What is the decimal part of the sixth power of the sum of two square roots?

[anemone]

Here is this week's POTW:

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Given that the decimal part of $X=\left(\sqrt{13}+\sqrt{11}\right)^6$ is $Y$, find the value of $X(1-Y)$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. Opalg

Solution from kaliprasad:

Spoiler

Let us take $Z = (\sqrt{13} - \sqrt{11})^6$

Now $ (\sqrt{13} - \sqrt{11}) < 1$ so $ Z < 1$

$\begin{align*}X+Z &= (\sqrt{13} + \sqrt{11})^6 + (\sqrt{13} - \sqrt{11})^6
\\&= 2((\sqrt{13})^ 6 + {6 \choose 2} (\sqrt{13})^4 (\sqrt{11})^2 + {6 \choose 4} (\sqrt{13})^2 (\sqrt{11})^4 + (\sqrt{11})^6)\\&= 2( 13^ 3 + {6 \choose 2} 13^4 * 11 ^2 + {6 \choose 4} 13^2 * 11 ^4 + 11^3)\,\,\text{which is an integer}\end{align*}$

As $Z < 1$ and $Y$ is decimal part of $X$ and $X+Z$ is an integer $\therefore Z = 1 - Y$ and

$\begin{align*}X(1-Y) &= XZ \\&= (\sqrt{13} + \sqrt{11})^6 * (\sqrt{13} - \sqrt{11})^6 \\&= (13-11)^6\\& = 2^6 \\&= 64\end{align*}$

Alternate solution from Opalg:

Spoiler

My little pocket calculator gives $\bigl(\sqrt{13} + \sqrt{11}\bigr)^6 = 110016$ (to 8 significant figures). So we can expect $X$ to be very close to an integer. This means that its fractional part $Y$ will be either very small or very close to $1$.

Square it first, to get $\bigl(\sqrt{13} + \sqrt{11}\bigr)^2 = 24 + 2\sqrt{143} = 2\bigl(12 + \sqrt{143}\bigr).$ Then cube that, getting $$X = \bigl(\sqrt{13} + \sqrt{11}\bigr)^6 = 8\bigl(1728 + 432\sqrt{143} + 5148 + 143\sqrt{143}\bigr) = 8\bigl(6876 + 575\sqrt{143}\bigr).$$ Write that as $X - 8\times6876 = 8\times575\sqrt{143}$, and square again: $$X^2 - 110016X + 64\times6876^2 = 64\times575^2\times143$$ (notice that the number $110016$ has made an appearance here!). Therefore $$X(110016-X) = 64(6876^2 - 143\times575^2).$$ A bit more arithmetic using the pocket calculator comes up with the pleasing result that $6876^2 - 143\times575^2 = 1.$ (For a deeper explanation of why this is so, notice that $\dfrac{6876}{575}$ is one of the convergents for the http://personal.maths.surrey.ac.uk/ext/R.Knott/Fibonacci/cfCALC.html of $\sqrt{143}$).

Therefore $X(110016-X) = 64$. Since we know that $X\approx 110016$, it follows that $110016 - X \approx \dfrac{64}{110016} \approx 0.0005817336$. This confirms that $X$ is indeed very slightly less than $110016$, its integer part being $110015$ and its fractional part $Y$ given by $1-Y = 110016 - X$, so that $Y \approx 1 - 0.0005817336 = 0.9994182664.$

Finally, $X(1-Y) = 64$.

Notice that, just by using the little pocket calculator, we get the result $X = 110015.9994182664$ correct to 16 significant digits.