# Prove that $r(2^{55555})$ divisible by $5^5$

In this question there is a comment by @amclade, that

$$r\left(2^{55555}\right) \equiv 0 \pmod{5^5},$$

where $r : \mathbb{N} \rightarrow \mathbb{N}$ function gives the reverse of a number.

I've checked this using Maple, and the statements seems to be true – verified by a CAS –, but how could we prove it analytically?

• Since it is likely an accident, I'm not sure how one would prove it except by pure computation. It's more like "there are five 9s in a row at digit $x$ of $\pi.$" – Thomas Andrews Oct 11 '14 at 22:34
• @ThomasAndrews I do not see it like an accident. Take a look at the accurate proof by Fengyang Wang down below. – user153012 Oct 11 '14 at 23:19
• That proof looks like an "accident" to me - it's just a computational result. – Thomas Andrews Oct 12 '14 at 0:01
• For example, there is a reason that $e^{\pi\sqrt{163}}$ is incredibly close to an integer. There is something underlying that result, some deep math that is being exposed. The 'reason' for this result is purely computational, not revealing any properties of number, nor giving you any clue how to find similar cases. – Thomas Andrews Oct 12 '14 at 0:04

It is easy to check that $5^5=3125$. Note that because $100000 \equiv 0 \mod 3125$, the divisibility of a decimal number by $3125$ can be checked simply by inspecting the last $5$ digits.

So it remains to compute the first $5$ digits of $2^{55555}$. First let's compute the number of digits of this number. This can be done with the computation $\lceil 55555 \log_{10} 2 \rceil= 16724$.

This means the relevant digits can be multiplied by $10^{16719}$ to approximate the number from below. So our computation becomes $\frac{2^{55555}}{10^{16719}}$, which will spit out the first $5$ digits of the number (decimal places can be discarded). Doing a trivial simplification leaves $\frac{2^{38836}}{5^{16719}}$.

Computation is inevitable, but only the first $5$ significant figures are needed. It is convenient to take the logarithm of this fraction.

$${\log \frac{2^{38836}}{5^{16719}}}={38836 \log 2 - 16719 \log{5}}$$

For good measure, we'll compute this to $10$ significant figures. The result is $10.8714462403$. Now taking $e^{10.8714462403}$, we have $52651$. Reversing this much smaller number, we identify it as a multiple of $3125$ and complete the proof.

The number of computations needed with this approach is much lower than the naive method of exponentiation. A more intelligent means of taking $2^{55555}$ to $5$ decimal places of precision (which is what the above accomplishes) may be more efficient for a computer to do:

$$2^{55555} = 2^{32768} 2^{16384} 2^{4096} 2^{2048} 2^{256} 2^2$$

The computer simply computes $2$ to sufficient precision, and squares the number each time. These operations are very fast and the result will be computed very quickly, likely faster than computing logarithms.

• Very nice! I was writing precisely the same argument now, except I was elaborating another way to get the first 5 digits, but I was eventually going to use the same technique as you. – Jonas Gomes Oct 11 '14 at 22:35
• Last fun fact, that $15625=5^6$, so $15625/5^5=5$. Thank you for your answer. Nice approach. – user153012 Oct 11 '14 at 23:15