divisibility question

[jc114]

If k and n are integers, is n divisible by 7?
(1) n-3= 2k
(2) 2k-4 is divisible by 7

can someone explain this to me pls?

[jayhawk2001]

If k and n are integers, is n divisible by 7?
(1) n-3= 2k
(2) 2k-4 is divisible by 7

can someone explain this to me pls?

1 and 2 are clearly insufficient by themselves.

Using 1, n/7 = (2k + 3)/7

Using 2, 2k-4 = 7p (where p is an integer) i.e. k = (7p+4)/2

Using 2 in 1, n/7 = (2k+3)/7 = (2*(7p+4)/2 + 3)/7 = p + 1

1 + 2 is sufficient. Hence C

[Stacey Koprince]

Can also try using real numbers, if that's easier for you to follow. This is a yes/no question, so you basically try to get both answers - if you can, it's insufficient. If you keep getting the same answer, though, then you can conclude it is insufficient.

(1) If k=2, n=7 and yes n is div by 7. If k=3, n=9 and no n is not div by 7. Insufficient.

(2) This does not give us a connection between k and n, so can conclude nothing about n. Insufficient.

(1) + (2) If k=2, 2k-4=0, which is div by 7. If k=2, n=7, so yes n is div by 7. If k=3, 2k-4=2, which is not div by 7, so k cannot equal 3. Try some others and see if you can figure out a pattern. If k=4, 2k-4=4, so can't use k=4. If k=5, 2k-4=6, so can't use k=5. Every time I increase k by one, I increase 2k-4 by 2. So the next one that will work is k=9. 2k-4=14, so I can use k=9. If k=9, n=21, so yes n is div by 7.

Note that using this method, you aren't 100% sure - it's more that you've tried enough numbers to find a pattern, so you extrapolate. But this can be easier than the official math way - so just have to make a choice.

[gviren]

If k and n are integers, is n divisible by 7?
(1) n-3= 2k
(2) 2k-4 is divisible by 7


(1) Insuffiecient ---> k= (n-3)/2

(2) Insufficient ---> (2k-4)/7 = I where I is an integer
K = (7I + 4)/2

(1) + (2) together n-3 = 7I +4

n= 7 (I + 4) ---> n/7 = I + 4 which is an integer

[800GMAT]

Thanks jayhawk for the simple, clear and concise explanation!!!

[rajt]

1. Insufficient
2. Insufficient

from 1, n-3= 2k
n-3-4 = 2k - 4 (add -4 on either side)
=> n= (2k-4) + 7
from (2) 2k-4 is divisible by 7 and 7 divides 7 1 time.

So C.