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j^2 – k^2 = 343 . . .

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j^2 – k^2 = 343 . . .

by Brent@GMATPrepNow » Sun Feb 22, 2009 8:41 am
If x and y are integers such that 0 < k < j < 75, and j^2 – k^2 = 343, then j =
A) 7
B) 14
C) 21
D) 28
E) 49
Please note that this is not an official GMAT question; it’s my attempt to create difficult (650+ level) GMAT-style questions for this forum.
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by sureshbala » Sun Feb 22, 2009 9:01 am
(j+k)(j-k) = 343.
(j+k)(j-k)= 49x7 (it can't be 343x1 because 0<k<j<75)
So j+k = 49
and j-k = 7

Adding these two equations we get j=28
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by bmlaud » Sun Feb 22, 2009 9:02 am
The question is difficult to solve in 2 mins time

(j+k) (j-k) = 7*49 =343

j+k=49
j-k=7 => 2j = 56 or j = 28
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by Brent@GMATPrepNow » Sun Feb 22, 2009 9:25 am
Nice work - the answer is D.
Here's my solution as well:
First recognize that 343 = 7×7×7

Factor both sides of the original equation to get (j + k)(j – k) = 7×7×7
Notice that we essentially have (?)(?) = 7×7×7

If the two values on the brackets must be integers, there aren’t many possible solutions. We could have (1)(343) = 343, (343)(1) = 343, (49)(7) = 343 or (7)(49) = 343
If 0 < k < j < 75, we can conclude that j + k = 49 and j – k = 7

We now have a system of two equations and two unknowns:
(1) j + k = 49
(2) j – k = 7

Add (1) and (2) to get 2j = 56, which means that j = 28

The correct answer is D.
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