Things I learned Late in Life about Repeating Decimals and Prime Divisors

In my last blog I mentioned that if you write out any three digit number and repeat it two times it will be divisible by 7, and 13. I didn't mention it, but it will also be divisible by 11. The question of course is a) why, and b) how does that help us understand divisibility and primes. The easy reason is that any such number, xyzxyz, is going to have a factor of 1001 (xyz) and since 7, 11, and 13 all are factors of 1001, they will divide into it evently. Now if you repeat a two digit number 3 times (xyxyxy) then you get a number that is divisible by 7 and 13, and 37, but not 11. I'm not sure it works for all numbers, but I have noticed some patterns in what I am going to call the "modular sequence" of prime numbers. Becuase it is short and easy to check on your calculator, I will use 7 and 13 for examples, and then show some extensions for others.... so here goes.

The modular sequence of 7 is 5,4,6,2,3,1... Ok.. what does that mean... look at the 2, it is in the hundreds place... Suppose I take a number, 243 and find the remainder on division by 7. (in this case 243 mod 7 == 5). If you increase the number by 100, the remainder will increase by 2.. (343 mod 7 == 5+2 == 0; and 443 mod 7 == 0+2==2); and similarly an increase in the value of the number by 1000 would increase the remainder by 6 (or reduce it by 1, which is the same thing.... imagine that 13 divided by 7 is one with a remainder of six, but you could think of it as two with a remainder of negative one).

Now how does that help us explain the fact that 7 divides xyxyxy and xyzxyz. One way to find the remainder that I mentioned earlier is to use the dot product of the vector x,y,x,y,x,y times the modular sequence, 5,4,6,2,3,1. You end up with 14x + 7y which is of course divisible by 7 because both the 14 in the first term and the 7 in the second are divisible by 7. In the same way, the dot product of xyzxyz and 546231 will give us 7x+7y+7z, which is also divisible by 7. And just as an added treat, xxxxxx will also be divisible by 7 because the totatl of all the numbers in the modular sequence is 21, so the dot product would be 21 x.

Ok, a little more about the modular sequence to help you see the patterns that I think are true. THe modular sequence is the same length as the repeat cycle of the unit fraction 1/p. For example, 1/7 and 1/13 both are repeating decimals of length 6. 1/7 = .142857.... and 1/13 = .076923...The modular sequence for 13 is 4, 3, 12, 9, 10, 1. If we add up all the numbers we see that the sum is 3 (13) = 39 [remember that the sequnce for 7 summed to 3(7)] So let's form a rule.. it works, you can check it later. THe sum of the terms in the modular sequence will be p(n)/2 where p is the prime number in question, and n is the length of the sequence. [A similar formula holds for the sum of the digits in the repeating cycle of 1/p, the sum of the digits will be 9d/2... every repeating cycle has a digit sum which is a multiple of nine].

If the length n is divisible by some number x, then the sum of every xth digit in the modular sequence will add to a multiple of p. This is what we saw with every second digit of the sequence for seven (5+6+3=14 and 4+2+1=7) . Since the length of 1/13 is also six digits, we can do the same with the alternate digits of the modular sequence for 13 ( 4, 3, 12, 9, 10, 1) 4+12+10 = 26 and 3+9+1=13... (not able yet to explain the 2:1 ratio, or if it persists in all even length sequences). If we add every third digit for seven we get 5+2=4+3=6+1. For 13 we get 4+9=3+10=12+1. A similar idea works in the repeat cycle of any prime for which the length of the cycle is divisible by two; all the numbers in the last half are the nines complement of the numbers in the first half... 1/7 = 142857... look at 142 and then 857; 1+8=9 and the same for each digit. That was a trick I showed earlier to complete long strings on your calculator.

Ok, how about a harder one.. 1/17 has a repeating decimal 16 digits long, .0588235294117647 (for decimals 1/p if you want to see the back half of the sequence, do (p-1)/p on your calculator... look at 16/17 on your calculatror for instance) First, confirm that the corresponding numbers in each half add up to nine. 0+9, 5+4, etc.. Now in the modular sequence for 17, they will add up to 17, so when we have the first half, we can just use compliments to get the second half. FOr 17 the sequence is 12, 8, 11, 13, 3, 2, 7, 16, 5, 9, 6, 4, 14, 15, 10, 1. Take a moment to see that the corresponding parts in each half add up to 17....

