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For how many integers pair (x,y) satisfies the result (1/x)+

Expert replies
by vipulgoyal » Mon Apr 27, 2015 6:16 pm
For how many integers pair (x,y) satisfies the result
(1/x)+((1/y)=1/12

a) 12
b) 6
c) 10
d) 16
e) 32

Kindly excuse me if this is not in proper format, bcz i copied it as it is from another thread , I had no clue for this, dont even know if this Q is flawed, experts plz consider if worth considering
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Source: — Problem Solving |

by regor60 » Tue Apr 28, 2015 6:03 am
nm
Last edited by regor60 on Tue Apr 28, 2015 7:27 am, edited 2 times in total.
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by GMATGuruNY » Tue Apr 28, 2015 7:12 am
I believe that the problem intends to ask the following:
vipulgoyal wrote:How many DISTINCT PAIRS OF INTEGERS satisfy the equation (1/x) + (1/y) = 1/12?

a) 12
b) 6
c) 10
d) 16
e) 32
This problem seems beyond the scope of the GMAT.
That said, here's one approach:

1/x + 1/y = 1/12

(x+y)/xy = 1/12

12x + 12y = xy

12x - xy + 12y = 0

(12x - xy + 12y) + (x-12)(y-12) = (x-12)(y-12)

(12x - xy + 12y) + (xy - 12x - 12y + 144) = (x-12)(y-12)

144 = (x-12)(y-12).

Implication:
(x-12) and (y-12) are FACTOR PAIRS OF 144.
Options:
x-12 = 1 and y-12 = 144 --> x=13, y=156
x-12 = 2 and y-12 = 72 --> x=14, y=84
x-12 = 3 and y-12 = 48 --> x=15, y=60
x-12 = 4 and y-12 = 36 --> x=16, y=48
x-12 = 6 and y-12 = 24 --> x=18, y=36
x-12 = 8 and y-12 = 18 --> x=20, y=30
x-12 = 9 and y-12 = 16 --> x=21, y=28
x-12 = 12 and y-12 = 12 --> x=24, y=24.
Total options = 8.

If (x-12) and (y-12) are NEGATIVE FACTOR PAIRS of 144, we get another 8 options:
x-12 = -1 and y-12 = -144 --> x=11, y=-132
x-12 = -2 and y-12 = -72 --> x=10, y=-60
x-12 = -3 and y-12 = -48 --> x=9, y=-36
x-12 = -4 and y-12 =-36 --> x=8, y=-24
x-12 = -6 and y-12 = -24 --> x=6, y=-12
x-12 = -8 and y-12 = -18 --> x=4, y=-6
x-12 = -9 and y-12 = -16 --> x=3, y=-4.
Total options = 7.

Thus:
Total distinct integer pairs = 8+7 = 15.

Since 15 is not among the answer choices, I would ignore this problem.
Last edited by GMATGuruNY on Mon May 04, 2015 1:58 am, edited 1 time in total.
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by Matt@VeritasPrep » Sun May 03, 2015 11:05 pm
Two issues here:

(1) It's not clear that "distinct" means that x = 13 and y = 156 is distinct from x = 156 and y = 13.

(2) Even if we assume that, we've still only got seven pairs in which exactly one of the numbers is negative. In the negatives, the only sets I see are {-132, 11}, {-60, 10}, {-36, 9}, {-24, 8}, {-12, 6}, {-6, 4}, and {-4, 3}. I think the mistaken assumption here is that the case corresponding to x = y in the positive set has a solution, but this gives x = 0, y = 0, and leads to division by zero.

So the answer, as far I can tell, should be either 15 or 29, depending on whether {x,y} is distinct from {y,x}. You know a problem is tough when its author doesn't even list the correct answer!
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