Question

$(7+4\sqrt{3}{)}^{100}$ = I + f, where I is the integral part and f is the fractional part of

$(7+4\sqrt{3}{)}^{100}$. Find the value of (I + f)(1 - f)

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Answer 1

$(7+4\sqrt{3}{)}^{100}$ = I + f

Consider

$(7-4\sqrt{3}{)}^{100}$

0 < $(7-4\sqrt{3}{)}^{100}$ < 1

Let $(7-4\sqrt{3}{)}^{100}$ = f'

I + f + f' = $(7+4\sqrt{3}{)}^{100}$ + $(7-4\sqrt{3}{)}^{100}$ = 2[${}^n{C}_0$ $7^{100}(4\sqrt{3}{)}^0$ + ${}^n{C}_1$$7^{99}(4\sqrt{3}{)}^1$..................]

⇒ 1 + f + f' is an even integer.

But we know 0 < f + f' < 2 ..............(1)

For I + f + f' to be an integer, f + f' should also be an integer.

The only integer value it can take is 1(from(1)).

⇒ f + f' = 1

⇒(1 - f) = f'

⇒(I + f)(1 - f) = (I + f)(f')

=$(7+4\sqrt{3}{)}^{100}$ $(7-4\sqrt{3}{)}^{100}$

= $\left[\right(7+4\sqrt{3}\left)\right(7-4\sqrt{3}){]}^{100}$

= $[1{]}^{100}$ = 1