The greatest interger by which `1+Sigma_(r=1)^(30) rxxr !` is divisible is

Options

A. composite number
B. odd number
C. divisible by 3
D. none of these

Explanation

To solve the problem of finding the greatest integer by which \( 1 + \sum_{r=1}^{30} r \cdot r! \) is divisible, we can follow these steps:

### Step 1: Rewrite the summation
We start with the expression:
\[ 1 + \sum_{r=1}^{30} r \cdot r! \]
We can rewrite \( r \cdot r! \) as \( (r+1)! - r! \). This gives us:
\[ \sum_{r=1}^{30} r \cdot r! = \sum_{r=1}^{30} ((r+1)! - r!) \]

### Step 2: Expand the summation
Now, we can expand the summation:
\[ \sum_{r=1}^{30} ((r+1)! - r!) = (2! - 1!) + (3! - 2!) + (4! - 3!) + \ldots + (31! - 30!) \]
Notice that this is a telescoping series, where all intermediate terms cancel out.

### Step 3: Simplify the expression
After cancellation, we are left with:
\[ = 31! - 1! \]
Since \( 1! = 1 \), we have:
\[ = 31! - 1 \]

### Step 4: Add 1 to the expression
Now, we add 1 to our result:
\[ 1 + (31! - 1) = 31! \]

### Step 5: Determine divisibility
Next, we need to find the greatest integer by which \( 31! \) is divisible. Since \( 31! \) is the product of all integers from 1 to 31, it is divisible by every integer less than or equal to 31.

### Conclusion
Thus, the greatest integer by which \( 1 + \sum_{r=1}^{30} r \cdot r! \) is divisible is:
\[ \boxed{31} \]