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Finding Remainder of Numbers of the Form a^b

The remainder...

Question

The remainder when $2^{2003}$ is divided by $17$ is

A

1

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B

2

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C

8

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D

11

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Solution

The correct option is C $8$
$2^{2003} = 2^{3} \times 2^{2000} = 8 (2^{4})^{500} = 8 (16)^{500} = 8 (17 - 1)^{500} = 8 [^{500} C_{0} \cdot 17^{500} -^{500} C_{1} \cdot 17^{499} + \dots +^{500} C_{500}] = 8 [17 k + 1], k \in N = 17 \times 8 k + 8$

Hence, the remainder is $8$ .

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Finding Remainder of Numbers of the Form a^b

Standard XII Mathematics

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