Question

The remainder obtained when \[1! + 2! + ...... + 50!\] is divided by \[30\] is
(A) \[3\]
(B) \[13\]
(C) \[11\]
(D) \[9\]

Answer 1

Hint: In order to solve this question, we must know about the concept of factorial i.e., \[n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot ..... \cdot 3 \cdot 2 \cdot 1\] .Here, in this question, we have to find the remainder of the given factorial sum when divided by \[30\] .For this, we know that the value of factorial after \[5!\] will leave the remainder zero when it is divided by \[30\] .So for calculating the remainder, we will calculate the remainder before that number.

Complete step by step answer:
So, first of all we will calculate the factorial till \[5!\] .
So, \[1! = 1\]
\[2! = 2 \times 1 = 2\]
\[3! = 3 \times 2 \times 1 = 6\]
\[4! = 4 \times 3 \times 2 \times 1 = 24\]
and if we see \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\] which is divisible by \[30\]
Now after \[5!\] we can write each factorial term as \[5!{\text{ }} \times {\text{ }}y\] where \[y\] is a positive number.
i.e., \[6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6 \times 5!\]
\[7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 7 \times 6 \times 5!\] and so on.
It means the value of factorial after \[5!\] will leave the remainder zero when it is divided by \[30\]
So, we will calculate the value remainder before \[5!\]
For this, add all the factorial values from \[1\] to \[4\]
\[ \Rightarrow 1! + 2! + 3! + 4!\]
Now on dividing by \[30\] we get
\[\dfrac{{1! + 2! + 3! + 4!}}{{30}}{\text{ }} - - - \left( 1 \right)\]
On substituting the values, we will get the equation \[\left( 1 \right)\] as
\[\dfrac{{1 + 2 + 6 + 24}}{{30}}{\text{ }} - - - \left( 2 \right)\]
Now adding the numerator of the equation \[\left( 2 \right)\] we get,
\[\dfrac{{33}}{{30}}\]
And now on dividing it by \[30\] we get the remainder as \[3\]
\[\left[ {\because 33 = 30 \times 1 + 3} \right]\]
Therefore, the remainder for the factorial \[1! + 2! + ...... + 50!\] when divided by \[30\] is \[3\]

Hence, the correct answer is option (A).

Note:
To solve this question, we don’t need to add all the factorial terms and divide it by \[30\] to obtain the answer. Because if we see \[30\] can be factored as a product of \[2,3\] and \[5\] . And, the least value of factorial which contains \[2,3\] and \[5\] is \[5!\] .So \[5!\] and any factorial greater than \[5\] always contains \[2,3\] and \[5\] and hence the remainder is going to be zero for any \[x!\] where \[x\] is either \[5\] or greater than \[5\] . So, we just calculate the remainder before \[5!\] and get the required result.