The Prime Factors of 999999

Repetends

The prime factorization of $999999$ is interesting. It is:

$999999 = 3 \times 3 \times 3 \times 7 \times 11 \times 13 \times 37. \tag{1}$

We can use this to find the repeating decimal representation of some common fractions. To see this, first note that we can write the reciprocal of $999999$ as

$\frac{1}{9999999} = \frac{0.\overline{999999}}{999999} = 0.\overline{000001}. \tag{2}$

Using this fact, here is the decimal representation of $1/7$:

$\frac{1}{7} = \frac{3 \times 3 \times 3 \times 11 \times 13 \times 37}{999999} = \frac{142857}{999999} = 0.\overline{142857}. \tag{3}$

Clearly, this idea works for any factor of $999999$. In the case of the prime factors above, we have:

$\begin{aligned} \\ \frac{1}{3} &= \frac{333333}{999999} = 0.\overline{333333}, \\\\ \frac{1}{7} &= \frac{142857}{999999} = 0.\overline{142857}, \\\\ \frac{1}{11} &= \frac{90909}{999999} = 0.\overline{090909}, \\\\ \frac{1}{13} &= \frac{62937}{999999} = 0.\overline{062937}, \\\\ \frac{1}{37} &= \frac{27027}{999999} = 0.\overline{027027}. \end{aligned} \tag{4}$

And of course, this idea works for non-prime factors. For example:

$\frac{1}{27} = \frac{7 \times 11 \times 13 \times 37}{999999} = \frac{37037}{999999} = 0.\overline{037037}. \tag{5}$

The repeated number in the decimal, e.g. $142857$, is called the repetend or reptend. In the case of $1/7$, the repetend is actually a cyclic number, and therefore we can easily compute other fractions such as $3/7$ by simply rotating the digits in the repetend appropriately, e.g.,

$\frac{3}{7} = 0.\overline{428571}. \tag{6}$

Furthermore, this gives us a nice trick to compute repeating decimals when the denominator is a multiple of seven, such as

$\frac{1}{14} = \frac{0.5}{7} = \frac{50}{7} \frac{1}{100} = 0.07\overline{142857}. \tag{7}$

Basically, you have to do some long-division in your head, but the moment you get a remainder you recognize, you’re essentially done.

For $1/11$, note that we can use our nines times table to easily compute other fractions repetends, such as

$\frac{4}{11} = 0.\overline{363636}. \tag{8}$

Of course, we are only scratching the mathematical surface here. A proper generalization for why this works lies in number theory. But I just wanted to document for myself this simple case, since it justifies some common rules used when computing fractions.

Prime factorization

One might ask: how could we compute the prime factorization of $999999$ “by hand”? The way I was taught was to observe that

$999999 = 1001 \times 999, \tag{9}$

and to then brute-force $1001$, which isn’t too hard since we only have to check odd numbers up to roughly $32$. But a more elegant way in my mind is to use the difference of two squares and the sum of two cubes:

$\begin{aligned} 999999 &= (10^6 - 1) \\ &= (10^3 - 1)(10^3 + 1) \\ &= (10^3 - 1)(10 + 1)(10^2 - 10 + 1) \\ &= 999 \times 11 \times 91. \end{aligned} \tag{10}$

Using times tables and basic divisibility rules, e.g. $999$ and $111$ are clearly divisible by three, we can easily factorize the remaining terms. I took this idea from here.