Sum of the last `2` digit in `sum_(r=0)^(101)(101-r)!`

Options

A. `4`
B. `8`
C. `0`
D. `5`

Explanation

To solve the problem of finding the sum of the last two digits in the expression \(\sum_{r=0}^{101} (101-r)!\), we can follow these steps:

### Step 1: Write the summation expression
The expression can be rewritten as:
\[ \sum_{r=0}^{101} (101-r)! = (101-0)! + (101-1)! + (101-2)! + \ldots + (101-101)! \]
This simplifies to:
\[ 101! + 100! + 99! + \ldots + 0! \]

### Step 2: Identify the factorial terms
We know that \(n!\) (for \(n \geq 10\)) will end with at least two zeros because \(10!\) and higher factorials contain the factors \(2\) and \(5\) multiple times. Therefore, we can ignore \(101!\), \(100!\), \(99!\), ..., \(10!\) when looking for the last two digits.

### Step 3: Calculate the last two digits of lower factorials
We need to calculate the last two digits of the factorials from \(9!\) down to \(0!\):
- \(9! = 362880\) (last two digits: 80)
- \(8! = 40320\) (last two digits: 20)
- \(7! = 5040\) (last two digits: 40)
- \(6! = 720\) (last two digits: 20)
- \(5! = 120\) (last two digits: 20)
- \(4! = 24\) (last two digits: 24)
- \(3! = 6\) (last two digits: 06)
- \(2! = 2\) (last two digits: 02)
- \(1! = 1\) (last two digits: 01)
- \(0! = 1\) (last two digits: 01)

### Step 4: Sum the last two digits of these factorials
Now we sum the last two digits:
\[ 80 + 20 + 40 + 20 + 20 + 24 + 06 + 02 + 01 + 01 \]

Calculating this step by step:
1. \(80 + 20 = 100\) (last two digits: 00)
2. \(00 + 40 = 40\)
3. \(40 + 20 = 60\)
4. \(60 + 20 = 80\)
5. \(80 + 24 = 104\) (last two digits: 04)
6. \(04 + 06 = 10\)
7. \(10 + 02 = 12\)
8. \(12 + 01 = 13\)
9. \(13 + 01 = 14\)

### Step 5: Find the sum of the last two digits
The last two digits of the total sum are \(14\). Now we find the sum of these two digits:
\[ 1 + 4 = 5 \]

### Final Answer
Thus, the sum of the last two digits in the expression is:
\[ \boxed{5} \]