quanthelp85 wrote:Given b is integer, is 3b+1 divisible by 10?
(1) b = 4c + 2
(2) b > 4
Dear
quanthelp85
This is a great question, and I am happy to help. First of all, let's just look at some values of b, and see which choices make 3b+1 divisible by 10 ---
b = 1 --> 3b+1 = 4, not divisible by 10
b = 2 --> 3b+1 = 7, not divisible by 10
b = 3 --> 3b+1 = 10, YES, divisible by 10
b = 4 --> 3b+1 = 13, not divisible by 10
b = 5 --> 3b+1 = 16, not divisible by 10
b = 6 --> 3b+1 = 19, not divisible by 10
It looks as if the answer for most b's is NO. The value b = 3 works, and leaping ahead, I see that the values of b = 13, b = 23, b = 33, etc. will also work. So, really, the prompt question is: does b have a units digit of 3?
Statement (1):
b = 4c + 2
Very tricky. We know, in order to get a YES answer to the prompt, we need b to have a units digit of 3, which makes b an odd number. If c were an integer, then 4c+2 would have to be even, and could never be odd. BUT, and this a huge BUT --- we have absolutely no reason to assume that c is an integer, and to fall into that trap, we would be making one of the most common mistakes on the entire GMAT Quant section. See:
https://magoosh.com/gmat/2013/how-to-plu ... questions/
We have no restrictions on c --- it could be any kind of number. For example, if c = 1/4, then b = 3, which gives a YES answer to the prompt, but if c = 1, then b = 6, which gives a NO answer. Different choices, nothing is
sufficient.
Statement (2):
b > 4
b = 13, we get a YES answer.
b = 20, we get a NO answer.
Different choices, nothing is
sufficient.
Combined: the problem is, statement #1 is really a piece of non-information: because c could be any kind of number, b = 4c + 2 could be made to equal absolutely any positive integer. This statement places absolutely no restriction on the value of b. Thus, even with combined statements, the only real restriction we have is #2, and we saw, that was hardly enough to determine an answer. Nothing is conclusive here, nothing is sufficient.
OA = [spoiler]
(E)[/spoiler]
Does this make sense?
Mike
