What is the unit digit of (217)^{413} times ( | vedclass

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(D) To find the unit digit of the product,we find the unit digit of each term individually.$1$. For $(217)^{413}$,the unit digit is $7$. The cycle of

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$8$

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(D) To find the unit digit of the product,we find the unit digit of each term individually.$1$. For $(217)^{413}$,the unit digit is $7$. The cycle of powers of $7$ is $(7^1=7, 7^2=9, 7^3=3, 7^4=1)$. The exponent $413$ divided by $4$ leaves a remainder of $1$. Thus,the unit digit is $7^1 = 7$.$2$. For $(819)^{547}$,the unit digit is $9$. The cycle of powers of $9$ is $(9^1=9, 9^2=1)$. Since $547$ is odd,the unit digit is $9$.$3$. For $(414)^{624}$,the unit digit is $4$. The cycle of powers of $4$ is $(4^1=4, 4^2=6)$. Since $624$ is even,the unit digit is $6$.$4$. For $(342)^{812}$,the unit digit is $2$. The cycle of powers of $2$ is $(2^1=2, 2^2=4, 2^3=8, 2^4=6)$. The exponent $812$ is divisible by $4$ (remainder $0$),so the unit digit is $6$.Now,multiply the unit digits: $7 	imes 9 	imes 6 	imes 6 = 63 	imes 36$. The product of the unit digits is $3 	imes 6 = 18$. Therefore,the unit digit of the entire expression is $8$.

What is the unit digit of $(217)^{413} \times (819)^{547} \times (414)^{624} \times (342)^{812} ?$

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