Question

What will be the remainder of the term $ 1! + 2! + 3! + 4! + ... + 100! $ when it is divisible by $ 240 $ respectively?
(a) None of these
(b) Cannot be determined
(c) $ 153 $
(d) $ 351 $

Answer 1

Hint: We will use the most eccentric concept of factorial. First of all, calculating the factorial from $ 6! $ respectively. As a result, finding the factorial of remaining five terms the desired value is obtained particularly. $ 6! $ is considered directly because it is the complete divisible of $ 6 $ i.e. reminder is ‘zero’.

Complete step-by-step answer:
The given sequence for the combinations of the equations can be written as,
 $ 1! + 2! + 3! + 4! + ... + 100! $
Predominantly the given sequence revolves in the concept of factorial which is the function or any sequence of series which is in terms of multiples of every number preceding to the next one.
As a result, considering this terminology,
It seems to be known that,
 $ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 $
Similarly,
 $ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 $
.
.
.
And so on!
Hence, it is presume that (for this sequence particularly) terms after $ 6 $ are all divisible by $ 240 $ respectively,
Now, only first five terms are remained to calculate the respective solution of the sequence,
Therefore, by definition by factorial, we get
 $ 1! + 2! + 3! + 4! + 5! $
Solving the sequence mathematically, we get
 $
  1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 \\
  1! + 2! + 3! + 4! + 5! = 153 \;
  $
 $ \Rightarrow $ The correct option is (c)
So, the correct answer is “Option c”.

Note: One must remember the concept of factorial as it is the efficient and best method to find the exact remainder of any complex sequence. how to find the factorial by multiplying with the number below it in the respective series or sequence. Basic knowledge of solving equations recommended for such problems. Algebraic identities play a significant role in solving this problem.