fitzpratik wrote:
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.
Alternative solution...
The question implicitly asks whether M or N or both are multiples of 3. So I approach it that way.
Also, observe that either of the given remainders (doesn't matter which) are odd after being divided by 6. Therefore, one of M or N must be odd, and the other must be even.
All even multiples of 3 are also multiples of 6. And all odd are multiples of 3 (duh). So let's say M may be the multiple of 6 (6m), and N may be the *ODD* multiple of 3 (3n, again note that n must be an odd number).
First consider whether M=6m. We see here that the second answer option gives a remainder of 3. The only possible way for M to be a multiple of 6 that, after being divided by 6, results in a remainder of 3, is if N=3n. However, under no circumstances will this ever result in a remainder of 5. Therefore, under no circumstances can M=6m.
So, can we be certain whether N=3n? Fist recall that N must be an even number and is NOT a multiple of 3. Here it IS possible that a remainder of 5 can be achieved. However, using the same prior logic, it's not possible to obtain a remainder of 3, after M is divided by 6, unless M=6m.
Therefore, using both A and B, we have sufficient info to show that neither M or N can be a multiple of 3. Sufficient.
...now rereading, I think my explanation is about as clear as mud....