If x, y text{ and } z are integers greater than 1, what is the value of x + y...

Understanding the Question

We need to find the specific value of \(\mathrm{x} + \mathrm{y} + \mathrm{z}\), where x, y, and z are integers greater than 1.

Given Information

• x, y, and z are integers
• Each must be greater than 1 (so \(\mathrm{x} \geq 2, \mathrm{y} \geq 2, \mathrm{z} \geq 2\))
• We need the exact sum \(\mathrm{x} + \mathrm{y} + \mathrm{z}\)

What "Sufficient" Means Here

Since this is a value question, we need enough information to determine exactly one possible value for \(\mathrm{x} + \mathrm{y} + \mathrm{z}\). If we can find multiple different sums that satisfy the given constraints, the information is not sufficient.

Analyzing Statement 1

Statement 1 tells us: \(\mathrm{xyz} = 70\)

Let's think about this step by step. We have three integers (each ≥ 2) that multiply to give 70. Here's the key insight:

First, let's find the prime factorization of 70:
\(70 = 2 \times 5 \times 7\)

Notice something special? This is already the product of exactly three prime numbers!

When we need to express a number as a product of exactly three integers (each > 1), and that number happens to be the product of exactly three distinct primes, we're forced to use those three primes. Why? Because if we try to combine any two primes (like making \(10 = 2 \times 5\)), we'd only have two factors total, and we'd need to use 1 as the third factor—but that violates our constraint that all factors must be greater than 1.

Therefore, x, y, and z must be 2, 5, and 7 in some order. Regardless of the order:
\(\mathrm{x} + \mathrm{y} + \mathrm{z} = 2 + 5 + 7 = 14\)

[STOP - Sufficient!]

Statement 1 is sufficient.

This eliminates choices B, C, and E.

Analyzing Statement 2

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

Statement 2 tells us: \(\frac{\mathrm{x}}{\mathrm{yz}} = \frac{7}{10}\)

This gives us a ratio: x is to (yz) as 7 is to 10. Let's explore what this means by testing different scenarios:

Scenario 1: What if \(\mathrm{yz} = 10\)?

• Then \(\mathrm{x} = 7\) (since \(\frac{7}{10} = \frac{7}{10}\) ✓)
• We need y and z such that \(\mathrm{yz} = 10\) with both > 1
• The only option is \(\mathrm{y} = 2, \mathrm{z} = 5\) (or vice versa)
• Sum: \(\mathrm{x} + \mathrm{y} + \mathrm{z} = 7 + 2 + 5 = 14\)

Scenario 2: What if \(\mathrm{yz} = 20\)?

• Then \(\mathrm{x} = 14\) (since \(\frac{14}{20} = \frac{7}{10}\) ✓)
• We need y and z such that \(\mathrm{yz} = 20\) with both > 1
• This could be \(\mathrm{y} = 2, \mathrm{z} = 10\) (or vice versa)
• Sum: \(\mathrm{x} + \mathrm{y} + \mathrm{z} = 14 + 2 + 10 = 26\)

We've found two different possible sums (14 and 26), which proves that Statement 2 does not give us a unique value.

Statement 2 is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Statement 1 alone forces a unique factorization of 70 into three factors greater than 1, giving us exactly one possible sum. Statement 2 alone allows multiple different sums.

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