topspin360 wrote:Can someone please advise on how to solve this algebraically?

If m>n, then is mn divisible by 3?

(1) The remainder when m + n is divided by 6 is 5
(2) The remainder when m - n is divided by 6 is 3

OA is C

Statement 1:
In other words, m+n is equal to 5 more than a multiple of 6:
m+n = 5, 11, 17, 23, 29...
If m+n=5, the following cases are possible:
Case 1:m=3 and n=2
In this case, mn=6, which is divisible by 3.
Case 2: m=4 and n=1
In this case, mn=4, which is not divisible by 3.
INSUFFICIENT.

Statement 2:
In other words, m-n is equal to 3 more than a multiple of 6:
m-n = 3, 9, 15, 21, 27...
If m-n=3, the following cases are possible:
Case 1: m=6 and n=3
In this case, mn=18, which is divisible by 3.
Case 2: m=5 and n=2
In this case, mn=10, which is not divisible by 3.
INSUFFICIENT.

Statements combined:
Case 1: m+n=5 and m-n=3
Adding the equations, we get:
2m=8
m=4, implying that n=1.
In this case, mn=4, which is not divisible by 3.

Case 2: m+n=23 and m-n=9
Adding the equations, we get:
2m=32
m=16, implying that n=7
In this case, mn=16*7, which is not divisible by 3.

In each case, mn is not divisible by 3.
Maybe one more case to be safe.

Case 3: m+n=29 and m-n=21
Adding the equations, we get:
2m=50
m=25, implying that n=4.
In this case, mn=25*4, which is not divisible by 3.

The 3 random cases above illustrate that -- when the statements are combined -- mn in every case will not be divisible by 3.
SUFFICIENT.

The correct answer is C.

For mn to be divisible by 3, we should have either one of two as a multiple of 3

Statement#1

m+n = 6x+5................1

Insufficient by itself

Statement#2

m-n = 6y-3.............2

Insufficient by itself

Combining statement 1 & 2,
Using equation 1&2 and solving for m we get
m= 3x+3y+1
where, 3x is multiple of 3, 3y is multiple of 3 but with addition of 1 at the end whole sum is NOT multiple of 3, hence m is not multiple of 3

Again using equation 1 & 2 and solving for n we get,
n=3x-3y+4
Using same logic as above, we can deduce that n is not a multiple of 3

And since both M and N are not multiple of 3, MN can not be multiple of 3

Combining 1 & 2 Sufficient

thanks all!