The units digit of sumlimits_{1

Question

(C) For $p \geq 5$,$p!$ ends in $0$ because it contains factors $2$ and $5$.Consider the first sum: $S_1 = \sum\limits_{1 < p < 100} {p!} = 2! + 3! +

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) For $p \geq 5$,$p!$ ends in $0$ because it contains factors $2$ and $5$.Consider the first sum: $S_1 = \sum\limits_{1 The units digit of $S_1$ is $2$.Consider the second sum: $S_2 = \sum\limits_{n = 1}^{50} {(2n)!} = 2! + 4! + \sum\limits_{n = 3}^{50} {(2n)!}$.Since $(2n)!$ for $n \geq 3$ contains $6! = 720$,which ends in $0$,all terms for $n \geq 3$ end in $0$.$S_2 = 2 + 24 + 0 = 26$.The units digit of $S_2$ is $6$.The units digit of $S_1 - S_2$ is the units digit of $32 - 26 = 6$.

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

Similar Questions

If $a + b + c = 50$ and $a, b, c$ are non-negative even integers,then the greatest value of $ab^2c$ is:

Numbers are to be formed between $1000$ and $3000$,which are divisible by $4$,using the digits $1, 2, 3, 4, 5$ and $6$ without repetition of digits. Then the total number of such numbers is.

Find the number of different signals that can be generated by arranging at least $2$ flags in order (one below the other) on a vertical staff,if five different flags are available.

The value of $\sum\limits_{r = 1}^{15} {{r^2} \left( \frac{^{15}C_r}{^{15}C_{r - 1}} \right)}$ is equal to

The number of ordered pairs $(m, n)$,where $m, n \in \{1, 2, 3, \ldots, 50\}$,such that $6^m + 9^n$ is a multiple of $5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module