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

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(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 $4$: $3^1 = 3$,$

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$5$

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(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 $n \in N$.

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