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If x and y are integers, which of the following CANNOT be th

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by Max@Math Revolution » Thu Jan 25, 2018 3:00 am
[GMAT math practice question]

If x and y are integers, which of the following CANNOT be the value of x^2+y^2?

A. 121
B. 122
C. 123
D. 125
E. 130
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by DrMaths » Thu Jan 25, 2018 7:41 am
The smallest Pythagorean triples are:
(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)
None of these include 11
Answer A is 11^2 = 121

As 11 is not in the list of triples, then A is impossible.
All the other values are not squares, so non-integer solutions are possible.
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by Brent@GMATPrepNow » Thu Jan 25, 2018 9:27 am
DrMaths wrote:The smallest Pythagorean triples are:
(3, 4, 5) (5, 12, 13) (8, 15, 17) (7, 24, 25)
None of these include 11
Answer A is 11^2 = 121

As 11 is not in the list of triples, then A is impossible.
All the other values are not squares, so non-integer solutions are possible.
'

Be careful; if x = 0 and y = 11, then we get: 0² + 11² = 121
Eliminate A

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by Brent@GMATPrepNow » Thu Jan 25, 2018 9:32 am
Max@Math Revolution wrote:[GMAT math practice question]

If x and y are integers, which of the following CANNOT be the value of x^2+y^2?

A. 121
B. 122
C. 123
D. 125
E. 130
Let's do this systematically:
0² = 0
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
11² = 121

A. 121 = 0² + 11² ELIMINATE A
B. 122 = 1² + 11² ELIMINATE B
C. 123 = can't eliminate
D. 125 = 10² + 5² ELIMINATE D
E. 130 = 9² + 7² ELIMINATE E

Answer: C

Cheers,
Brent
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by GMATGuruNY » Thu Jan 25, 2018 11:02 am
Max@Math Revolution wrote:[GMAT math practice question]

If x and y are integers, which of the following CANNOT be the value of x^2+y^2?

A. 121
B. 122
C. 123
D. 125
E. 130
Eliminate the four answer choices that CAN be the value of x²+y², where x and y are integers.
A: 121 = 11² + 0²
B: 122 = 11² + 1²
D: 125 = 11² + 2²
E: 130 = 11² + 3²
Eliminate A, B, D and E.

The correct answer is C.
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by Max@Math Revolution » Sun Jan 28, 2018 5:39 pm
=>
Squares of even integers, (2k)^2 = 4k^2 have the remainder 0, when they are divided by 4.

Squares of odd integers, (2k+1)^2 = 4k^2 + 4k + 1 have the remainder 1, when they are divided by 4.

Hence, squares of integers can have remainders of 0 or 1 only, when they are divided by 4. So, the sum of two squares of integers cannot have the remainder of 3 when it is divided by 4.

Thus, 123 cannot be the value of x^2+y^2.
Therefore, the answer is C.
Answer: C
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by Jeff@TargetTestPrep » Mon Jan 29, 2018 9:42 am
Max@Math Revolution wrote:[GMAT math practice question]

If x and y are integers, which of the following CANNOT be the value of x^2+y^2?

A. 121
B. 122
C. 123
D. 125
E. 130

Let's test our answer choices:

A) 121

121 = 0^2 + 11^2

B) 122

122 = 1^1 + 11^2

C) 123

There are no two perfect squares that sum to 123.

Answer: C

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