# Similarities in the digits of the powers of 2 and 5

Many may have noticed that the negative powers of 5 contain the same digits as the positive powers of 2: This pattern intrigued me. I started to wonder if it exists in different number bases. I soon realized that, for a base n, such a pattern seems to exist when you pick a pair of factors of n. As an example, in base 6, the digits will be the same for powers of 2 and 3: For number bases where n is a square number, that means there is a symmetry in the digits for the powers of √n. For example, let's look at the powers of 3 in base 9:

And, finally if n has multiple pairs of factors, the pattern works for each pair individually. Let's take base 12, with pairs of factors (2,6) and (3,4): I wonder why this pattern seems to universally emerge in these scenarios - and if the pattern is real at all! Is there anyone who has an idea on this?

• It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. Commented Aug 31, 2023 at 15:29
• Note that dividing by a power of $10$ is the same as moving the decimal place over (similarly for other bases).
– lulu
Commented Aug 31, 2023 at 15:29
• And that $10^n=2^n\cdot5^n$. Commented Aug 31, 2023 at 15:30

This is a consequence of the fact that $$10 = 2 \times 5$$, and thus:
$$5^{-n} = \frac{1}{5^n} = \frac{2^n \times 5^n \times 10^{-n}}{5^n} = 2^n \times {10}^{-n}$$
IOW, a negative power of $$5$$ is a positive power of $$2$$ scaled by the same power of $$10$$ (which means having the same digits, but shifting the decimal point). And vice versa.
Of course, this pattern generalizes to base 12 with $$12 = 2 \times 6 = 3 \times 4$$, and to any factorable (non-prime) base.
For base $$n$$, and any pair of $$ab = n$$, $$b^{-k} =\left ({n\over a}\right)^{-k} = n^{-k} a^k$$. The part $$n^{-k}$$ basically is just shifting period left or right in base $$n$$ system.