KCET · Maths · Application of Derivatives
The remainder obtained when \( 1 !+2 !+3 !+\ldots+11 ! \) is divided by \( 12 \) is
- A \( 09 \)
- B \( 08 \)
- C \( 7 \)
- D \( 06 \)
Answer & Solution
Correct Answer
(A) \( 09 \)
Step-by-step Solution
Detailed explanation
Given that, \(1 !+2 !+3 !+\cdot s+11 !\)
Since \(4 !=4 \times 3 \times 2 \times 1\) is divisible by 12
\(5 !=5 \times 4 \times 3 \times 2 \times 1\) is divisible by 12
And so on.
\(11 !=11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
is divisible by 12
So, reminder is \(1 !+2 !+3 !=1+2+6=9\)
Since \(4 !=4 \times 3 \times 2 \times 1\) is divisible by 12
\(5 !=5 \times 4 \times 3 \times 2 \times 1\) is divisible by 12
And so on.
\(11 !=11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
is divisible by 12
So, reminder is \(1 !+2 !+3 !=1+2+6=9\)
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