Question

162. For an integer n, n! = n(n - 1) (n - 2)............ 3.2.1.
(P.C.S., 2008)
Then, 1! + 2! + 3! + ......... + 100! when divided by
5 leaves remainder
(a) 0
(b) 1
(C) 2.
(d) 3
163. The number of prime factors in the expression​

Answer 1

Answer:

d

Step-by-step explanation:

5! +6!+.... all end digits are zero

remaining in expression is 1!+2!+3!+4!= 1+2+6+24=33 which gives remainder 3 when divided by 5

Answer 2

Answer:

• Option (d) 3

On dividing 1! + 2! + 3! + ....... + 100! by 5 remainder will be 3

Explanation:

• For an Integer n , n! = n (n - 1) (n - 2) ...... 3 . 2 . 1

We need to find that,

• When , 1! + 2! + 3! + ........ 100! will be divided by 5 , then what will be the remainder

so,

Each term from 5! to 100! in the given Expression [ 1! + 2! + 3! + .......100! ] will contain 5. hence, will be multiple of 5 and completely divisible of 5.

Hence calculating sum of terms left,

→ Calculating : 1! + 2! + 3! + 4!

→ = ( 1 ) + ( 2 × 1 ) + ( 3 × 2 × 1 ) + ( 4 × 3 × 2 × 1 )

→ = 1 + 2 + 6 + 24

→ = 33

On dividing 33 by 5

33 = 5 × 6 + 3

We got the remainder = 3

Therefore,

• On Dividing 1! + 2! + 3! + .......... + 100! by 5 we will get Remainder 3

Hence,

Option (d) 3 is Correct.