The subject of repeating decimals has fascinated me since I was introduced to them in fourth grade. I may have mentioned before that my parents always sought out math enrichment activities for me, and during my fourth grade year, I enrolled in the Russian School of Mathematics. During one of our classes, we discussed the reciprocals of 2, 3, 4, 5, 6, 7, 8, 9, and 10, and how to derive them using long division (remember that a “reciprocal” of a number takes 1 as a dividend and the number as a divisor and determining the quotient). Why not we take a look at what they actually are…
| Fraction Representation | Decimal Representation |
| 1/2 | 0.5 |
| 1/3 | 0.333… |
| 1/4 | 0.25 |
| 1/5 | 0.2 |
| 1/6 | 0.1666… |
| 1/7 | 0.142857142… |
| 1/8 | 0.125 |
| 1/9 | 0.111… |
| 1/10 | 0.1 |
So notice how 1/2, 1/4, 1/5, 1/8, and 1/10 have decimals that terminate, while the other four go on forever. Furthermore, 1/7 is possibly the craziest of them all, being that it is equal to 0.142857142857142857…
Why don’t we examine why. Recall how when I talked about bases/radixes that our number system is base/radix-10, and that 10 = 2 x 5. First consider this: 4 = 2 x 2 and 8 = 2 x 2 x 2. Then, consider how 6 = 2 x 3 and 9 = 3 x 3. So why don’t we take a look at the next ten reciprocals (11-20)…
| Fraction Representation | Decimal Representation |
| 1/11 | 0.090909… |
| 1/12 | 0.08333… |
| 1/13 | 0.0769230769… |
| 1/14 | 0.07148257148… |
| 1/15 | 0.0666… |
| 1/16 | 0.0625 |
| 1/17 | 0.0588235… |
| 1/18 | 0.0555… |
| 1/19 | 0.0526315789… |
| 1/20 | 0.05 |
Here, only two reciprocals terminated: 1/16 and 1/20. Let’s look at their prime factorization: 16 = 2 x 2 x 2 x 2, while 20 = 2 x 2 x 5. The other primes obviously do not contain a 2 or a 5, while the composite numbers 12, 14, and 18 carry at least one other prime that isn’t 2 or 5. You can probably conclude that a reciprocal will have a terminating decimal representation if and only if its prime factorization is composed of 2s and/or 5s, but with no other prime numbers.
So let us take a look at the ones that DO repeat (I can talk about the terminating ones another time). As seen with 1/3, for example, the 3s will go on forever, and there will be no other digits. In 1/7, however, the pattern seems to be that “142857” repeats over and over and over again. This “142857” is referred to as a “repetend”, and the amount of digits that repeat is referred to as the “period”. In the case of 1/3, the period is 1 as the repetend is a mere “3”, while in the case of 1/7, the period is 6. In the case of 1/13, the 0 that recurs is included in the repetend, meaning that the repetend for 1/13 is actually “076923”, thereby meaning that the period is 6 just like with 1/7. The repetends for 1/17 and 1/19, however, are far more chaotic. The former’s is “0588234294117647” (16 digits), while the latter’s is “052631578947368421” (18 digits). In fact, many other reciprocals will have longer periods, which I will discuss later.
So notice how with 1/13, 1/17, and 1/19, we include the 0 that starts the decimal, but in 1/6, we do not as there is no 0 that repeats. Notice how this is also the case with 1/12, 1/14, 1/15, and 1/18. This is due to how with the prime factorization of the denominators, there are 2s and or 5s. If we want to examine numbers that include a 0 in its repetend as the first digit, then we should examine numbers that have a prime factorization which do not contain a 2 or 5. The lowest such number is 21, and 1/21 has a repetend of “047619”.
