2

Can you find the first digit of

$2^{2^{2^{2^{2^2}}}}$?

Basically, it is $2$ to the $2$ to the $65536$ power.

You cannot use a computer, but are allowed to use a calculator.

Good luck!

Daniel Mathias
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A Math guy
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2 Answers2

9

I presume we are looking for the leftmost digit.

I don't think this is doable with a calculator.

For comparison and by way of warm-up, an easier, solvable problem is to find the first digit of $A = 2^{65536}$.

Simply use logarithms base 10.

$\log A = 65536 \times \log 2 = 65536 \times 0.301029995664... = 19728.3017958...$

Taking the antilog (exponentiation to the power 10), we get:

$A = 2.00352977045... \times 10^{19278}$

So the first digit is $2$. It's only the decimal places of $\log A$ which tell us this: the whole number part $19728$ only affects the magnitude. I.e. we needed to compute $\log A$ to at least $6$ significant figures. Easily done with a calculator.

However, if we turn to $B = 2^A = 2^{2^{65526}}$ and try a similar game, we get:

$\log B = (2.00352977... \times 10^{19728}) \times \log 2 = 0.603122557976... \times 10^{19728}$

So in order to find the first digit of $B$, we need to know the term $0.603122557976...$ to at least $19729$ significant figures. This is beyond the scope of even the most prodigious calculators.

EDIT: Adding info provided in a useful comment by Daniel Mathias (thank you, thank you): OEIS has $20001$ digits for $\log 2$ (luckily this is greater than the required $19729$!), so computing (by computer) the exact value of $A=2^{65536}$ allows us to find $\log B$ with sufficient precision. The fractional part is $≈0.32634379468$ so $B≈2.12003872881×10^{10^{19728}}$

Laska
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    More accurately, $A\approx 2.003529930406846465 \times 10^{19728}$ and $\log B \approx 0.603122606263029537 \times 10^{19728}$ – Daniel Mathias Dec 09 '23 at 16:45
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    These considerations are exactly why I have a comment on the question asking whether the OP knows the thing is doable. Maybe there's some ingenious trick that makes it possible even though the "obvious" route is impassable, but I rather doubt it. – Gareth McCaughan Dec 09 '23 at 18:27
  • Yes @GarethMcCaughan it was your earlier comment that set me off on this path of thinking. – Laska Dec 09 '23 at 22:54
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    OEIS has 20001 digits for $\log 2$, so computing the exact value of $A = 2^{65536}$ allows us to find $\log B$ with sufficient precision. The fractional part is $\approx 0.32634379468$ and $B\approx 2.12003872881\times 10^{\text{I stayed up way too late for this}}$ – Daniel Mathias Dec 10 '23 at 06:11
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    How do you access OEIS without a computer? – msh210 Dec 10 '23 at 11:50
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    @msh210 With a cellphone. – enzo Dec 14 '23 at 19:38
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If we can choose

to answer in something other than base 10, then we can give our answer in base 2, in which case the first (leftmost) digit is 1. (and the rightmost digit is 0.)

The full solution

in base 2 is $10000...000_2$, which is a 1 followed by $2^{2^{2^{2^2}}}$ $0$s, or a 1 followed by $2^{65536}$ $0$s

$2 = 2_{10} = 10_2 $, or a 1 followed by 1 zero.
$2^2 = 4_{10} = 100_2$, or a 1 followed by 2 zeros.
$2^{2^2} = 16_{10} = 10000_2$, or a 1 followed by $2^2$ zeros.
$2^{2^{2^2}} = 65536_{10} = 10000000000000000_2$, or a 1 followed by $2^{2^2}$ zeros.
$2^{2^{2^{2^2}}}$ is too large to write, but following the pattern we know it's a 1 followed by $2^{2^{2^2}}$ zeros.
$2^{2^{2^{2^{2^2}}}}$ is far too large to write, but following the pattern we know that it's a 1 followed by $2^{2^{2^{2^2}}}$ zeros.

I believe this answer is incomplete, however, because

base 10 was used for the problem so is implicitly expected for the solution.

We could convert the full base 2 answer to base 10, but I'm not sure of a way to do that for such a large number which can be done on a calculator. Perhaps there's a clever way that solves it digit by digit.

jla
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