So here we can use value x = 1, since 1 divided by any number besides 1 will leave a remainder of 1. However, the next issue is finding the next value of X that will leave a remainder of 1 for all possible values of 2,4,6,8.

When dividing by 2, we know the number must be odd, when dividing by 4, we know the number must be above a multiple of (3,4), when dividing by 6 we know the number must be above a common multiple of (3,6), and when dividing by 8 we can follow the same pattern of (3,8). The common lowest multiple all 3 share is 24.

So if X = 25, all will leave a remainder of 1.

Leading us to the answer choice of D, as there are multiple values which can work.

The answer choice here can't be D. There are multiple choices for D which are all GREATER than 24. Carcass, it should be ideally A.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

Since x leaves remainder 1 when divided by 2,4,6 , and 8 , we need:


\(x \equiv 1 \quad(\bmod 2), \quad x \equiv 1 \quad(\bmod 4), \quad x \equiv 1 \quad(\bmod 6), \quad x \equiv 1 \quad(\bmod 8)
\)

The least common multiple of 2,4,6,8$ is 24 , so:


\(x=24 k+1\)


where k is a nonnegative integer.

So possible values of x include:


\(x=1,25,49, \ldots\)


If x=1$, then x<24 .
If x=25 , then x>24 .


Thus the relationship between Quantity A and Quantity B cannot be determined.

D The relationship cannot be determined from the information given.

damn, I missed the x=1 case, but it is pretty straight forward. 8 can divided be 8, 4, 2 but not 6, how ever 6 is 2*3 so we only need 8*3 to get the LCM so 25 is another case. but 1 < 24 then D