Returning to what we had earlier with 1/3, notice how 1/9 is 0.111… . Notice how when you multiply 0.111… by 3, you have 0.333…, which is 1/3. This is consistent with how 1/9 multiplied by 3 gives 1/3. But what happens when you have 0.111… multiplied by 9? In theory, you should have 0.999… , yet 1/9 multiplied by 9 should give 1. Indeed, this should be enough to conclude that the infamous number 0.999… is actually just equal to 1. There are many ways of proving that 0.999… is equal to 1, but that’s for another time.
Why I introduce 0.999… is that it will aid in my main point of discussion. We all know that 3 x 3 = 9, and using our earlier discussion on 1/9, consider how 0.333… multiplied by 3 is equal to 0.999…, or 1, which is consistent with how 1/3 x 3 = 1. But what happens when we take our repetends multiplied by the original numbers? Let us analyze using the numbers whose prime factorizations do not include a 2 or 5, along with a few others:
| Prime No. | Repetend of Reciprocal | Product | # of digits in Product |
| 3 | 3 | 9 | 1 |
| 7 | 142857 | 999999 | 6 |
| 9 | 1 | 9 | 1 |
| 11 | 09 | 99 | 2 |
| 13 | 076923 | 999999 | 6 |
| 17 | 0588234294117647 | 9999999999999999 | 16 |
| 19 | 052631578947368421 | 999999999999999999 | 18 |
| 21 | 047619 | 999999 | 6 |
| 23 | 0434782608695652173913 | 9999999999999999999999 | 22 |
| 27 | 037 | 999 | 3 |
| 29 | 0344827586206896551724137931 | 9999999999999999999999999999 | 28 |
Here, notice how the number of digits in the product is equal to the period of the repetend. Additionally, notice how the product is composed entirely of 9s as its digits. This is the case for all such numbers. Notice how the original numbers end in 1, 3, 7, and 9, as it cannot be even (to exclude 2) nor end in 5 (to exclude 5). So when you consider how the repetend repeats over and over in the actual decimal, and that when multiplying by the original number, you are essentially building 0.999…, which as mentioned before, is 1.
Notice how there’s quite a bit of variance with the periods thus far. Are there prime numbers that produce periods in between 1 and 28? Well as it turns out, yes…
| Period | Smallest Prime | Repetend |
| 1 | 3 | 1 |
| 2 | 11 | 09 |
| 3 | 37 | 027 |
| 4 | 101 | 0099 |
| 5 | 41 | 02439 |
| 6 | 7 | 142857 |
| 7 | 239 | 0041841 |
| 8 | 73 | 01369863 |
| 9 | 333667 | 000002997 |
| 10 | 9091 | 0001099989 |
| 11 | 21649 | 00004619151 |
| 12 | 9901 | 000100999899 |
| 13 | 53 | 0188679245283 |
| 14 | 909091 | 00000109999989 |
| 15 | 31 | 032258064516129 |
| 16 | 17 | 0588234294117647 |
| 17 | 2071723 | 00000048269001213 |
| 18 | 19 | 052631578947368421 |
| 19 | 1111111111111111111 | 0000000000000000009 |
| 20 | 3541 | 00028240609997175939 |
| 21 | 43 | 023255813953488372093 |
| 22 | 23 | 0434782608695652173913 |
| 23 | 11111111111111111111111 | 00000000000000000000009 |
| 24 | 99990001 | 000000010000999999989999 |
| 25 | 21401 | 0000467267884678286061399 |
| 26 | 859 | 00116414435389988358556461 |
| 27 | 757 | 001321003963011889035667107 |
| 28 | 29 | 0344827586206896551724137931 |
So consider this scenario: what if we take two distinct prime numbers that are not 2 and 5 and then examine the reciprocal of their product (you can confirm the following results using this online calculator)?
