If (27)^{999} is divided by 7,then the remain | vedclass

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(D) We need to find the remainder when $(27)^{999}$ is divided by $7$.We can write $27$ as $(28 - 1)$.So,$(27)^{999} = (28 - 1)^{999}$.Using the Binom

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$6$

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(D) We need to find the remainder when $(27)^{999}$ is divided by $7$.We can write $27$ as $(28 - 1)$.So,$(27)^{999} = (28 - 1)^{999}$.Using the Binomial Theorem,$(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.$(28 - 1)^{999} = \binom{999}{0} (28)^{999} (-1)^0 + \binom{999}{1} (28)^{998} (-1)^1 + \dots + \binom{999}{999} (28)^0 (-1)^{999}$.Every term except the last one contains a factor of $28$,which is divisible by $7$.$(27)^{999} = 7k + (-1)^{999}$,where $k$ is an integer.$(27)^{999} = 7k - 1$.To find the positive remainder,we write $7k - 1 = 7(k - 1) + 7 - 1 = 7(k - 1) + 6$.Thus,the remainder is $6$.

If $(27)^{999}$ is divided by $7$,then the remainder is

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