It is very easy to show that there exists a power of 3 - 3ⁿ - 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 (3ⁿ - 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:

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 you remember?  = 3.14159 26535... Then, there exists a power of, say, 2 that starts with 314 and another that starts with 14159. There also exists a power of 7 that in the octal system starts with 26535.

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·bk  N < (S+1)·bk,

where b is the base of the selected system. Replacing N with aⁿ and taking logarithms in base b gives

k + logbS n·logba < k + logb(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

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

Copyright © 1996-2010 Alexander Bogomolny