Find the remainder when `1xx2+2xx3+3xx4+…+99xx100` is divisible by 101.

Answer

To solve the problem of finding the remainder when \(1 \times 2 + 2 \times 3 + 3 \times 4 + \ldots + 99 \times 100\) is divided by 101, we can follow these steps:

### Step 1: Understand the Series
The expression can be rewritten as:
\[ S = 1 \times 2 + 2 \times 3 + 3 \times 4 + \ldots + 99 \times 100 \]
This can be expressed in a more general form:
\[ S = \sum_{n=1}^{99} n(n+1) \]

### Step 2: Simplify the Expression
We can expand the term \(n(n+1)\):
\[ n(n+1) = n^2 + n \]
Thus, we can rewrite the sum:
\[ S = \sum_{n=1}^{99} (n^2 + n) = \sum_{n=1}^{99} n^2 + \sum_{n=1}^{99} n \]

### Step 3: Use the Formulas for the Sums
We can use the formulas for the sum of the first \(n\) natural numbers and the sum of the squares of the first \(n\) natural numbers:
- The sum of the first \(n\) natural numbers is given by:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]
- The sum of the squares of the first \(n\) natural numbers is given by:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Calculate Each Part
For \(n = 99\):
1. Calculate \(\sum_{n=1}^{99} n\):
\[ \sum_{n=1}^{99} n = \frac{99 \times 100}{2} = 4950 \]

2. Calculate \(\sum_{n=1}^{99} n^2\):
\[ \sum_{n=1}^{99} n^2 = \frac{99 \times 100 \times 199}{6} = 328350 \]

### Step 5: Combine the Results
Now we can combine these results:
\[ S = \sum_{n=1}^{99} n^2 + \sum_{n=1}^{99} n = 328350 + 4950 = 333300 \]

### Step 6: Find the Remainder When Divided by 101
Now we need to find \(333300 \mod 101\):
1. Calculate \(333300 \div 101\):
\[ 333300 \div 101 \approx 3297.0297 \quad \text{(take the integer part, which is 3297)} \]
2. Multiply back to find the product:
\[ 3297 \times 101 = 332997 \]
3. Subtract from \(333300\) to find the remainder:
\[ 333300 - 332997 = 303 \]

### Final Answer
Thus, the remainder when \(1 \times 2 + 2 \times 3 + 3 \times 4 + \ldots + 99 \times 100\) is divided by 101 is:
\[ \boxed{303} \]