The last digit of the number 7^{886} is | vedclass

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(A) To find the last digit of $7^{886}$,we observe the pattern of the last digits of powers of $7$: $7^{1} = 7$ $7^{2} = 49$ (last digit $9$) $7^{3} =

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$9$

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(A) To find the last digit of $7^{886}$,we observe the pattern of the last digits of powers of $7$: $7^{1} = 7$ $7^{2} = 49$ (last digit $9$) $7^{3} = 343$ (last digit $3$) $7^{4} = 2401$ (last digit $1$) The cycle of last digits is $(7, 9, 3, 1)$ with a period of $4$. We divide the exponent $886$ by $4$: $886 = 4 	imes 221 + 2$ Thus,$7^{886} = (7^{4})^{221} 	imes 7^{2}$. The last digit of $(7^{4})^{221}$ is $1^{221} = 1$. The last digit of $7^{2}$ is $9$. Therefore,the last digit of $7^{886}$ is $1 	imes 9 = 9$.

The last digit of the number $7^{886}$ is

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