If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit...

Understanding the Question

We need to determine if the hundreds digit of x equals the sum of the hundreds digits of y and z, given that x, y, and z are three-digit positive integers and \(\mathrm{x = y + z}\).

Let's translate this: When we add two 3-digit numbers, does the hundreds digit of their sum equal the sum of their individual hundreds digits?

What We Need to Determine: Whether hundreds digit of x = hundreds digit of y + hundreds digit of z

The Key Insight: This question is really asking whether there's carrying from the tens place to the hundreds place during addition. If there's no carrying from tens to hundreds, the answer is YES. If there is carrying, the answer is NO (because the hundreds digit of x would be 1 more than the sum of the hundreds digits).

For this yes/no question to be sufficient, we need to determine definitively whether carrying occurs from tens to hundreds.

Analyzing Statement 1

Statement 1 tells us: The tens digit of x equals the sum of the tens digits of y and z.

This statement reveals something crucial about carrying. When we add two numbers:

• If the tens digits of y and z sum to less than 10, there's no carrying to hundreds
• If they sum to 10 or more, there would be carrying to hundreds, and the tens digit of x would be the sum minus 10

Since Statement 1 says the tens digit of x equals the direct sum of the tens digits (not reduced by 10), this can only happen if the tens digits sum to less than 10. Therefore, there's no carrying to the hundreds place.

With no carrying to hundreds, the hundreds digit of x MUST equal the sum of the hundreds digits of y and z.

The answer to our question is definitively YES.

[STOP - Statement 1 is Sufficient!]

Statement 1 alone 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: The units digit of x equals the sum of the units digits of y and z.

This means there's no carrying from the units place to the tens place (since if units digits summed to 10 or more, the units digit of x would be the sum minus 10).

However, this tells us nothing about what happens with the tens digits. The tens digits of y and z could still:

• Sum to less than 10 (no carrying to hundreds) → Answer is YES
• Sum to 10 or more (carrying to hundreds) → Answer is NO

Let's verify with concrete examples:

Example 1: \(\mathrm{y = 234, z = 145}\), then \(\mathrm{x = 379}\)

• Units check: \(\mathrm{4 + 5 = 9}\) ✓ (matches x's units digit)
• Tens: \(\mathrm{3 + 4 = 7}\) (less than 10, no carrying)
• Hundreds: \(\mathrm{2 + 1 = 3}\) ✓ (Answer: YES)

Example 2: \(\mathrm{y = 264, z = 185}\), then \(\mathrm{x = 449}\)

• Units check: \(\mathrm{4 + 5 = 9}\) ✓ (matches x's units digit)
• Tens: \(\mathrm{6 + 8 = 14}\) (≥ 10, causes carrying)
• Hundreds: \(\mathrm{2 + 1 = 3}\), but x has 4 due to carrying (Answer: NO)

Since we get different answers in different scenarios, Statement 2 alone cannot definitively answer our question.

Statement 2 alone is NOT sufficient. This eliminates choices B and D.

The Answer: A

Statement 1 alone tells us there's no carrying to the hundreds place, which definitively answers our question as YES.

Statement 2 alone cannot determine whether carrying occurs from tens to hundreds.

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