First slide

Binomial theorem for positive integral Index

Question

If {x} denotes the fractional part of x, then $\{\frac{2^{2003}}{17}\}$ is

Moderate

A

2/17
B

4/17
C

8/17
D

16/17

Solution

$\begin{array}{l} 2^{2003} = {(2^{4})}^{500} \cdot 2^{3} \\ \Rightarrow 2^{2003} = 8 (16)^{500} \\ \Rightarrow 2^{2003} = 8 (17 - 1)^{500} \\ \Rightarrow 2^{2003} = 8 [(17)^{500} -^{500} C_{1} (17)^{499} + \dots -^{500} C_{499} (17) + 1] \end{array}$
$\Rightarrow \frac{2^{2003}}{17} = 8 k + \frac{8}{17}$ ,
where $k = (17)^{499} -^{500} C_{1} (17)^{498} + \dots +^{500} C_{499}$
such that k is an integer
$∴ \{\frac{2^{2003}}{17}\} = \frac{8}{17}$

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