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Positive integer n is equal to the difference of the squares
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25 Jun 2020, 23:31
25 Jun 2020, 23:31
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70% (02:06) correct
30% (01:49) wrong based on 20 sessions
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Positive integer n is equal to the difference of the squares of x and y, where x and y are integers. If n is equal to 21, then which of the following could be the sum of x and y?
A. 0
B. 11
C. -1
D. -11
E. 121
A. 0
B. 11
C. -1
D. -11
E. 121
Re: Positive integer n is equal to the difference of the squares
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26 Jun 2020, 07:57
26 Jun 2020, 07:57
Expert Reply
(x^2-y^2) = 21
=> (x+y)(x-y) = 21
Since x and y are integers
we can have following possibilities
7*3
3*7
21*1
1*21
-1*-21
-21*-1
out of the given options (-1) is the only value that fits into the solution.
=> (x+y)(x-y) = 21
Since x and y are integers
we can have following possibilities
7*3
3*7
21*1
1*21
-1*-21
-21*-1
out of the given options (-1) is the only value that fits into the solution.
Re: Positive integer n is equal to the difference of the squares
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26 Jun 2020, 08:46
26 Jun 2020, 08:46
1
Farina wrote:
Positive integer n is equal to the difference of the squares of x and y, where x and y are integers. If n is equal to 21, then which of the following could be the sum of x and y?
A. 0
B. 11
C. -1
D. -11
E. 121
A. 0
B. 11
C. -1
D. -11
E. 121
n= x2 - y2 (or y2-x2 but it wont make a difference in this case)
x2-y2 = (x-y)(x+y)
21= (x-y)(x+y)
x-y and x+y must be integers (since x and y are integers)
therefore,
21/(x+y) must produce an integer
only possible value for x+y to produce an integer from the given choices is: x+y=-1
