What ist he unit digit of ` 1^(5) + 2^(5) + 3^(5) + . . . + 20 ^(5)` ?

Options

A. 0
B. 5
C. 2
D. 4

Answer

To find the unit digit of the expression \( 1^5 + 2^5 + 3^5 + \ldots + 20^5 \), we can follow these steps:

### Step 1: Identify the unit digits of powers
First, we need to determine the unit digit of \( n^5 \) for \( n = 0, 1, 2, \ldots, 9 \) since the unit digits repeat every 10 numbers.

- \( 0^5 = 0 \) (unit digit is 0)
- \( 1^5 = 1 \) (unit digit is 1)
- \( 2^5 = 32 \) (unit digit is 2)
- \( 3^5 = 243 \) (unit digit is 3)
- \( 4^5 = 1024 \) (unit digit is 4)
- \( 5^5 = 3125 \) (unit digit is 5)
- \( 6^5 = 7776 \) (unit digit is 6)
- \( 7^5 = 16807 \) (unit digit is 7)
- \( 8^5 = 32768 \) (unit digit is 8)
- \( 9^5 = 59049 \) (unit digit is 9)

### Step 2: Calculate the unit digits for \( n = 10 \) to \( 20 \)
The unit digits for \( n = 10 \) to \( 20 \) will be the same as those for \( n = 0 \) to \( 9 \) because they share the same unit digits.

- \( 10^5 \) (unit digit is 0)
- \( 11^5 \) (unit digit is 1)
- \( 12^5 \) (unit digit is 2)
- \( 13^5 \) (unit digit is 3)
- \( 14^5 \) (unit digit is 4)
- \( 15^5 \) (unit digit is 5)
- \( 16^5 \) (unit digit is 6)
- \( 17^5 \) (unit digit is 7)
- \( 18^5 \) (unit digit is 8)
- \( 19^5 \) (unit digit is 9)
- \( 20^5 \) (unit digit is 0)

### Step 3: Sum the unit digits
Now we sum the unit digits from \( 1^5 \) to \( 20^5 \):

- From \( 1^5 \) to \( 9^5 \): \( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \) (unit digit is 5)
- From \( 10^5 \) to \( 19^5 \): \( 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 \) (unit digit is 5)
- For \( 20^5 \): unit digit is 0.

### Step 4: Combine the results
Now, we combine the unit digits:
- Total unit digit from \( 1^5 \) to \( 19^5 \) is \( 5 + 5 = 10 \) (unit digit is 0).
- Adding the unit digit from \( 20^5 \) (which is 0) does not change the result.

### Final Answer
Thus, the unit digit of \( 1^5 + 2^5 + 3^5 + \ldots + 20^5 \) is **0**.

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