The digit in the unit place of the number (18 | vedclass

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

Question

(A) We know that $n!$ ends in $0$ for all $n \geq 5$. Since $183 \geq 5$,the unit digit of $183!$ is $0$.Next,we find the unit digit of $3^{183}$. The

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) We know that $n!$ ends in $0$ for all $n \geq 5$. Since $183 \geq 5$,the unit digit of $183!$ is $0$.Next,we find the unit digit of $3^{183}$. The powers of $3$ follow a cycle of $4$: $3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81$. The unit digits are $3, 9, 7, 1$ respectively.We divide the exponent $183$ by $4$: $183 = 4 	imes 45 + 3$.Therefore,the unit digit of $3^{183}$ is the same as the unit digit of $3^3$,which is $7$.Thus,the unit digit of $(183!) + (3^{183})$ is $0 + 7 = 7$.

The digit in the unit place of the number $(183!) + (3^{183})$ is

Similar Questions

If $^n{P_r} = 840$ and $^n{C_r} = 35$,then $n$ is equal to

$n$ objects are distributed at random among $n$ persons. The number of ways in which this can be done so that at least one of them will not get any object is

Five balls of different colors are to be placed in three boxes of different sizes,where each box can hold all five balls. In how many ways can the balls be placed such that no box remains empty?

If $\alpha$ represents the number of arrangements of $p$ men and $q$ women in a row such that all men are together and $\beta$ represents the number of circular arrangements of the same people with the same condition,then $\alpha: \beta$ is

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module