Bunuel wrote:
If a and b are positive integers, 2 < a, and 0 < b < 10, is the hundreds digit of x a perfect cube?
(1) x = (5a+b)(5a-b)
(2) a^2 + 84 = 20a
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTIONSolution: C.
Start by simplifying the question itself. The only three perfect cubes which could be expressed as a single digit are 0, 1 and 8, so the question is really asking us whether x = 0, 1 or 8.
Now consider the statements. Statement (1) is a difference of squares equation that gives us 25(a^2) - b^2 = x. Trying a few values for a and b shows that x could have many different hundreds digits: if a = 3 and b = 4, for example, then x = 225 - 16 = 209, but if a = 3 and b = 9, then x = 225 - 81 = 144; INSUFFICIENT. Statement (2) tells us nothing about x at all; INSUFFICIENT.
Statement (2) does give us two possible values for a, however: if we subtract 20a from both sides we have a quadratic equation -- a^2 - 20a + 84 = 0 -- with solutions of a = 6 and a = 14, respectively. Taking the two statements together, then, we have either:
a = 6, so x = (30+b)(30-b) = 900 - b^2
a = 14, so x = (70+b)(70-b) = 4900 - b^2
Since b is a single digit integer, b^2 must be between 0 and 100. This means that 900 - b^2 must be between 800 and 900, so its hundreds digit is 8; much the same is true of 4900 - b^2, which must be between 4800 and 4900, so its hundreds digit is also 8. SUFFICIENT; C.
for the explanation provided for STMT 2 , we do not what is 'x' in terms of 'a' and 'b' so this option is straight forward out.
Please let me know if i am missing something.