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

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(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

$7$

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(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

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