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\(x^{2^{2013}}\)

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

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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 · 1 year, 9 months ago

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@Daniel Liu 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 · 1 year, 9 months ago

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@Pranshu Gaba 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 · 1 year, 9 months ago

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

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

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@Lolly Lau Indeed. Thanks for your observation. Isam Dimacutac · 1 year, 7 months ago

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

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47 Raman Rai · 1 year, 9 months ago

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