# First digit of a large number

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!

• Oh yes,it should be six twos.Sorry for the confusion. Dec 9, 2023 at 3:08
• By first digit, do you mean right-most or left-most digit?
– jla
Dec 9, 2023 at 10:32
• Do you know that this is actually doable? Dec 9, 2023 at 12:15
• The question is "Can you find it?". The answer is "Nope.". Dec 9, 2023 at 22:04

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 $$B≈2.12003872881×10^{10^{19728}}$$

• More accurately, $A\approx 2.003529930406846465 \times 10^{19728}$ and $\log B \approx 0.603122606263029537 \times 10^{19728}$ Dec 9, 2023 at 16:45
• 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. Dec 9, 2023 at 18:27
• 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 $B\approx 2.12003872881\times 10^{\text{I stayed up way too late for this}}$ Dec 10, 2023 at 6:11
• How do you access OEIS without a computer? Dec 10, 2023 at 11:50
• @msh210 With a cellphone.
– enzo
Dec 14, 2023 at 19:38

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.