The remainder, when 1 + (1 × 2) + (1 × 2 × 3) +......+ (1 × 2 × 3 × .... × 500) is divided by 8, is 

The correct answer is Option 1

Key PointsTo find the remainder when 
⇒ S=1+(1×2)+(1×2×3)+…+(1×2×3×…×500) is divided by 8, so we can analyze the terms in the series.
⇒ The nth term of the series is  n! (n factorial). We need to find the sum of factorials from  1! to 500! and then find the remainder when this sum is divided by 8.
⇒ Calculating the factorials modulo 8:

• 1!=1≡1 mod 8
• 2!=2≡2mod8
• 3!=6≡6 mod 8
• 4!=24≡0 mod8

⇒  For  n≥4 n! will always be divisible by 8 (since  4! and higher factorials include the factors 2 and 4). Thus, we only need to

consider the first three terms:
⇒  S≡1+2+6 mod 8
Calculating this:
⇒  S≡1+2+6=9 ≡1 mod 8
Therefore, the remainder when 
S is divided by 8 is  1.