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04 Jan 2025, 19:16
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (01:44) correct
66%
(01:32)
wrong
based on 152
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are integers such that \(x^2 + y^2 = 100\), how many different pairs of (x,y) satisfy this equation?
A. 2
B. 4
C. 8
D. 6
E. 12
A. 2
B. 4
C. 8
D. 6
E. 12
05 Jan 2025, 05:06
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HarshaBujji
Solving this question by brute force is easier. Note that \(x^2\) and \(y^2\) are non-negative integers. As the sum of the two terms is 100, each term must not exceed 10.
For each pair of (x,y) that satisfies the condition, we need to consider the positive and negative values.
So, let's start
1. x = 0
If x = 0 ⇒ \(y^2 = 100\)
\(y = \pm 10\)
Hence we have two pairs of (x,y) ⇒ (0,10) & (0,-10)
When the positions are swapped, we have two more pairs of (x,y) ⇒ (10, 0) and (-10, 0)
Therefore, 4 pairs of (x,y) until now.
2. x = 1
\(x^2 + y^2 = 100\)
\(y^2 = 99\)
y is not an integer. Ignore.
3. x = 2
\(x^2 + y^2 = 100\)
\(y^2 = 96\)
y is not an integer. Ignore.
4. x = 3
\(x^2 + y^2 = 100\)
\(y^2 = 91\)
y is not an integer. Ignore.
5. x = 4
\(x^2 + y^2 = 100\)
\(y^2 = 84\)
y is not an integer. Ignore.
6. x = 5
\(x^2 + y^2 = 100\)
\(y^2 = 75\)
y is not an integer. Ignore.
7. x = 6
\(x^2 + y^2 = 100\)
\(y^2 = 64\)
If \(x^2 = 36\), we can have two values of x, i.e. \(\pm 6\) ; similarly, y can have two values, \(\pm 8\)
Hence, we have total of 4 pairs here
(6, 8) ; (6, -8) ; (-6, 8) ; (-6, -8)
The same holds true for when \(x^2 = 64\) (i.e. swap the positions of x, and y)
(8, 6) ; (8, -6) ; (-8, 6) ; (-8, -6)
We don't need to test further as all the cases have been covered.
Total number of pairs of (x,y) = 8 + 4 = 12
Option E
07 Jan 2025, 14:06
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Agree with gmatophobia 's explanation!
Highlighting key traps to be careful of in the question:
1. Don't forget to test case where x or y can be 0 - 0 is very well an integer, and square of 0 is 0.
2. Don't forget that x and y can be negative - x^2 and y^2 will always be positive, however we have to find values of x and y
Highlighting key traps to be careful of in the question:
1. Don't forget to test case where x or y can be 0 - 0 is very well an integer, and square of 0 is 0.
2. Don't forget that x and y can be negative - x^2 and y^2 will always be positive, however we have to find values of x and y
