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Do there exist integers $r,s,t\ge100$ such that the decimal representation of $(r+\sqrt{s})^t$ is of the form $~~~\ldots\ldots2015\,.\,2016\ldots\ldots$?

(In other words: the four digits before the decimal point are 2015, and the first four digits after the decimal point are 2016.)

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The answer is yes, this is possible. Fix $t = 100$. A big key is found in this excellent puzzle:

The quantity $(r + \sqrt{s})^{100} + (r - \sqrt{s})^{100}$ is an integer, since all the terms with an odd $\sqrt{s}$ power cancel.

Notice that $\{r - \sqrt{s}\}_{r, s \geq 100}$ is a dense subset of $\mathbb{R}^{>0}$. To see this, note that the quantity $\epsilon_m := m - \sqrt{m^2 - 1} = \frac{1}{m + \sqrt{m^2 - 1}}$ goes to zero for large $m$. We can then take $k \epsilon_m$ to get close to any value we choose.

Hence $\{(r - \sqrt{s})^{100}\}_{r, s \geq 100}$ is also dense in $\mathbb{R}^{>0}$. In particular, we can find an $r$ and $s$ such that $(r - \sqrt{s})^{100}$ has any particular finite substring of decimal digits after the decimal that we want. This allows us to force $(r + \sqrt{s})^{100}$ to have any digits (after the decimal) that we want.

Thus, we force $(r + \sqrt{s})^{100}$ to have the substring of digits $20152016$ appear roughly a googol places after the decimal point. Then take $(10r + \sqrt{100s})^{100}$ and we will have our answer.

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  • $\begingroup$ Might as well be precise and say a googol-4 places. Other then that +1. $\endgroup$ – DrunkWolf Dec 29 '15 at 14:56
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    $\begingroup$ I don't follow the $\epsilon_m$ part: how does $m - \sqrt{m^2-1} = \frac{1}{m - \sqrt{m^2-1}}$? $\endgroup$ – Lawrence Dec 30 '15 at 7:33
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    $\begingroup$ @Lawrence it says : $m - \sqrt{m^2 - 1} = \frac{1}{m + \sqrt{m^2 - 1}}$ which can be seen quite easily from: $(m - \sqrt{m^2 - 1})(m + \sqrt{m^2 - 1})=m^2-(m^2-1)=1$ $\endgroup$ – DrunkWolf Dec 30 '15 at 11:53
  • $\begingroup$ @DrunkWolf Thanks for the explanation! Brilliant construction also with the difference of squares summing to 1. $\endgroup$ – Lawrence Dec 30 '15 at 13:07
  • $\begingroup$ Very nice solution! $\endgroup$ – Fimpellizieri Dec 30 '15 at 22:29
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By brute-force search, yes. I started by searching over all tuples $(r,s,t)$ less than $1000$, stopping at the first example I found:

$$\left(138 + \sqrt{320}\right)^{570} \approx 10^{1249.9041}$$

In order to find the smallest example, I used the following strategy to search all examples smaller than the previous best (which I'll call $x$). Since the inside of the parenthesis $(r+\sqrt{s})$ must be at least $100+\sqrt{100}=110$, we can find the largest exponent we need to search by taking the logarithm:

$$ x = (r+\sqrt{s})^{t} > (110)^{t} \\ \log x > \log\left(110^t\right) \\ t < \frac{\log x}{\log 110} $$

Thus we can start our search with the exponent at $\lfloor \log x / \log 110 \rfloor$ and search all exponents down to $100$.

For a given exponent $t$, we then need a way to search all pairs $(r,s)$. To put this mathematically:

$$ (r+\sqrt{s})^t < x \\ r + \sqrt{s} < x^{1/t} $$

Starting with $r$, we start our search at $r=100$ and end it at:

$$ r + \sqrt{100} < x^{1/t} \\ r = \left\lfloor x^{1/t} - 10\right\rfloor $$

Then, for each $r$ we search over $s$ from $s=100$ to:

$$ r + \sqrt{s} < x^{1/t} \\ s = \left\lfloor \left(x^{1/t} - r\right)^2\right\rfloor $$

Every time we find a tuple $(r,s,t)$ with the desired property, we set $x$ to that new value (to reduce the number of tuples searched on further iterations). Note that any exponents we've already searched don't need to be searched again with the smaller limit.

Programmatically, this looks something like the following:

x = (138 + Sqrt[320])^570

For[t = Floor[Log[110, x]], t >= 100, t--,
  For[r = 100, r < x^(1/t) - 10, r++,
    For[s = 100, s < (x^(1/t) - r)^2, s++,
      If[First@RealDigits[(r + Sqrt[s])^t, 10, 8, 3] == {2, 0, 1, 5, 2, 0, 1, 6},
        x = (r + Sqrt[s])^t
      ]
    ]
  ]
]

(My actual code is a little different, but the strategy is the same.)

After a letting my search run for about a day, I realized that the search space was shrinking enough with each new $x$ that I could determine the smallest value in a reasonable timeframe. I let the search run for three more days, and the smallest value I found was:

$$ \left(140+\sqrt{27\ 027}\right)^{102} \\ \small \begin{split} = \phantom{} & 20473626230261221907121652389205011207101608771896002935841545009658234973292645 \\ & 17085491200709344560082101956950241075706483286182828328461907335813082258790620 \\ & 99769785021312880153909437562183881164238660035585722314971992368862943216420220 \\ & 1161896814\underline{2015.2016}08038424667587735557301174134230740327977685729631456117821\ldots \end{split} \\ \approx 10^{253.311194\ldots} $$

Rerunning the search produced no results, so I'm fairly confidant that this is the smallest example. There should be about 20 million smaller $(r,s,t)$, a fairly reasonable number to check by computer, so I would appreciate if someone could double-check my results!

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  • $\begingroup$ So has @TylerSeacrest found the smallest solution? $\endgroup$ – ghosts_in_the_code Dec 30 '15 at 11:32
  • $\begingroup$ @ghost_in_the_code: this is a much smaller solution than mine. My $r$ and $s$ would be roughly a googol. This is surprisingly small. $\endgroup$ – Tyler Seacrest Dec 30 '15 at 17:02
  • $\begingroup$ @TylerSeacrest: It's not too surprising, considering that the three numbers multiplied together are on the same order of magnitude as 20152016. $\endgroup$ – Joe Z. Mar 1 '16 at 5:07

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