$x^{2^{2013}}$

Let $x$ be a real number so that $x+ \frac{1}{x} = 3$ . Find the last two digits of $x^{2^{2013}} + \frac{1}{x^{2^{2013}}}.$

#Algebra

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Note that given $x^{2^k}+\dfrac{1}{x^{2^k}}=n$ then $x^{2^{k+1}}+\dfrac{1}{x^{2^{k+1}}}=\left(x^{2^k}+\dfrac{1}{x^{2^k}}\right)^2-2=n^2-2$

Thus we can define a sequence $\{a_k\}$ to be $a_0=3$ and $a_{n+1}=a_n^2-2$ which will give us $a_k=x^{2^k}+\dfrac{1}{x^{2^k}}$

We want to find the last two digits of $a_{2013}$ . Note that when $k\ge 1$ , the units digit of $a_k$ is $7$ , so brute force calculation of the period of this recursion won't be that bad: $a_0\equiv 3\pmod{100}$ $a_1\equiv 7\pmod{100}$ $a_2\equiv 47\pmod{100}$ $a_3\equiv 7\pmod{100}$ $\vdots$

And thus we see that the last two digits of $a_k$ are periodic with period $2$ . It remains to find the last two digits of $a_{2013}$ , which we see is clearly $\boxed{7}$ .

Daniel Liu - 5 years, 12 months ago

Actually, $a_0 = 3$ and $a_1 = 7$ , so $a_{2013} \equiv 7 \pmod{100}$ . Also, is it possible to find an explicit formula for the sequence?

Pranshu Gaba - 5 years, 12 months ago

Thanks for the correction.

As for an explicit formula, just note that $x+\dfrac{1}{x}=3\iff x^2-3x+1=0\iff x=\dfrac{3\pm\sqrt{5}}{2}$

Thus, we have $a_n=x^{2^n}+\dfrac{1}{x^{2^n}}=\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}+\dfrac{1}{\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}}$

Daniel Liu - 5 years, 12 months ago

@Daniel Liu – https://oeis.org/A005248 saves a lot of calculating effort :)

Lolly Lau - 5 years, 10 months ago

raman rai - 5 years, 12 months ago

hmm i did the same . Answer is 47.

Sarthak Gupta - 5 years, 12 months ago

Isn't this one of your problems :\

Lolly Lau - 5 years, 10 months ago

Indeed. Thanks for your observation.

Benedict Dimacutac - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

x22013x^{2^{2013}}x22013

Comments

$x^{2^{2013}}$