Question

(1!)^1!+(2!)^2!+(3!)^3!+.....+(100!)^100! is divided by 5. find the remainder​

Answer 1

Answer:

2

Step-by-step explanation:

All the terms from (5!)^5! onwards are multiples of 5 themselves since 5 is a factor of n! when n≥5.

So modulo 5 we have

(1!)^1! + (2!)^2! + ... + (100!)^100!

≡ (1!)^1! + (2!)^2! + (3!)^3! + (4!)^4!

≡ 1¹ + 2² + 6⁶ + 24²⁴

≡ 1 + 4 + 1⁶ + (-1)²⁴          [ since 6 ≡ 1 and 24 ≡ -1 modulo 5 ]

= 1 + 4 + 1 + 1

= 7

≡ 2 (mod 5)

Answer 2

Answer:

2

Step-by-step explanation:

(1!)^1!+(2!)^2!+(3!)^3!+...............(100!)^100! / 5

= 1 + 2^2 + 6^6 + 24^24 + [(5!)^5! + (6!)^6! + .....+ (100!)^100!] / 5

from (5!)^5! to 100!^100! each term is divisible by 5 & gives reminder = 0

required remainder = 1 + 2^2 + 6^6 + 24^24 / 5 => remainder = 7/5 = 2

[unit digit of 1 + 2^2 + 6^6 + 24^24 = 1+4 +6+6 = 7]