If a and b are positive integer, is sqrt[3]{ab} an integer? sqrt(a) is an integer b=sqrt(a)...

Understanding the Question

We need to determine whether \(\sqrt[3]{\mathrm{ab}}\) is an integer, where \(\mathrm{a}\) and \(\mathrm{b}\) are positive integers.

For the cube root of \(\mathrm{ab}\) to be an integer, the product \(\mathrm{ab}\) must be a perfect cube. This means \(\mathrm{ab}\) must equal some integer cubed (like \(2^3 = 8\) or \(5^3 = 125\)).

What makes a statement sufficient? We need enough information to definitively answer YES (it's always an integer) or NO (it's not always an integer, or never is).

Key Insight

A perfect cube has a special property: when you break it down into prime factors, each prime appears a multiple of 3 times. For example:

• \(8 = 2^3\) (the prime 2 appears 3 times)
• \(216 = 2^3 \times 3^3\) (each prime appears 3 times)

Analyzing Statement 1

Statement 1 tells us: \(\sqrt{\mathrm{a}}\) is an integer

This means \(\mathrm{a}\) is a perfect square. We can write \(\mathrm{a} = \mathrm{k}^2\) for some positive integer \(\mathrm{k}\).

But here's the crucial question: What about \(\mathrm{b}\)? Since \(\mathrm{b}\) can be any positive integer, let's test whether different values of \(\mathrm{b}\) lead to different answers.

Testing specific cases:

• Case 1: Let \(\mathrm{a} = 4\) (since \(\sqrt{4} = 2\)) and \(\mathrm{b} = 2\)
  • Then \(\mathrm{ab} = 4 \times 2 = 8 = 2^3\)
  • So \(\sqrt[3]{8} = 2\) ✓ (integer)
• Case 2: Let \(\mathrm{a} = 4\) and \(\mathrm{b} = 1\)
  • Then \(\mathrm{ab} = 4 \times 1 = 4 = 2^2\)
  • So \(\sqrt[3]{4} = \sqrt[3]{2^2} \approx 1.587...\) ✗ (not an integer)

We found cases where the answer is YES and cases where it's NO.

Statement 1 alone is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Important: Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: \(\mathrm{b} = \sqrt{\mathrm{a}}\)

This creates an elegant relationship. Since both \(\mathrm{a}\) and \(\mathrm{b}\) are positive integers, and \(\mathrm{b} = \sqrt{\mathrm{a}}\), this means:

• \(\mathrm{b}\) must be a positive integer
• \(\mathrm{a}\) must be a perfect square
• Specifically, \(\mathrm{a} = \mathrm{b}^2\)

Now watch what happens when we calculate \(\mathrm{ab}\):

\(\mathrm{ab} = \mathrm{b}^2 \times \mathrm{b} = \mathrm{b}^3\)

This is beautiful! We're multiplying \(\mathrm{b}\) by itself three times, which automatically creates a perfect cube.

Therefore:

\(\sqrt[3]{\mathrm{ab}} = \sqrt[3]{\mathrm{b}^3} = \mathrm{b}\)

Since we know \(\mathrm{b}\) is a positive integer, the cube root is always an integer.

[STOP - Sufficient!] Statement 2 guarantees a YES answer every time.

Statement 2 alone is sufficient.

This eliminates choices C and E.

The Answer: B

Statement 2 alone provides sufficient information to answer the question with a definitive YES, while Statement 1 alone does not.

Answer Choice B: "Statement 2 alone is sufficient, but Statement 1 alone is not sufficient."