Find the unit digit of `1^1 + 2^2 + 3^3 + …… 10^(10)`.

Answer

To find the unit digit of the expression \(1^1 + 2^2 + 3^3 + \ldots + 10^{10}\), we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the unit digit of each term**:
- **For \(1^1\)**: The unit digit is **1**.
- **For \(2^2\)**: \(2^2 = 4\), so the unit digit is **4**.
- **For \(3^3\)**: \(3^3 = 27\), so the unit digit is **7**.
- **For \(4^4\)**: \(4^4 = 256\), so the unit digit is **6**.
- **For \(5^5\)**: The unit digit is always **5**.
- **For \(6^6\)**: The unit digit is always **6**.
- **For \(7^7\)**: The unit digits of powers of 7 cycle every 4: \(7, 9, 3, 1\). Since \(7 \mod 4 = 3\), the unit digit is **3**.
- **For \(8^8\)**: The unit digits of powers of 8 cycle every 4: \(8, 4, 2, 6\). Since \(8 \mod 4 = 0\), the unit digit is **6**.
- **For \(9^9\)**: The unit digits of powers of 9 cycle every 2: \(9, 1\). Since \(9 \mod 2 = 1\), the unit digit is **9**.
- **For \(10^{10}\)**: The unit digit is always **0**.

2. **Summing the unit digits**:
Now we add the unit digits we calculated:
\[ 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 \]

3. **Perform the addition**:
- First, add \(1 + 4 = 5\)
- Then, \(5 + 7 = 12\)
- Next, \(12 + 6 = 18\)
- Then, \(18 + 5 = 23\)
- Next, \(23 + 6 = 29\)
- Then, \(29 + 3 = 32\)
- Next, \(32 + 6 = 38\)
- Finally, \(38 + 9 = 47\)
- The unit digit of \(47\) is **7**.

### Final Answer:
The unit digit of \(1^1 + 2^2 + 3^3 + \ldots + 10^{10}\) is **7**.