The remainder, when 1 + (1 × 2) + (1 × 2 × 3) + … + (1 × 2 × 3 × … × 500) is divided by 8, is
1
2
3
4
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2024
Let’s compute the first few factorials modulo 8:
$1! = 1 \equiv 1 \pmod 8$
$2! = 2 \equiv 2 \pmod 8$
$3! = 6 \equiv 6 \pmod 8$
$4! = 24 = 3 \times 8 \equiv 0 \pmod 8$
$5! = 5 \times 4!$. Since $4!$ is a multiple of 8, $5!$ is also a multiple of 8. $5! \equiv 0 \pmod 8$.
In general, for any integer $n \ge 4$, $n!$ includes $4!$ as a factor. Since $4!$ is a multiple of 8, $n!$ is a multiple of 8 for all $n \ge 4$.
The sum modulo 8 is:
$S \pmod 8 = (1! + 2! + 3! + 4! + … + 500!) \pmod 8$
$S \pmod 8 = (1! \pmod 8 + 2! \pmod 8 + 3! \pmod 8 + 4! \pmod 8 + … + 500! \pmod 8) \pmod 8$
$S \pmod 8 = (1 + 2 + 6 + 0 + 0 + … + 0) \pmod 8$
$S \pmod 8 = (1 + 2 + 6) \pmod 8$
$S \pmod 8 = 9 \pmod 8$
$S \pmod 8 = 1$.
The remainder is 1.