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Integers n for which n+19 is a multiple of n+3

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by Brent@GMATPrepNow » Sun Mar 30, 2014 6:51 pm
gmattesttaker2 wrote: How many positive integers n are there such that n+19 is a multiple of n+3?

A) 0
B) 1
C) 2
D) 3
E) 4
Given that the answer choices are so small, I'd just start testing positive (integer) values of n.

n = 1: Is (1+19) a multiple of (1+3)? YES
n = 2: Is (2+19) a multiple of (2+3)? NO
n = 3: Is (3+19) a multiple of (3+3)? NO
n = 4: Is (4+19) a multiple of (4+3)? NO
n = 5: Is (5+19) a multiple of (5+3)? YES
n = 6: Is (6+19) a multiple of (6+3)? NO
n = 7: Is (7+19) a multiple of (7+3)? NO
n = 8: Is (8+19) a multiple of (8+3)? NO
n = 9: Is (9+19) a multiple of (9+3)? NO
n = 10: Is (10+19) a multiple of (10+3)? NO
n = 11: Is (11+19) a multiple of (11+3)? NO
n = 12: Is (12+19) a multiple of (12+3)? NO
n = 13: Is (13+19) a multiple of (13+3)? YES
n = 14: Is (14+19) a multiple of (14+3)? NO
At this point, we should recognize that n+3 will always be more than half of n+19. As such, n+19 can not be a multiple of n+3 for values of n that are greater than 13. So, we can stop checking values.

Altogether, there are 3 values of n that meet the given condition.

Answer: D

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by GMATGuruNY » Sun Mar 30, 2014 8:28 pm
gmattesttaker2 wrote:Hello,

Can you please assist with this:

How many positive integers n are there such that n+19 is a multiple of n+3?

A) 0
B) 1
C) 2
D) 3
E) 4

OA: D
n+19 is a multiple of n+3.
Thus:
(n+19) / (n+3) = integer
[(n+3) + 16] / (n+3) = integer
(n+3)/(n+3) + 16/(n+3) = integer
1 + 16/(n+3) = integer.

For the resulting equation to be valid, (n+3) must be a factor of 16.
Options:
n+3 = 4, in which case n=1.
n+3 = 8, in which case n=5.
n+3 = 16, in which case n=13.
Thus, there are 3 possible values for n.

The correct answer is D.
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