# A Property of a^{n}

It is very easy to show that there exists a power of 3 - 3^{n} - that ends with 001. A quick glance at the proof will reveal that this is also true of powers of other odd numbers (except those divisible by 5.) As a different extension, we may select arbitrary long sequences of zeros that end with a single 1 - for each of them there exists a power of three that ends with this sequence. And, again, the same applies to every odd number excluding multiples of 5.

The differences of powers like (3^{n} - 3^{m}) also have a more generic (though a simpler) property that for any integer A, there exist two powers of 3 whose difference is divisible by A. Both facts are most easily proved with the help of the Pigeonhole Principle.

The latter is also used to prove the following astonishing property:

(*) | For every sequence of decimal digits S, there exists a power of 3 that starts with the sequence S. |

The same is true for other integers instead of 3. And there is no reason to restrict oneself to the base 10. Any *positional* system will serve the same purpose.

Say, how many digits of the number π you remember?

We do not know how long the power will be but, regardless of how many digits is takes to write a number, the fact that N starts with a sequence S can be expressed as

S·b^{k} ≤ N < (S + 1)·b^{k},

where b is the *base* of the selected system. Replacing N with a^{n} and taking logarithms in base b gives

k + log_{b}S ≤ n·log_{b}a < k + log_{b}(S+1)

The proof of this fact takes a few pages but can be shortened if one employs a lemma we used while shredding the torus.

## Remark

Of course it's not true that there exists a power of 3 that ends with 3141592 in the decimal system.

## References

- R. Honsberger,
*Ingenuity in Mathematics*, MAA, New Math Library, 1970

|Contact| |Front page| |Contents| |Algebra| |Up|

Copyright © 1996-2018 Alexander Bogomolny