Question

When $1! + 2! + 3! + ..... + 25!$ is divided by $7$, what will be the remainder?
A) $0$
B) $5$
C) $1$
D) None of the above

Answer 1

We will have to find all the factorials from $1$ to $25$and then add all the factorials and divide it by $7$ , but first we will have to check whether all the terms of addition are factors of $7$ or not which can be done by calculating the sum of factorials and dividing it.

Complete step-by-step solution:
As we know, the terms after $7!$ are all divisible by $7$ as $7$ is one of the factors. The remainder that is left after adding from $7! + 8! + 9! + 10! + 11! + 12! + 13! + 14! + 15! + 16! + 17! + 18! + 19! + 20! + 21! + 22! + 23! + 24! + 25!$ will be $0$.
So, now we consider only the terms from $1!$ to $6!$ ,
The sum of factorials from $1!$ to $6!$ will be as follows,
$1! + 2! + 3! + 4! + 5! + 6!$
$= 1 + 2 + 6 + 24 + 120 + 720$
$= 873$
Now we will divide the sum $873$ by $7$,
$\dfrac{{873}}{7} = 124$& Remainder=5
The remainder that we will get after dividing is $5$ .
So, the correct answer is option $B.$ $5$.

Note: If we add $1! + 2! + .... + n!$ by a number then the factors in the sum after the number we are dividing will give us a remainder of $0$. For example, in the above question , all the terms after $7!$ dividing it by $7$ will give us a remainder of $0$. And if we divide the sum of factorials by a number $12$ then the sum of factorials from $12!$ to $n!$ dividing by $12$ will be $0$, as $12$ is a factor of $12$. This is a property of factors and is very useful in finding the quotient and remainder of the sum of factorials divided by a natural number.