The remainder,when 3^{2003} is divided by 28, | vedclass

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Question

(C) We need to find the remainder when $3^{2003}$ is divided by $28$. First,observe that $3^3 = 27 = 28 - 1$. We can write $3^{2003} = 3^2 \cdot 3^{20

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(C) We need to find the remainder when $3^{2003}$ is divided by $28$. First,observe that $3^3 = 27 = 28 - 1$. We can write $3^{2003} = 3^2 \cdot 3^{2001} = 9 \cdot (3^3)^{667}$. Substituting $3^3 = 28 - 1$,we get $9 \cdot (28 - 1)^{667}$. Using the Binomial Theorem,$(28 - 1)^{667} = \sum_{k=0}^{667} \binom{667}{k} (28)^{667-k} (-1)^k$. This expression is of the form $28n + (-1)^{667} = 28n - 1$ for some integer $n$. So,$3^{2003} = 9(28n - 1) = 28(9n) - 9$. Since the remainder must be positive,we write $-9$ as $28 - 37$ or simply adjust: $28(9n - 1) + 19$. Thus,the remainder is $19$.

The remainder,when $3^{2003}$ is divided by $28$,is

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