# Nines in Quinary

In my article “Nines in Binary”, I proved the following: positive integers of the form 10n-1, that is, integers made up of n digits of 9s, have binary representations with exactly n digits of trailing 1s. Pat Ballew made a clever observation, adapting my result to prove an equivalent statement for base 5 (quinary): positive integers of the form 10n-1 have quinary representations that have exactly n digits of trailing 4s. For example, 9999 in decimal is 304444 in quinary.

In “Nines in Binary”, I derived an expression for 10n – 1 that shows its structure as a binary number:

10n – 1 = (5n – 1) 2n + (2n – 1)

Pat derived a similar expression for 10n – 1 that shows its structure as a quinary number:

10n – 1 = (2n – 1) 5n + (5n – 1)

In essence, he swapped the 2s and 5s, making it the “dual” of my formula, if you will.

I’ll show the details of the derivation and prove why the formula works.

## Deriving the Expression

To derive the formula for the “binary expression” of 10n – 1, I worked backwards from the patterns I saw in its binary representation:

(5n – 1) 2n + (2n – 1) = 5n2n – 2n + 2n – 1 = 10n – 1

Pat derived the formula for the “quinary expression” of 10n – 1 without having to look at its representation in quinary — he simply modified the algebra that I presented. He observed that – 2n + 2n could be replaced with – 5n + 5n, since both equal 0. This allowed him to rearrange the expression as follows:

10n – 1 = 2n5n – 5n + 5n – 1 = (2n – 1) 5n + (5n – 1)

## What the Expression Shows

The quinary expression has a leading part, 2n – 1 shifted left n places in base 5, and a trailing part, 5n – 1. The trailing part shows that there are at least n trailing 4s, since 5n – 1 is n 4s in quinary (like 2n – 1 is n 1s in binary, 10n – 1 is n 9s in decimal, etc.). The leading part shows that there are exactly n trailing 4s, since 2n – 1 does not end with the digit 4 in quinary. Let’s see why.

To show that 2n – 1, when expressed in base 5, does not end in 4, we can use modular arithmetic. Let’s start by analyzing the possible ending digits of the positive powers of two:

This shows the numbers 2n (n ≥ 1) mod 5 cycle through the ending digits 2, 4, 3, and 1. This means that the numbers 2n – 1 cycle through the ending digits 1, 3, 2, and 0. The digit 4 is not on the list!

## Nonnegative Powers of Ten in Quinary

In my article “A Pattern in Powers of Ten and Their Binary Equivalents”, I showed that the trailing digits of nonnegative powers of ten, in binary, match the power of ten; in other words, a 1 is followed by n trailing 0s. For example, 10010 = 11001002.

A similar, though slightly weaker property, holds for nonnegative powers of ten in quinary: 10n ends with n 0s. A nonnegative power of ten is a multiple of a power of two and a power of five: 10n = 2n * 5n. A power of two in base 5 ends in 1, 2, 3, or 4 — never 0 — as I showed above. When you multiply a power of two by a power of five, you shift the power of two left by n places in base 5, which adds n trailing 0s. So the quinary representation ends with a nonzero digit followed by n 0s — exactly n trailing 0s!