink] New post   16 Aug 2021, 08:37
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SUPER HARD

\(2^x + 2^y = x^2 + y^2\), where \(x\) and \(y\) are non-negative integers.

Quantity A
Quantity B
Greatest possible value of \(|x - y|\)
2


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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A
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KarunMendiratta
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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink] New post   18 Aug 2021, 20:53
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\(2^x + 2^y = x^2 + y^2\)

For \(|x - y|\) to be maximum, we must make \(x\) maximum and \(y\) minimum
So we can take \(y = 0\)

\(2^x + 2^0 = x^2 + 0^2\)
\(2^x + 1 = x^2\)

Put \(x = 1\)
\(2^1 + 1 = 1^2\)
\(3 ≠ 1\)

Put \(x = 2\)
\(2^2 + 1 = 2^2\)
\(5 ≠ 4\)

Put \(x = 3\)
\(2^3 + 1 = 3^2\)
\(9 = 9\)

Check no more!

Col. A: |3 - 0| = 3
Col. B: 2

Hence, option A
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2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink] New post   16 Aug 2021, 09:30
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given question's stem, compare Quantity A with Quantity B for 4 choices greater/less/equal/different always

\(2^x+2^y+2xy=(x+y)^2\)
this is possible when x=y=2 ---> \(4+4+2*2*2=(2+2)^2=16\) making Quantity A < Quantity B
and, when x=4, y=2 ---> \(16+4+2*4*2=(4+2)^2=36\) making Quantity A = Quantity B
Hence, answer must be D
KarunMendiratta wrote:
SUPER HARD

\(2^x + 2^y = x^2 + y^2\), where \(x\) and \(y\) are non-negative integers.

Quantity A
Quantity B
Greatest possible value of \(|x - y|\)
2


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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KarunMendiratta
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Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3277 [1]
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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink] New post   18 Aug 2021, 20:47
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motion2020 wrote:
given question's stem, compare Quantity A with Quantity B for 4 choices greater/less/equal/different always

\(2^x+2^y+2xy=(x+y)^2\)
this is possible when x=y=2 ---> \(4+4+2*2*2=(2+2)^2=16\) making Quantity A < Quantity B
and, when x=4, y=2 ---> \(16+4+2*4*2=(4+2)^2=36\) making Quantity A = Quantity B
Hence, answer must be D
KarunMendiratta wrote:
SUPER HARD

\(2^x + 2^y = x^2 + y^2\), where \(x\) and \(y\) are non-negative integers.

Quantity A
Quantity B
Greatest possible value of \(|x - y|\)
2


A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


I think you are missing out on - Greatest possible value!
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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink] New post   14 Nov 2024, 01:33
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers. [#permalink]
14 Nov 2024, 01:33
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