×

# $$x^{2^{2013}}$$

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

Note by Isam Dimacutac
1 year, 9 months ago

Sort by:

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 $$\boxed{7}$$. · 1 year, 9 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? · 1 year, 9 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 $$a_n=x^{2^n}+\dfrac{1}{x^{2^n}}=\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}+\dfrac{1}{\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}}$$ · 1 year, 9 months ago

https://oeis.org/A005248 saves a lot of calculating effort :) · 1 year, 7 months ago

Isn't this one of your problems :\ · 1 year, 7 months ago

Indeed. Thanks for your observation. · 1 year, 7 months ago

hmm i did the same . Answer is 47. · 1 year, 9 months ago