Statement 1:
n-3=2k
-> n=2k+3

if k = 2, n = 7 YES
if k = 3, n = 9 NO

Insufficient

Statement 2:
2k-4 is divisible by 7

Insufficient since we don't learn anything about n.

Together:
n=2k+3 and 2k-4 is divisible by 7

Now see that when we add 7 to 2k-4, we get 2k+3.
So 2k-4, which is divisible by 7, plus 7, which will also be divisible by 7, is 2k+3.

Since n = 2k+3, it will be divisible by 7.

Sufficient.

(Example: If 2k-4 = 14, then 2k+3=21 and both are divisible by 7.)

tmmyc wrote:Statement 1:


Now see that when we add 7 to 2k-4, we get 2k+3.
So 2k-4, which is divisible by 7, plus 7, which will also be divisible by 7, is 2k+3.

Since n = 2k+3, it will be divisible by 7.

Sufficient.

(Example: If 2k-4 = 14, then 2k+3=21 and both are divisible by 7.)

THANK YOU! I finally understand the way you explained it! The book had similar explanation but I couldn't figure out what they were trying to say.

We know that individually each is insufficient.

Consider both statements.

n = 2k + 3 --------- (1)

2k - 4 is divisible by 7

2k - 4 = 7a (for some integer a)

=> 2k = 7a +4 ------------------(2)

substitute 2 in 1
n = 7a +4 +3
n= 7a +7
n = 7(a+1) a+1 is another integer say b.
n = 7b

This implies n is divisible by 7. So C.

IMO C:

n= 2k + 3
n= 5 , 7, 11,

Not suff

2. 2k-4 = n
n = 0, 2, 4, ..

Not suff


But from 1 and 2

n = 7( m+1)

Hence n multiple of 7