The least positive integer n such that `2015^(n) + 2016^(n) + 2017^(n)` is divisible by 10 is

Answer

To find the least positive integer \( n \) such that \( 2015^n + 2016^n + 2017^n \) is divisible by 10, we need to analyze the last digits of each term in the expression.

### Step 1: Identify the last digits of the bases
- The last digit of \( 2015 \) is \( 5 \).
- The last digit of \( 2016 \) is \( 6 \).
- The last digit of \( 2017 \) is \( 7 \).

### Step 2: Determine the last digits of the powers
1. **Last digit of \( 2015^n \)**:
- Any power of a number ending in \( 5 \) will also end in \( 5 \).
- Therefore, the last digit of \( 2015^n \) is \( 5 \) for all \( n \).

2. **Last digit of \( 2016^n \)**:
- Any power of a number ending in \( 6 \) will also end in \( 6 \).
- Therefore, the last digit of \( 2016^n \) is \( 6 \) for all \( n \).

3. **Last digit of \( 2017^n \)**:
- The last digit of \( 2017^n \) depends on \( n \):
- \( n = 1 \): last digit is \( 7 \)
- \( n = 2 \): last digit is \( 9 \)
- \( n = 3 \): last digit is \( 3 \)
- \( n = 4 \): last digit is \( 1 \)
- \( n = 5 \): last digit is \( 7 \) (and the pattern repeats every 4 terms)

### Step 3: Calculate the last digit of the sum
Now we need to find \( n \) such that the last digit of \( 2015^n + 2016^n + 2017^n \) is \( 0 \).

The last digit of the sum can be expressed as:
\[ \text{Last digit} = \text{Last digit of } 5 + \text{Last digit of } 6 + \text{Last digit of } 2017^n \]
\[ = 5 + 6 + \text{Last digit of } 2017^n \]
\[ = 11 + \text{Last digit of } 2017^n \]

To make this sum end in \( 0 \), we need:
\[ 11 + \text{Last digit of } 2017^n \equiv 0 \mod 10 \]
This simplifies to:
\[ \text{Last digit of } 2017^n \equiv -1 \equiv 9 \mod 10 \]

### Step 4: Find the smallest \( n \)
From our earlier analysis of \( 2017^n \):
- The last digit of \( 2017^n \) is \( 9 \) when \( n = 2 \).

Thus, the least positive integer \( n \) such that \( 2015^n + 2016^n + 2017^n \) is divisible by \( 10 \) is:
\[ \boxed{2} \]