If the fractional part of the number frac{2^{ | vedclass

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(B) We need to find the remainder when $2^{403}$ is divided by $15$.We can write $2^{403} = 2^3 	imes 2^{400} = 8 	imes (2^4)^{100} = 8 	imes (16)^

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$8$

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(B) We need to find the remainder when $2^{403}$ is divided by $15$.We can write $2^{403} = 2^3 	imes 2^{400} = 8 	imes (2^4)^{100} = 8 	imes (16)^{100}$.Since $16 = 15 + 1$,we have $2^{403} = 8(15 + 1)^{100}$.Using the Binomial Theorem,$(15 + 1)^{100} = \sum_{r=0}^{100} \binom{100}{r} 15^r (1)^{100-r} = 1 + 100(15) + \binom{100}{2} 15^2 + \dots + 15^{100}$.Thus,$2^{403} = 8[1 + 15(100 + \binom{100}{2} 15 + \dots)] = 8 + 15 	imes [8(100 + \binom{100}{2} 15 + \dots)]$.Therefore,$\frac{2^{403}}{15} = \frac{8}{15} + 8(100 + \binom{100}{2} 15 + \dots)$.The fractional part is $\frac{8}{15}$,which is given as $\frac{k}{15}$.Hence,$k = 8$.

If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$,then $k$ is equal to

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