Bunuel wrote:
If a and b are integers and \((\sqrt[3]{a}*\sqrt{b})^6 = 500\), then a + b could equal
A. 2
B. 3
C. 4
D. 5
E. 6
Before people complain that the answer they have found is not in the suggestions of the problem prompt,
they need to remember that an integer can be either positive and negative. This was, in my opinion, the main trap with this problem.
We have: \((\sqrt[3]{a}*\sqrt{b})^6 = 500\) (A)
This equality is too complex for my tastes, let's simplify it while keeping in mind that:
1/ \(\sqrt{x} = x^\frac{1}{2}\)
2/ \(\sqrt[y]{x} = x^\frac{1}{y}\)
3/ \((a*b)^c = a^c * b^c\)
Therefore, using the first 3 properties above, A becomes:
\((\sqrt[3]{a}*\sqrt{b})^6 = (a^\frac{1}{3} * b^\frac{1}{2})^6 = a^2 * b^3 = 500\) (B)
If we decompose 500 into its prime components, we get \(500 = 2^2*5^3\)
Thus B can be written as: \(a^2 * b^3 = 2^2*5^3\)
At this stage, we're tempted to say that \(a = 2\) and \(b = 5\) and thus complain that \(a+b = 7\) isn't in the choices.
Or, we can remember that
even exponents remove the negative sign from negative integers and deduce that the only solution possible is \(a = -2\) and \(b = 5\) (since odd exponents keep the negative sign of negative integers and since 500 is positive, it follows that b can never be negative).
Thus, \(a + b = 3\) i.e. answer B.
close
65% (02:13) correct 
