multiple of 7

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by gmatnewbe » Mon May 10, 2010 5:00 am
Is 5a-6b is a multiple of 7 ?? a & b are integers
1 : 3a+5b is a multiple of 7
2 : 5a + 3b is a multiple of 7

For me, the ans is C. I welcome vetran members to verify.
Most important part here is a and b both are integers.
1: not sufficient alone as we dont know the values of a and b. we can fit numbers though but there are multiple options like a=1,b=5.
2: Again not sufficient alone as we dont know the values of a and b.
Combine 1+2: when 2 numbers are multiple of a number then there sum and their differences too are multiples of that number. you can fit examples and it is independent of the negative or positive sign.
Add both eq to get: 8a+8b-> 8(a+b). now 8 is not multiple of 7 so a+b must be the multiple of 7.
Subtract eq to get: 2a-2b-> 2(a-b), again here a-b is the multiple of 7 as 2 is not.
Again same fundamental and we can add and substract the new eq so (a+b) + (a-b) = 2a--> a is multiple of 7, (a+b)-(a-b)=2b--> b is multiple of 7, so all m*a+/-n*b will be multiple of 7.

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by debmalya_dutta » Mon May 10, 2010 4:34 pm
IMO C
using both the statements, it can be derived that a & b are both multiples of 7
=> 5a-6b is a multiple of 7

s1:3a+5b = 7k1
s2:5a + 3b= 7k2

s2-s1

5a + 3b - (3a+5b) = 7(k2-k1)
=> 2a - 2b = 7(k2-k1) or 7k3
=> a-b = 7 * constant

now from the above derived equation
5a-5b=7*5 * constant
5a-5b=7*constant -call this eq 1

Statement 2 - eq 1
5a + 3b - (5a-5b) = 7k2 - [7*constant]
8b = 7 * k3 (say a constant)
=> b is a multiple of 7
similarly you can derive that a is a multiple of 7

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by 2011mbaspirant » Wed May 12, 2010 1:47 pm
I think the answer should be E.
@debmalya- If a+b is a multiple of 7 then all values of a and b that sum together as multiple of 7 should satisfy the condition for 5a-6b to be a multiple of 7 as well. But that is not the case.
Please correct me if I am wrong. We need an expert response to this tricky question.