The remainder when `1^3+2^3+3^3+…+999^3+1000^3` is divided by 13 is :

Options

A. 7
B. 11
C. 12
D. none of these

Answer

To find the remainder when \(1^3 + 2^3 + 3^3 + \ldots + 999^3 + 1000^3\) is divided by 13, we can follow these steps:

### Step 1: Use the formula for the sum of cubes
The sum of the cubes of the first \(n\) natural numbers is given by the formula:
\[ \left(\frac{n(n+1)}{2}\right)^2 \]
For \(n = 1000\), we can calculate:
\[ \frac{1000 \times 1001}{2} = 500500 \]
Thus, the sum of cubes becomes:
\[ (500500)^2 \]

### Step 2: Find \(500500 \mod 13\)
Next, we need to find the remainder of \(500500\) when divided by \(13\). We can do this by first simplifying \(500\) and \(1001\) modulo \(13\):
\[ 500 \mod 13 = 500 - (38 \times 13) = 500 - 494 = 6 \]
\[ 1001 \mod 13 = 1001 - (77 \times 13) = 1001 - 1001 = 0 \]
So,
\[ 500500 \mod 13 = (500 \mod 13) \times (1001 \mod 13) \div 2 \mod 13 = (6 \times 0) \div 2 \mod 13 = 0 \]

### Step 3: Calculate \((500500)^2 \mod 13\)
Since \(500500 \equiv 0 \mod 13\), we have:
\[ (500500)^2 \mod 13 = 0^2 \mod 13 = 0 \]

### Conclusion
The remainder when \(1^3 + 2^3 + 3^3 + \ldots + 999^3 + 1000^3\) is divided by \(13\) is:
\[ \boxed{0} \]