Binomial Theorem Ques 52
If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$, then $k$ is equal to
(2019 Main, 9 Jan I)
(a) 14
(b) 6
(c) 4
(d) 8
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Answer:
Correct Answer: 52.(d)
Solution:
- Consider,
$ \begin{aligned} 2^{403} & =2^{400+3}=8 \cdot 2^{400}=8 \cdot\left(2^{4}\right)^{100}=8(16)^{100}=8(1+15)^{100} \\ & =8\left(1+{ }^{100} C_{1}(15)+{ }^{100} C_{2}(15)^{2}+\ldots+{ }^{100} C_{100}(15)^{100}\right) \end{aligned} $
[By binomial theorem,
$ \left.(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots{ }^{n} C_{n} x^{n}, n \in N\right] $
$ =8+8\left({ }^{100} C_{1}(15)+{ }^{100} C_{2}(15)^{2}+\ldots+{ }^{100} C_{100}(15)^{100}\right) $
$=8+8 \times 15 \lambda$
where $\lambda{ }^{100} C_{1}+\ldots . .+{ }^{100} C_{100}(15)^{99} \in N$
$\therefore \frac{2^{403}}{15}=\frac{8+8 \times 15 \lambda}{15}=8 \lambda+\frac{8}{15}$
$ \Rightarrow \quad (\frac{2^{403}}{15})=\frac{8}{15} $
(where { . } is the fractional part function)
$ \therefore \quad k=8 $
Alternate Method
$2^{403}=8 \cdot 2^{400}=8(16)^{100}$
Note that, when $16$ is divided by $15$ , gives remainder $1$ .
$\therefore$ When $(16)^{100}$ is divided by $15$ , gives remainder $1^{100}=1$ and when $8(16)^{100}$ is divided by $15$ , gives remainder $8$ .
$\therefore \quad (\frac{2^{403}}{15})=\frac{8}{15}$.
(where { . } is the fractional part function)
$ \Rightarrow \quad k=8 $
BETA
