Q105 / 180 · KCET 2015 · Mathematics

The remainder obtained when \( 1 !+2 !+3 !+\ldots+11 ! \) is divided by \( 12 \) is

  • Correct answer
    A
    \( 09 \)
  • B
    \( 08 \)
  • C
    \( 7 \)
  • D
    \( 06 \)

Solution

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 $1 !+2 !+3 !=1+2+6=9$

Hence, the correct answer is option: 'A'.

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