Find the remainder when (1!)^2 + (2!)^2 + (3! | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We need to find the remainder of $S = \sum_{n=1}^{100} (n!)^2$ when divided by $100$.Calculating the first few terms:$(1!)^2 = 1^2 = 1$$(2!)^2 = 2

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) We need to find the remainder of $S = \sum_{n=1}^{100} (n!)^2$ when divided by $100$.Calculating the first few terms:$(1!)^2 = 1^2 = 1$$(2!)^2 = 2^2 = 4$$(3!)^2 = 6^2 = 36$$(4!)^2 = 24^2 = 576 \equiv 76 \pmod{100}$$(5!)^2 = 120^2 = 14400 \equiv 0 \pmod{100}$For all $n \ge 5$,$n!$ contains at least two factors of $5$ and two factors of $2$,so $n!$ is a multiple of $10$. Thus,$(n!)^2$ is a multiple of $100$.Therefore,the sum $S \equiv 1 + 4 + 36 + 76 \pmod{100}$$S \equiv 117 \pmod{100}$$S \equiv 17 \pmod{100}$The remainder is $17$.

Find the remainder when $(1!)^2 + (2!)^2 + (3!)^2 + \dots + (100!)^2$ is divided by $10^2$.

Similar Questions

$A$ linguistic club consists of $6$ girls and $4$ boys. $A$ team of $4$ members is to be selected from this group including the selection of a leader (from among these $4$ members) for the team. If the team has to include at most one boy,the number of ways of selecting the team is

From $6$ different novels and $3$ different dictionaries,$4$ novels and $1$ dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

The number of integers lying between $1000$ and $10000$ such that the sum of all the digits in each of those numbers becomes $30$ is:

The units digit of $\sum\limits_{1 < p < 100} {p! - \sum\limits_{n = 1}^{50} {(2n)!} }$ is:

$A$ letter lock consists of three rings with $15$ different letters each. If $N$ denotes the number of ways in which it is possible to make unsuccessful attempts to open the lock, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module