The remainder when (2023)^{2023} is divided b | vedclass

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(A) We need to find the remainder of $(2023)^{2023}$ when divided by $35$.First,note that $2023 = 35 	imes 57 + 28$,so $2023 \equiv 28 \equiv -7 \pmo

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) We need to find the remainder of $(2023)^{2023}$ when divided by $35$.First,note that $2023 = 35 	imes 57 + 28$,so $2023 \equiv 28 \equiv -7 \pmod{35}$.Thus,$(2023)^{2023} \equiv (-7)^{2023} \pmod{35}$.We can write $(-7)^{2023} = -7 	imes 7^{2022} = -7 	imes (7^2)^{1011} = -7 	imes (49)^{1011}$.Since $49 \equiv 14 \pmod{35}$,this does not simplify easily. Let's use $49 = 35 + 14$.Alternatively,$7^{2022} = (7^2)^{1011} = 49^{1011} = (35 + 14)^{1011} \equiv 14^{1011} \pmod{35}$.Since $14^2 = 196 = 35 	imes 5 + 21$ and $14^3 = 14 	imes 21 = 294 = 35 	imes 8 + 14$,we see a cycle: $14^1 \equiv 14$,$14^2 \equiv 21$,$14^3 \equiv 14 \pmod{35}$.For odd powers $n$,$14^n \equiv 14 \pmod{35}$.Thus,$7^{2022} = (7^2)^{1011} = 49^{1011} \equiv 14^{1011} \equiv 14 \pmod{35}$.Therefore,$(2023)^{2023} \equiv -7 	imes 14 = -98 \pmod{35}$.Since $-98 = -3 	imes 35 + 7$,we have $-98 \equiv 7 \pmod{35}$.The remainder is $7$.

The remainder when $(2023)^{2023}$ is divided by $35$ is $..........$.

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