If `a_n` is the digit in the unit place of the number `1!+2!+3!+……n!` then
`a_8+a_9+a_10+…..+a_16=`

Options

A. 9
B. 18
C. 27
D. none of these

Answer

To solve the problem, we need to find the unit digit of the sum of factorials from \(1!\) to \(n!\) for \(n = 8\) to \(n = 16\) and then sum those unit digits.

### Step-by-Step Solution:

1. **Calculate the unit digits of factorials from \(1!\) to \(8!\)**:
- \(1! = 1\) → unit digit = 1
- \(2! = 2\) → unit digit = 2
- \(3! = 6\) → unit digit = 6
- \(4! = 24\) → unit digit = 4
- \(5! = 120\) → unit digit = 0
- \(6! = 720\) → unit digit = 0
- \(7! = 5040\) → unit digit = 0
- \(8! = 40320\) → unit digit = 0

From \(5!\) onwards, all factorials have a unit digit of 0.

2. **Sum the unit digits for \(n = 8\)**:
\[ a_8 = 1 + 2 + 6 + 4 + 0 + 0 + 0 + 0 = 13 \]
The unit digit of \(13\) is \(3\). Thus, \(a_8 = 3\).

3. **Calculate \(a_9\)**:
\[ a_9 = 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! \]
Since \(9! = 362880\) (unit digit = 0), the unit digit remains the same as \(a_8\):
\[ a_9 = 3 \]

4. **Calculate \(a_{10}\)**:
\[ a_{10} = 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! \]
Since \(10! = 3628800\) (unit digit = 0), the unit digit remains the same:
\[ a_{10} = 3 \]

5. **Repeat for \(n = 11\) to \(n = 16\)**:
- \(a_{11} = 3\)
- \(a_{12} = 3\)
- \(a_{13} = 3\)
- \(a_{14} = 3\)
- \(a_{15} = 3\)
- \(a_{16} = 3\)

6. **Sum \(a_8 + a_9 + a_{10} + \ldots + a_{16}\)**:
\[ a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 \]
There are \(9\) terms of \(3\):
\[ = 9 \times 3 = 27 \]

### Final Answer:
\[ a_8 + a_9 + a_{10} + \ldots + a_{16} = 27 \]