Which is the last digit of `1+2+3+……+ ` n if the last digit of `1 ^(3) + 2 ^(3) + ….. + n ^(3)` is 1 ?

Answer

To solve the problem, we need to find the last digit of the sum \( S = 1 + 2 + 3 + \ldots + n \) given that the last digit of the sum of cubes \( 1^3 + 2^3 + \ldots + n^3 \) is 1.

### Step 1: Understand the formula for the sum of cubes
The formula for the sum of the cubes of the first \( n \) natural numbers is given by:
\[ 1^3 + 2^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
This means that the sum of cubes can be expressed as the square of the sum of the first \( n \) natural numbers.

### Step 2: Find the last digit of the sum of cubes
We are given that the last digit of \( 1^3 + 2^3 + \ldots + n^3 \) is 1. Since \( 1^3 + 2^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2 \), we can deduce that the last digit of \( \left( \frac{n(n+1)}{2} \right)^2 \) must also be 1.

### Step 3: Determine the last digit possibilities
For a number \( x^2 \) to end with the digit 1, \( x \) must end with either 1 or 9. Therefore, the last digit of \( \frac{n(n+1)}{2} \) must be either 1 or 9.

### Step 4: Analyze the last digit of \( S \)
The sum \( S = 1 + 2 + 3 + \ldots + n \) can be expressed as:
\[ S = \frac{n(n+1)}{2} \]
Thus, the last digit of \( S \) is directly related to the last digit of \( \frac{n(n+1)}{2} \).

### Step 5: Find the last digit of \( S \) based on the last digit of \( \frac{n(n+1)}{2} \)
1. If the last digit of \( \frac{n(n+1)}{2} \) is 1, then the last digit of \( S \) is also 1.
2. If the last digit of \( \frac{n(n+1)}{2} \) is 9, then the last digit of \( S \) is 9.

### Conclusion
Since the last digit of \( \frac{n(n+1)}{2} \) can be either 1 or 9, the last digit of \( S \) can also be either 1 or 9. Therefore, the last digit of \( S \) could be either 1 or 9.

### Final Answer
The last digit of \( S \) could be **1 or 9**.

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