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1/X+1/Y=36/323. Find X-Y. (1) X and Y are primes (2)

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1/X+1/Y=36/323. Find X-Y. (1) X and Y are primes (2) [#permalink]  20 Jul 2003, 01:02
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1/X+1/Y=36/323. Find X-Y.

(1) X and Y are primes
(2) X>Y>1
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Re: DS: COMPLEX 6 [#permalink]  20 Jul 2003, 02:49
stolyar wrote:
1/X+1/Y=36/323. Find X-Y.

(1) X and Y are primes
(2) X>Y>1

Restating the question stem, we get:

(X + Y)/(XY) = 36/323

(1) restricts X and Y to prime integers. With little experimentation, we can discover that 323 = 19 * 17, two prime numbers. Just by inspection, we can see that X and Y can be 19 and 17 or 17 and 19 respectively and satisfy the equation. Moreover, since X and Y are primes, they are the ONLY solutions to the equation. However, we don't know which one is which so we cannot ascertain what X-Y is. Hence (1) is NOT sufficient and we must eliminate A and D.

(2) tells us that X > Y and both are positive. However, it does not restrict the solution to prime number, or even integers. Hence, there are probably an infinite number of solutions to the equation and (2) is also not sufficient and we must eliminate B.

Using both, (1) tells us that X and Y must be 17 & 19 or 19 and 17, and (2) tells us that X is larger, so we now know which solution is the correct one and we can answer the question. Hence, the correct answer is C.
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[#permalink]  20 Jul 2003, 22:13
It took 10 minutes to find the prime numners?
What shoud one do to bring the timing to 2 or 3 minutes?
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[#permalink]  21 Jul 2003, 09:20
That is why I call it a complex problem.
36 is a sum of two primes.
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[#permalink]  21 Jul 2003, 15:01
It took 10 minutes to find the prime numners?
What shoud one do to bring the timing to 2 or 3 minutes?

It shouldn't have taken you that long if you think logically and use a systematic method of ellimination. Just using commonly known rules of divisibility (or using quick long division for 7 and 11), you could quickly eliminate everything up trhough 11.

20*20 is too high so we know that at least one factor must be 13, 17, or 19. The last digit is 3 and 17 x 19 would have a last digit of 3 so the would be my first guess. Took about 45seconds.
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[#permalink]  22 Jul 2003, 23:43
Sorry guys, but I vote for 'B'

Logic is as follows:
simplifying 1/x+1/y=36/323 yields (x+y)/xy=36/323. We can solve it using a system of equations: a) x+y=36, b)xy=323.
From the system of equations four solutions are possible: x=17,19; y=17,19.

Now, from 'B' it follows that 1<y<x, therefore, it is sufficient.
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[#permalink]  22 Jul 2003, 23:52
Lynov Konstantin wrote:
Sorry guys, but I vote for 'B'

Logic is as follows:
simplifying 1/x+1/y=36/323 yields (x+y)/xy=36/323. We can solve it using a system of equations: a) x+y=36, b)xy=323.
From the system of equations four solutions are possible: x=17,19; y=17,19.

Now, from 'B' it follows that 1<y<x, therefore, it is sufficient.

Sorry, but (x+y)/xy = 36/323 does NOT mean a) x+y=36, b)xy=323. You are breaking up ONE equation and TWO unknowns into ONE PARTICULAR solution of TWO equations and TWO unknowns.

Example: a/b = 4/5. We cannot say that a=4 and b=5.

X and Y have unlimited solutions unless constrained to prime integers.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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[#permalink]  23 Jul 2003, 06:29
You got me Akamai. Looks like the skies are on your side today.
But choice B is so tempting...
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KL

[#permalink] 23 Jul 2003, 06:29
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