Question

The remainder obtained when $1! + 2! + 3! + 4!{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. + 11!}}$ is divides by 12 is
A. 9
B. 8
C. 7
D. 6

Answer 1

Hint: Consider every factorial one by one and check it whether it is divisible by 12 or not, by checking factors of 12 in it. Then add all the remainder to get the answer.

Complete step-by-step answer:
In the question, we have been given a sum which is represented as $1! + 2! + 3! + 4!{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. + 11!}}$ Now, it is further said, that it is divided by 12. So, after this division, we have to find its remainder.
Now, we will find the remainder of each division which is done like we will check division of 11! by 12, then 10! by 12 and like this up to 0! and then add it all together if it is less than 12 then it is fine and if more than 12 then we will once again subtract it by the nearest multiple of 12 to get the answer.
So, first on dividing 12 by 11!, 12 completely divides it as 11! consists of products of 4 and 3, thus, remainder is 0.
Now, for 10!, 12 divides it completely as 10! contains products of 4 and 3, thus, remainder is 0.
Now, similarly we will do it for 9!, 12 divides it completely.
For the same reason 12 also divides 8!, 7!, 6!, 5! and till 4!.
For 3!, the value is 6 which is not divisible by 12. The same holds for value 2! and 1! which is 2 and 1 respectively.
So, the sum of remainders is $1 + 2 + 6 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 \Rightarrow 9$
Hence, the correct option is ‘A’.

Note: We can also do the same question, in another way by finding out each value of factorial like 1! as 1, 2! as 2, 3! as 6 and so on up to 11! as 39916800 and then add it all and finally divide by 12 to get the remainder. This method is very tedious and lengthy, so it is preferred not to be used.