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What is 1's digit of $((183)!+3^{183})$ Can this be said without using a calculator

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closed as off-topic by user21820, max_zorn, callculus, José Carlos Santos, Trần Thúc Minh Trí Apr 3 at 11:48

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Last digit of $(183)!$ is $0$.

Last digit of $3^{183}$ is $7$.

Hence the last digit of $((183)!+3^{183})$ = $0+7$ = $7$

Logic : Last digit of $n!$, for $n>4$ is always $0$.And the last digits of powers of any integer repeat themselves in cycles of $4$. In this case, last digits of the powers of $3$ are: $$3^1 = 3$$ $$3^2=9$$ $$3^3=27$$ $$3^4=81$$ So the last digits are always: $3,9,7,1$.

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$$((183)!+3^{183})=(1\cdot 2\cdot 3\cdots 10\cdots 183 +27\cdot3^{180})\equiv 0+7\cdot3^{180} \mod{10}$$

$$\equiv 7\cdot 9^{90} \mod{10}\equiv 7\cdot (-1)^{90} \mod{10}\equiv 7\cdot 1 \mod{10}$$

$$\equiv 7 \mod{10}$$

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HINT :

The $1$'s digit (should be rightmost) of $183!$ is $0$. (Do you know why?)

Also, $3, 3^2=9, 3^3=27, 3^4=81, 3^5=243$

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