knight247 wrote:Firstly, make it easier on the eyes by multiplying throughout by 12. So we have,
2k+3m=t
(1)k is a multiple of 3 so 2k has to be a multiple of 3. Remember one ground rule,
If two multiples of a certain number are added/subtracted then the resultant number is also a multiple of the same number. So we have Multiple of 3+Multiple of 3=Multiple of 3. Hence, t is a multiple of 3. And, 12 is also a multiple of 3. So, t and 12 do have a common factor greater than 1 i.e. 3. Sufficient
(2)m is a multiple of 4 so 3m is a multiple of 4. Remember another ground rule, When one multiple of a certain number is added/subtracted to a non multiple the resultant number is NEVER a multiple of the first number.
So we have, 2k+Some multiple of 4=t . Now, 2k might have another 2 in it or it might not. We can't say for certain. All we know is that 2k is multiple of 2 and NOTHING MORE. So we can assume that t would be a multiple of 4 if 2k was a multiple of 4.But t would definitely not be a multiple of 4 if 2k is not a multiple of 4. Since, we get conflicting possibilities from b this statement is insufficient.
Hence A
P.S. The ground rule I mentioned in statement 2 has an exception to it. The number 2. When two odd number are added the resultant number is even which is obviously divisible by 2.
thanks for the post..
I have one doubt.
If an even number (2k) and multiple of 4 added , the result will be an even number which will definitely be a multiple of 2, how can we say option 2 is insufficient?
Am I understanding that exception to the rule in a proper sense?
________
Samuel
