We have n>k>4Abdulla wrote:OA is A
We know that k is not a factor of n; in other words, n is not a multiple of k, and so we know that n does not evenly divide k. But we are dividing n by k, and we need to figure out whether the remainder will be bigger than 2.
(1) The greatest common factor of k and n is 4
If 4 is their greatest common factor but n is not divisible by k, then the remainder will always be 4.
Pick some numbers to see this.
Let k = 8, and let n = 12. Note that we can't let n =16 b/c then 8 would be their greatest common factor (and we would be violating the statement). If we divide 12 by 8, then the remainder is 4. Or, let k = 8, and n = 20. Then 20/8 gives a remainder of 4. Or, let k = 8, and n = 28, then 28/8 gives a remainder of 4.
At this point, you will have convinced yourself that the statement is sufficient.
(2) The least common multiple of k and n is 84.
This means that both k and n are factors of 84. The prime factorization of 84 is: 2^2 * 3 * 7
k and n can be any two numbers that we are able to "make" from this prime factorization, so long as their least comon multiple is 84 (and so long as n>k).
For example, n can be 3*7 = 21 while k can be 2^2 * 3 = 12. Then, 21/12 gives a remainder of 9, and the answer to the question is "yes". But n can be 2*7 = 14 while K can be 2^2 * 3 = 12. Then, 14/12 gives a remainder of 2, and the answer to the question is "no". Because we can get both a yes and no answer, the second statement is insufficient.
The first statement is independently sufficient while the second one is not.
Choose A.
