When 2^{23} is divided by 10,the remainder wi | vedclass

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(D) To find the remainder when $2^{23}$ is divided by $10$,we need to find the unit digit of $2^{23}$.Observe the pattern of the unit digits of powers

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$8$

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(D) To find the remainder when $2^{23}$ is divided by $10$,we need to find the unit digit of $2^{23}$.Observe the pattern of the unit digits of powers of $2$:$2^1 = 2$$2^2 = 4$$2^3 = 8$$2^4 = 16$ (unit digit is $6$)$2^5 = 32$ (unit digit is $2$)The unit digits repeat in a cycle of $4$: $(2, 4, 8, 6)$.Divide the exponent $23$ by the cycle length $4$: $23 = 4 	imes 5 + 3$.The remainder is $3$,which means the unit digit of $2^{23}$ is the same as the unit digit of $2^3$.Since $2^3 = 8$,the unit digit of $2^{23}$ is $8$.When any number is divided by $10$,the remainder is equal to its unit digit.Therefore,the remainder when $2^{23}$ is divided by $10$ is $8$.

When $2^{23}$ is divided by $10$,the remainder will be

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