Suppose F is the fractional part of M = (13+11)6then value of M (1– F) is

A
128
B
64
C
32
D
16

Solution:

Let $N = (\sqrt{13} - \sqrt{11})^{6}$

As $(\sqrt{13} - \sqrt{11}) = \frac{2}{\sqrt{13} + \sqrt{11}}$ we get $0 < N < 1$

Also, $M + N = (\sqrt{13} + \sqrt{11})^{6} + (\sqrt{13} - \sqrt{11})^{6}$

$\begin{array}{l} = 2 [^{6} C_{0} (\sqrt{13})^{6} +^{6} C_{2} (\sqrt{13})^{4} (\sqrt{11})^{2} + \\ ^{6} C_{4} (\sqrt{13})^{2} (\sqrt{11})^{4} +^{6} C_{6} (\sqrt{11})^{6}] \end{array}$

$M + N$ is an integer, say, $J$

Let, $M = K + F$ where $K$ is the greatest integer contained in $M$ .

We have

As $0 < N < 1, 0 < F < 1$ we get, $0 < N + F < 2$

Also, $N + F$ is an integer

$\Rightarrow N + F = 1 \Rightarrow 1 - F = N$

Thus $M (1 - F) = (\sqrt{13} + \sqrt{11})^{6} (\sqrt{13} - \sqrt{11})^{6} = 2^{6} = 64$

Suppose $F$ is the fractional part of $M$ = $(\sqrt{13} + \sqrt{11})^{6}$
then value of $M (1 - F)$ is

Solution:

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