If x and y are positive integers, is xy a multiple of 8 ? The greatest common divisor of x...

Understanding the Question

We need to determine: Is \(\mathrm{xy}\) a multiple of 8?

Since \(8 = 2^3\), this question is really asking: Does the product \(\mathrm{xy}\) contain at least three factors of 2?

For sufficiency, we need a definitive "yes" or "no" answer - either \(\mathrm{xy}\) is always a multiple of 8, or it never is, based on the given information.

Key Insight

There's a fundamental relationship between gcd and lcm: \(\mathrm{gcd}(x,y) \times \mathrm{lcm}(x,y) = \mathrm{xy}\).

This formula will be crucial if we need to combine statements.

Analyzing Statement 1

Statement 1 tells us that \(\mathrm{gcd}(x,y) = 10 = 2 \times 5\).

This means both \(x\) and \(y\) are multiples of 10. We can write:

• \(x = 10a\) for some positive integer \(a\)
• \(y = 10b\) for some positive integer \(b\)
• where \(\mathrm{gcd}(a,b) = 1\) (since 10 is already their greatest common divisor)

Therefore: \(\mathrm{xy} = (10a)(10b) = 100ab = 2^2 \times 5^2 \times ab\)

We already have two factors of 2 from the 100. To reach three factors of 2 (making \(\mathrm{xy}\) divisible by 8), we need at least one more factor of 2 from the product \(ab\).

Let's test specific cases:

• Case 1: If \(a = 1\) and \(b = 1\)
  • Then \(x = 10\), \(y = 10\)
  • \(\mathrm{xy} = 100 = 2^2 \times 5^2\) → Only 2 factors of 2 → NOT divisible by 8
• Case 2: If \(a = 2\) and \(b = 1\)
  • Then \(x = 20\), \(y = 10\)
  • \(\mathrm{xy} = 200 = 2^3 \times 5^2\) → Has 3 factors of 2 → IS divisible by 8

Since we get different answers ("no" in Case 1, "yes" in Case 2), Statement 1 alone is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

Starting fresh - let's analyze Statement 2 independently.

Statement 2 tells us that \(\mathrm{lcm}(x,y) = 100 = 2^2 \times 5^2\).

The lcm gives us the highest powers of each prime that appear in either \(x\) or \(y\). However, it doesn't tell us:

• How these prime factors are distributed between \(x\) and \(y\)
• What their gcd is

Let's test different pairs that have lcm = 100:

• Example 1: \(x = 4 = 2^2\) and \(y = 25 = 5^2\)
  • \(\mathrm{lcm}(4,25) = 2^2 \times 5^2 = 100\) ✓
  • \(\mathrm{xy} = 4 \times 25 = 100 = 2^2 \times 5^2\) → Only 2 factors of 2 → NOT divisible by 8
• Example 2: \(x = 20 = 2^2 \times 5\) and \(y = 100 = 2^2 \times 5^2\)
  • \(\mathrm{lcm}(20,100) = 2^2 \times 5^2 = 100\) ✓
  • \(\mathrm{xy} = 20 \times 100 = 2000 = 2^4 \times 5^3\) → Has 4 factors of 2 → IS divisible by 8

Different pairs with the same lcm give different answers to our question. Statement 2 alone is NOT sufficient.

This eliminates choice B.

Combining Statements

Now we know both:

• \(\mathrm{gcd}(x,y) = 10 = 2 \times 5\)
• \(\mathrm{lcm}(x,y) = 100 = 2^2 \times 5^2\)

Using the fundamental relationship: \(\mathrm{gcd}(x,y) \times \mathrm{lcm}(x,y) = \mathrm{xy}\)

We can calculate exactly:
\(\mathrm{xy} = 10 \times 100 = 1000 = 2^3 \times 5^3\)

Since 1000 contains exactly three factors of 2, we have:
\(1000 = 8 \times 125\)

Therefore, \(\mathrm{xy}\) IS a multiple of 8. [STOP - Sufficient!]

With both statements together, we get a definitive "yes" answer. The statements combined are sufficient.

This eliminates choice E.

The Answer: C

Neither statement alone allows us to determine whether \(\mathrm{xy}\) is divisible by 8, but together they give us the exact value \(\mathrm{xy} = 1000\), which is divisible by 8.

Answer Choice C: "Both statements together are sufficient, but neither statement alone is sufficient."