If a and b are integers, is ab/8 an integer? (1) 2a=b (2) ab/4 is an

Very nicely designed question, though I think it would be more realistic if it said a and b were positive integers, since almost all GMAT divisibility questions are restricted to positive integers.

The answer is quickly C or E. From Statement 1, we know a is divisible by one more '2' than b is. It's a bit like knowing "Azim has 1 more apple than Brenda". The total number of 2's in ab (or apples that Azim and Brenda have in total) then must be odd. Statement 2 tells us we can divide ab by 2^2, so ab contains at least two 2's, but if it also contains an odd number of 2's, it must contain at least three 2's, and must be divisible by 8.

Less abstractly, if b = 2a, substituting into Statement 2 we learn 2a^2/4 is an integer, so 2a^2 is divisible by 4, and a^2 is divisible by 2. If a^2 is divisible by 2, then a is divisible by 2, and if b = 2a, b must be divisible by 4. So ab must be divisible by 8, and the answer is C.

If a and b are integers, is ab/8 an integer?

Stat1: 2a=b
Not sufficient, as a = 1, b = 2 and ab/8 won't be an integer in some cases. Not sufficient

Stat2: ab/4 is an integer.
a = 1 and b=4, still ab/8 won't be an integer in some cases. Not sufficient

Combining both, if a = 2, b =4, ab/8 will be an integer sufficient

So, I think C.

1)\(\frac{2a^2}{8 }= integer?\)
If a = 2, Yes!
If a = 3, No! - Insufficient!
2)ab/4 = integer!
if ab = 8, Yes!
If ab = 12, No! - Insufficient!
Combining (1) & (2):
\(\frac{2a^2}{4} = integer\)
For the above to be an integer, the minimum value of a has to be 2.
So, this will always be a divisible as 2a^ will be 8. Sufficient!
IMO, (C)!

Very tricky question. I got this wrong but realized my mistake. Sharing my learning here.

S1: 2a=b [Insufficient]

If we plug the value of \(b\) into \(\frac{ab}{8}\) then we get \(\frac{2a^2}{8}\) or \(\frac{a^2}{4}\). Now if \(a = 1\) then the entire expression is not an integer but if \(a = 4\) then it is

S1: ab/4 is an integer [Insufficient]

\(\frac{ab}{4} = k\) where \(k\) is an integer. We can multiply both sides with \(\frac{1}{2}\) to get \(\frac{ab}{8} = \frac{k}{2}\). Now if \(k = 1\) then the equation is not an integer but if \(k = 4\) then it is

S1 + S2 [Sufficient]

Plugging \(b = 2a\) in \(\frac{ab}{8} = \frac{k}{2}\) we get \(\frac{2a^2}{8} = \frac{k}{2}\) which upon further simplification gives \(a = \sqrt{2k}\)

Now since \(a\) is an integer \(\sqrt{2k}\) must be an integer so \(k =\) \(2\), \(8\) etc.

Taking the smallest value \(k = 2\) gives \(a = 2\) and \(b = 4\) (from \(b = 2a\))

Plugging these values into \(\frac{ab}{8}\) gives \(\frac{2 * 4}{8} = 1\) which is an integer

Ans. C

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