If x, y, and z are consecutive integers, is `x+y+z` divisible by 3?

Answer

To determine if the sum of three consecutive integers \(x\), \(y\), and \(z\) is divisible by 3, we can analyze the problem step by step.

### Step 1: Define the consecutive integers
Let’s define the three consecutive integers. We can represent them in two ways:
1. **Increasing Order**: Let \(x = a\), \(y = a + 1\), and \(z = a + 2\).
2. **Decreasing Order**: Let \(x = a\), \(y = a - 1\), and \(z = a - 2\).

### Step 2: Calculate the sum for increasing consecutive integers
For the increasing consecutive integers:
\[ x + y + z = a + (a + 1) + (a + 2) \]
Simplifying this, we get:
\[ x + y + z = a + a + 1 + a + 2 = 3a + 3 \]

### Step 3: Calculate the sum for decreasing consecutive integers
For the decreasing consecutive integers:
\[ x + y + z = a + (a - 1) + (a - 2) \]
Simplifying this, we get:
\[ x + y + z = a + a - 1 + a - 2 = 3a - 3 \]

### Step 4: Check divisibility by 3
Now, we need to check if both sums are divisible by 3.

1. **For increasing integers**:
\[ 3a + 3 = 3(a + 1) \]
This expression is clearly divisible by 3.

2. **For decreasing integers**:
\[ 3a - 3 = 3(a - 1) \]
This expression is also clearly divisible by 3.

### Conclusion
Since both cases (increasing and decreasing consecutive integers) yield sums that are divisible by 3, we conclude that the sum \(x + y + z\) is indeed divisible by 3 for any set of three consecutive integers.

### Final Answer
Yes, \(x + y + z\) is divisible by 3.

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