The last digit of (3^P + 2) is,where P = 3^{4 | vedclass

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

Question

(D) We need to find the last digit of $(3^P + 2)$ where $P = 3^{4n}$ and $n \in N$.The last digit of powers of $3$ follows a cycle of $4$: $3^1 = 3$,$

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) We need to find the last digit of $(3^P + 2)$ where $P = 3^{4n}$ and $n \in N$.The last digit of powers of $3$ follows a cycle of $4$: $3^1 = 3$,$3^2 = 9$,$3^3 = 27$ (last digit $7$),$3^4 = 81$ (last digit $1$).Since $P = 3^{4n}$,$P$ is a multiple of $4$ (specifically,$P$ is a power of $3$ which is always odd,but $3^{4n}$ is a multiple of $81$,so $P$ is a multiple of $4$ is incorrect; rather,$3^{4n}$ is of the form $(3^4)^n = 81^n$).Any number ending in $1$ raised to any power $n$ still ends in $1$. Thus,$3^{4n} = (3^4)^n = 81^n$ ends in $1$.Therefore,$3^P = 3^{(3^{4n})}$. Since $3^{4n}$ is a multiple of $4$,let $3^{4n} = 4k$ for some integer $k$.Then $3^P = 3^{4k} = (3^4)^k = 81^k$,which ends in $1$.Finally,the last digit of $(3^P + 2)$ is $1 + 2 = 3$ is incorrect. Let us re-evaluate: $P = 3^{4n}$. $3^{4n}$ is a power of $3$ which is always of the form $4k+1$ or $4k+3$. Actually,$3^{4n} = (3^4)^n = 81^n$. Since $81^n$ ends in $1$,$3^{4n}$ is of the form $4k+1$.Thus $3^P = 3^{4k+1} = 3^{4k} 	imes 3^1$. The last digit of $3^{4k}$ is $1$,so the last digit of $3^P$ is $1 	imes 3 = 3$.Therefore,the last digit of $(3^P + 2)$ is $3 + 2 = 5$.

The last digit of $(3^P + 2)$ is,where $P = 3^{4n}$ and $n \in N$.

Similar Questions

The greatest integer which divides the number $101^{100} - 1$ is

For any natural number $n$,$(15 \times 5^{2n}) + (2 \times 2^{3n})$ is divisible by

If $[x]$ denotes the greatest integer function of $x$,then the number of positive integral divisors of $\left[(2+\sqrt{3})^5\right]$ is

If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$,then $k$ is equal to

The sum $1! + 4! + 7! + 10! + 12! + 13! + 15! + 16! + 17!$ is divisible by

Vedclass Test Series

Exam Paper Generator

Online Exam Module