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Joined: 16 Apr 2020
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Location: India
WE:Education (Education)
2^x + 2^y = x^2 + y^2, where x and y are non-negative integers.
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16 Aug 2021, 08:37
16 Aug 2021, 08:37
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Question Stats:
21% (02:29) correct
78% (02:31) wrong based on 147 sessions
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SUPER HARD
\(2^x + 2^y = x^2 + y^2\), where \(x\) and \(y\) are non-negative integers.
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
\(2^x + 2^y = x^2 + y^2\), 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
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: 2^x + 2^y = x^2 + y^2, where x and y are non-negative integers.
[#permalink]
18 Aug 2021, 20:53
18 Aug 2021, 20:53
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3
Bookmarks
\(2^x + 2^y = x^2 + y^2\)
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
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
General Discussion
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
2^x + 2^y = x^2 + y^2, where x and y are non-negative integers.
[#permalink]
16 Aug 2021, 09:30
16 Aug 2021, 09:30
3
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
\(2^x+2^y+2xy=(x+y)^2\)
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.
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
\(2^x + 2^y = x^2 + y^2\), 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