| Pair | Periods of Pair’s Reciprocals | Product | Period of Reciprocal of Product |
| 3 x 11 | 1, 2 | 33 | 2 |
| 3 x 7 | 1, 6 | 21 | 6 |
| 11 x 7 | 2, 6 | 77 | 6 |
| 7 x 13 | 6, 6 | 91 | 6 |
| 11 x 37 | 2, 3 | 407 | 6 |
| 37 x 7 | 3, 6 | 259 | 6 |
| 11 x 41 | 2, 5 | 451 | 10 |
| 101 x 7 | 4, 6 | 707 | 12 |
So there’s quite a bit to process here. Notice how with the first two rows, the periods seem to fit a multiplication statement: 1 x 6 = 6 and 1 x 2 = 2. This also is the case with the fifth and seventh rows. However, this fails with the third and fourth rows. The fourth row in particular is peculiar in that the periods are all identical. In the third and sixth rows, one period in the pair is also the period of the product. The most interesting is probably the last row. Is there a way to relate 4, 6, and 12?
Let us consider different examples, and this time, each one chosen will be deliberate (again, you can confirm the results with the link above):
| Pair | Periods of Pair’s Reciprocals | Product | Period of Reciprocal of Product |
| 7 x 73 | 6, 8 | 511 | 24 |
| 7 x 31 | 6, 15 | 211 | 30 |
| 7 x 43 | 6, 21 | 301 | 42 |
| 7 x 17 | 6, 16 | 119 | 48 |
| 7 x 23 | 6, 22 | 161 | 66 |
| 7 x 67 | 6, 33 | 469 | 66 |
| 7 x 29 | 6, 28 | 203 | 84 |
| 31 x 19 | 15, 18 | 589 | 90 |
By now, it should be apparent that the relation between the three numbers goes beyond a simple multiplication statement. Perhaps we should consider the divisibility of the larger number by the two smaller numbers. Indeed, this checks out: for example, 24 is divisible by both 6 and 8, as is 84 by both 6 and 28. With 589, 90 is divisible by both 15 and 18. This then begs the question: if we were to consider the multiples of the smaller two, at what point will we see the larger number? Let us try with 6 and 8 first:
| 6 | 12 | 18 | 24 | 30 | … |
| 8 | 16 | 24 | 32 | 40 | … |
As you can see, not only does this confirm that 24 is a common multiple as shown with how 24 is divisible by both numbers, it also their LEAST common multiple. Doesn’t that terminology sound familiar?
So let us see with the much earlier example with 4 and 6 and see for ourselves…
| 4 | 8 | 12 | 16 | 20 | … |
| 6 | 12 | 18 | 24 | 30 | … |
Once again, we find that 12 is the Least Common Multiple of 4 and 6.
This now leads me to the dilemma that I hinted at from the title of this blog. These results led me to pose the following theorem…
“Suppose that a function ‘f’, with the natural numbers excluding 1 acting as the domain and whole numbers acting as the co-domain, returns the period of the reciprocal of a number. Given two distinct prime numbers p and q such that neither p or q are equal to 2 or 5, f(pq) = LCM(f(p), f(q)).”
For example, f(7) = 6, f(101) = 4, f(707) = 12, and LCM(6, 4) = 12. Notice how with the domain, we do not want to consider f(1). Additionally, the co-domain MUST be the whole numbers as f(2) = f(5) = 0. Note how f(4), f(8), and f(10), among others, also produce 0 (these represent the numbers whose reciprocals have terminating decimals). Also, as a side note, because of how we’ve defined f, f is a surjective function (props to you if you understand this minor detail).
I have thus far been trying to determine how to prove this theorem, but my efforts have thus far been rather fruitless. I’ve attempted to use the period-repetend relationship and acting on an example (such as 7, 101, and 707), but this has also led nowhere. I do hope to achieve some breakthrough in the future, though, although WHEN is the question.
I also wish to elaborate more on some of my findings later on, as I have yet to discuss powers of primes, for example. I have also done a full dissection of the natural numbers from 1 to 1000 whose reciprocals produce periods of 99 or less, and I do wish to share some findings from that as well (in addition to some of the ones I’ve talked about in this blog).
Math Toolbox 