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

Group $A$ consists of $7$ boys and $3$ girls,while group $B$ consists of $6$ boys and $5$ girls. The number of ways,$4$ boys and $4$ girls can be invited for a picnic if $5$ of them must be from group $A$ and the remaining $3$ from group $B$,is equal to:

The difference between an integer $n$ and its cube $n^3$ is always divisible by which of the following?

If $^n{P_4} = 30 \times {^n}{C_5}$,then $n = $

The sum of all three-digit numbers,formed using non-zero digits,where all the digits are perfect squares of a natural number,is

The number of ordered pairs $(x, y)$ of integers such that their product $xy$ is a positive integer less than $100$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module