If { p } denotes the fractional part of the n | vedclass

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(A) We need to find the fractional part of $\frac{3^{200}}{8}$.$\frac{3^{200}}{8} = \frac{(3^2)^{100}}{8} = \frac{9^{100}}{8}$.We can write $9$ as $(1

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$\frac{1}{8}$

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(A) We need to find the fractional part of $\frac{3^{200}}{8}$.$\frac{3^{200}}{8} = \frac{(3^2)^{100}}{8} = \frac{9^{100}}{8}$.We can write $9$ as $(1 + 8)$.So,$\frac{9^{100}}{8} = \frac{(1 + 8)^{100}}{8}$.Using the Binomial Theorem,$(1 + 8)^{100} = 1 + \binom{100}{1}8 + \binom{100}{2}8^2 + \dots + \binom{100}{100}8^{100}$.Dividing this by $8$,we get $\frac{(1 + 8)^{100}}{8} = \frac{1}{8} + \binom{100}{1} + \binom{100}{2}8 + \dots + \binom{100}{100}8^{99}$.Since $\binom{100}{1} + \binom{100}{2}8 + \dots + \binom{100}{100}8^{99}$ is an integer,let it be $k$.Then $\frac{3^{200}}{8} = k + \frac{1}{8}$.The fractional part $\{ p \}$ is defined as $p - \lfloor p \rfloor$.Therefore,$\left\{\frac{3^{200}}{8}\right\} = \left\{ k + \frac{1}{8} \right\} = \frac{1}{8}$.

If $\{ p \}$ denotes the fractional part of the number $p$,then $\left\{\frac{3^{200}}{8}\right\}$ is equal to

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