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IMO B

143 = 11 X 13

this means x + y = 13 (or 11)
x - y = 11 (or 13)

Equating either one, we get the value of "Y" as -1 and +1

Hence Answer is B ie 0


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(x-y)(x+y)=143
(13)(11)= 143; (x-y)=13; (x+y)=11, solving both equations, x=12; y=-1
OR
(11)(13)=143; (x-y)=11; (x+y)=13, solving both equations, x=12; y =-1

In both cases, y+y = -1+-1 = -2.

Why does this logic not work here? Why is it assumed that y=+1 or -1?
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Bunuel
If (x − y)(x + y) = 143, and (x − y) and (x + y) are prime, what is the value of the sum of all possible values of y?

A. -1
B. 0
C. 1
D. 12
E. 24

Given (x-y) and (x+y) are prime, we could have (13)(11) = 143 or (11)(13) = 143.


\(x - y = 11\)
\(x + y = 13\)
\(x = 12, y = - 1, 1\)

\(x - y = 13\)
\(x + y = 11\)
\(x = 12, y = -1, 1\)

\(-1 + 1 = 0\)

Answer is B.
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yashvgupte
(x-y)(x+y)=143
(13)(11)= 143; (x-y)=13; (x+y)=11, solving both equations, x=12; y=-1
OR
(11)(13)=143; (x-y)=11; (x+y)=13, solving both equations, x=12; y =-1

In both cases, y+y = -1+-1 = -2.

Why does this logic not work here? Why is it assumed that y=+1 or -1?

\((x-y)=11\); \((x+y)=13\)

x = 12; y = 1

\((x-y)=13\); \((x+y)=11\)

\(x = 12\); \(y = -1\)

\(y = 1, -1\)
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Bunuel
If (x − y)(x + y) = 143, and (x − y) and (x + y) are prime, what is the value of the sum of all possible values of y?

A. -1
B. 0
C. 1
D. 12
E. 24
\((x − y)(x + y) = 143\)

Or, \((x − y)(x + y) = 11*13\)

Or, \((12 − 1)(12 + 1) = 11*13\) {x = 12 , y =1}

Hence, y can be -1 (As in 12 - 1 ) & y can be 1 (As in 12 + 1 )

Sum of all possible values of y = 0, Answer must be (B)
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Bunuel
If (x − y)(x + y) = 143, and (x − y) and (x + y) are prime, what is the value of the sum of all possible values of y?

A. -1
B. 0
C. 1
D. 12
E. 24
Solution:

Since 143 = 11 * 13, we could have 1) x - y = 11 and x + y = 13 OR 2) x - y = 13 and x + y = 11. In either case, let’s subtract the first equation from the second:

Case 1: 2y = 2 → y = 1

Case 2: 2y = -2 → y = -1

We see that the sum of the two possible values of y is 1 + (-1) = 0.

Answer: B
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