Find the unit's digit in the sum of the fifth powers of the first 100 natural numbers.

### Concept & Logic This question relies on the **Cyclicity of Unit Digits**. For any integer $n$, raising it to a power creates a repeating pattern in its unit digit. The fundamental rule of cyclicity is that every number repeats its unit digit cycle every 4th power. $$n^{4k+1} \equiv n^1 \pmod{10}$$ This means the unit digit of $n^5$ is identical to the unit digit of $n^1$. ### Step-by-Step Solution * **Given:** The sum $1^5 + 2^5 + 3^5 + ... + 100^5$. * **Calculation:** According to the cyclicity rule ($n^5$ has the same unit digit as $n^1$), we can simplify the expression if we only care about the final unit digit. * The unit digit of the entire sum will be exactly the same as the unit digit of the simplified sum: $$1^1 + 2^1 + 3^1 + ... + 100^1$$ * This is simply the sum of the first 100 natural numbers. Use the standard formula for the sum of an arithmetic progression: $$\text{Sum} = \frac{n(n + 1)}{2}$$ * Substitute $n = 100$: $$\text{Sum} = \frac{100(101)}{2}$$ $$\text{Sum} = 50 \times 101 = 5050$$ * The total sum is 5050. The unit digit of 5050 is 0. ### Exam Strategy & Shortcut Memorize the cyclicity of 5: any number to the 5th power ends in the same digit as the base number itself. The moment you see "fifth powers", mentally delete the exponent. The question instantly becomes "Find the unit digit of the sum of the first 100 numbers." $50 \times 101$ obviously ends in a zero. The whole process takes less than 15 seconds mentally. ### Common Pitfall A common trap is trying to evaluate the first few fifth powers ($1, 32, 243, 1024...$) and looking for a new pattern within those sums, which burns time and increases the chance of a simple arithmetic mistake. ### Final Answer Therefore, the correct answer is **0**.