Question

The remainder obtained when 1!+2!+.........+95! is divided by 15 is?????????

Answer 1

for the given series to be divisible by 15 each term should contain 3 and 5. the terms without 3 and 5 are the remainders
so 1!+2!+3!+4!=33 is the remainder
from 5!=1×2×3×4×5 all terms contain 3 and 5
i hope this is the ans and if ur satisfied with my ans plz mark me as brainliest

Answer 2

As it forms an AP :-

1 + 2 + 3....... + 95

Here,

a = 1

d = 2 - 1                           (a₂ - a₁ = d)
   
   = 1

aₓ = 95

Then to find x - term.

aₓ = a + (x - 1)d

95 = 1 + (x - 1)1

(x - 1) = 95 -1 

x = 95th term                              (cancelling -1 on both sides)

Also, To find the Sum

Sₓ = x/2 (a+aₓ)

     = 95/2 (1 + 95)

     = 95 × 96 / 2

     = 4560

Dividing 15 on the result according to the question.

  = 4560 / 15

  = 304

And so there will be no remainder R = 0


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