If a and b are integers and (3√a*√b)^6 = 500, then a + b could equal

By solving LHS, we get a^2*b^3=500
We know 500 = 2^2*5^3

Hence, a=2 b=5
a+b = 7 (Not in the options)

a^2 * b^3 = (-2)^2 * 5^3 = 2^2 * 5^3

a + b = 3/7

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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.

We can rewrite the expression:

[(a^1/3) x (b^1/2)]^6 = 500

a^2 x b^3 = 500

a^2 x b^3 = 2^2 x 5^3

Thus a could be 2 and b could be 5, and a + b = 7. However, notice that a^2 = 2^2 = 4, so a could be -2 also. In that case, a + b = -2 + 5 = 3.

Answer: B

The given equation can be written as,

\(a^2 * b^3 = 5^3 * 2^2 = 5^3 * (-2)^2\)

So, \(a + b = 7\) or \(3.\) Hence, option B.

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\((\sqrt[3]{a}*\sqrt{b})^6 = [\sqrt[3]{a}]^6 * [\sqrt{b}]^6 = [a]^{\frac{6}{3}} * [b]^{\frac{6}{2}} = a^2 * b^3 = 500.\)

\(500 = (2)^2 * (5)^3 = (-2)^2 * (5)^3\)

\((a+b)\) could equal \((2+5) = 7\), or \((-2+5) = 3\). Ans - B.

\(500 = 5^3* 2^2\)
Also\(( a^(1/3) *b^(1/2))^6 = a^2*b^3\)
Hence a can be 2, b can be 5.
But a+b =7.
It is not in the option.

Bingo: we have forgot the fact that even power changes negative to positive.
Actually a can be +/- 2
Now considering a = -2, b =5
Hence a+b = -5+2 =3. Answer B

\((\sqrt[3]{a}*\sqrt{b})^6 = 500\)

Or,\( a^2 *b^3 = 2^2*5^3\)

Hence, \(a =\) +\(2\) & \(b = 5\)

Thus, \(a + b = 3\) , Answer must be (B)

a^2 b^3=5^3 2^2
a=2 or-2, b=5
a+b=7 or 3

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