What is the unit digit of `1^(6)+2^(6)+....3^(6)+...+20^(6)`?

Options

A. 0
B. 1
C. 2
D. 4

Answer

To find the unit digit of the expression \(1^6 + 2^6 + 3^6 + \ldots + 20^6\), we will first determine the unit digits of \(n^6\) for \(n = 1\) to \(20\) and then sum them up.

### Step-by-Step Solution:

1. **Calculate the unit digit of \(n^6\) for \(n = 1\) to \(10\)**:
- **For \(n = 1\)**: \(1^6 = 1\) → Unit digit is **1**
- **For \(n = 2\)**: \(2^6 = 64\) → Unit digit is **4**
- **For \(n = 3\)**: \(3^6 = 729\) → Unit digit is **9**
- **For \(n = 4\)**: \(4^6 = 4096\) → Unit digit is **6**
- **For \(n = 5\)**: \(5^6 = 15625\) → Unit digit is **5**
- **For \(n = 6\)**: \(6^6 = 46656\) → Unit digit is **6**
- **For \(n = 7\)**: \(7^6 = 117649\) → Unit digit is **9**
- **For \(n = 8\)**: \(8^6 = 262144\) → Unit digit is **4**
- **For \(n = 9\)**: \(9^6 = 531441\) → Unit digit is **1**
- **For \(n = 10\)**: \(10^6 = 1000000\) → Unit digit is **0**

2. **List the unit digits from \(1\) to \(10\)**:
- The unit digits are: \(1, 4, 9, 6, 5, 6, 9, 4, 1, 0\)

3. **Calculate the sum of these unit digits**:
- \(1 + 4 + 9 + 6 + 5 + 6 + 9 + 4 + 1 + 0 = 45\)

4. **Determine the unit digit of the sum**:
- The unit digit of \(45\) is **5**.

5. **Since the unit digits repeat every \(10\) numbers**, we can conclude that the unit digits for \(11\) to \(20\) will be the same as for \(1\) to \(10\):
- Therefore, the unit digits for \(11\) to \(20\) are also \(1, 4, 9, 6, 5, 6, 9, 4, 1, 0\).

6. **Calculate the sum of the unit digits from \(11\) to \(20\)**:
- This will also sum to \(45\).

7. **Add the sums from \(1\) to \(10\) and \(11\) to \(20\)**:
- Total sum of unit digits = \(45 + 45 = 90\)

8. **Determine the unit digit of the total sum**:
- The unit digit of \(90\) is **0**.

### Final Answer:
The unit digit of \(1^6 + 2^6 + 3^6 + \ldots + 20^6\) is **0**.