Now my conjecture is that if the length n is divisible by d, then any number of d digits repeated enough times to make d digits will be divisible by p, in this case, 17. If this is true then in the modulus the numbers located d units apart must add up to a multiple of p. OK, so 2 divides 16, so every other number should add up to a multple of 17; 12+ 11+3+7+5+6+14+10=68 which is 4x17, and 8+13+2+16+9+4+15+1=68 also. Ok, it works for two, but four also divides 16, so every fourth number should add up to a multple of 17 as well. The first set would be 12+ 3+ 5+14=34 which is 2x17 (are YOU seeing a pattern?)... the second is 8+2+ 9+ 15= 34.. you try the other two.

If you want to play around with this some more, as I do, Here is some information of interest:

The length of the divisors, and some of the repeating digits and modular sequences that I have done so far out of the first few primes

n a(n) repeat seq mod seq


2 0

3 1 .3 1

5 0

6 1 6 4

7 6 142857 5,4,6,2,3,1


11 2 09 10,1

13 6 076923 4, 3,12,9,10,1

17 16 0588235294117647 2, 8, 11, 13, 3, 2, 7, 16, 5, 9, 6, 4, 14, 15, 10, 1

19 18 052631578947368421 2, 4, 8, 16, 13, 7, 14, 9, 18, 17, 15, 11, 3, 6, 12, 5,10,1

23 22 0434782608695652173913

29 28 0344827586206896551724137931

31 15 032258064516129

37 3 027 26,10,1

41 5 02439 37,16,18,10,1

43 21

47 46 0212765957446808510638297872340425531914893617

53 13

NOtice that the numbers whose sequence lengths are not divisible by two (such as 31 or 41) do not yield complimentary halfs, but they should still obey the division rules. (I hope that pqrstpqrstpqrst will be divisible by 31 since it repeats a sequence of five digits three times,) Have fun...

Out of the Mouth of a Donkey?

Got an e-mail a couple of days ago from an Indian Mathematician named Sajeev Singh. Sanjeev asked me to let people know about his work at Total Gadha where he posts notes about mathematical ideas with some comments etc. I took a look and found some really well done posts on some interesting ideas about math that is accessible to high-school and college age folks. Gadha, by the way, if my memory serves me in old age, means Donkey, but I'm not sure how that reflects in the local culture. One more reason to visit India some day.

One of the notes he posted took me back, yet again, to the question about the 6! or 720 six-digit numbers that can be made using only the digits 1,2,3,4,5, and 6 once each. I realized that with a single brief mental calculation you could confirm that a) none of them were prime, and b) none of them were perfect squares. Why NOT?? (think, then read)

The reason has to do with the digit root; what used to be called casting-out-nines(and some of us still do call it that). If you add the digits, 1+2+3+4+5+6 the total is 21, and if you add 2+1 you get 3, so the remainder when any of the 720 numbers is divided by nine is going to be three, try it. 243561, for example, when divided by nine has a quoatient of 27062 with a remainder of three. Any number that has a digit root of three is divisible by three. All 720 of the six-digit arrangements are divisible by three. But how do I know none of them is a perfect square???? Same answer, the digit root is 3. THe digit root of a square number is the square of the digit root of the square root. Let's look at some numbers and their squares:

1² = has digit root of one

2²=has digit root of four

3²=digit root of nine

4²=16, so digit root of seven

5²=25 so digit root of seven

6²=36 so digit root of nine

7²=49 so digit root of four

8²=64 so digit root of one

9²=81 so digit root is nine

Keep goin.. these are the only four numbers you get for the digit root of a square,... ever...square 37?? digit root is one... square 155?... digit root is four.. you do these in your head.. To find the digit root of the square of 23 you add the 2 and 3 to get five, so the digit root of 23 is five. That means the digit root of 23 squared is the digit root of 5 squared.... seven. 23 ² = 529. Add the digits you get 16 and add those digits you get...yep, 7. What does NOT ever happen is a digit root of 3; so none of the 720 numbers are square numbers. Can you prove than none of them are perfect CUBES????

Two things better than being reminded of something you once knew are to learn something you never knew, and even better, to be stimulated by a question to discover something new on your own.... and both those things happened after going to TotalGahda. Rather than throw all that at you and once though, I will give you a chance to experiment with one.

Think of a two digit number, now write it down three times xyxyxy and divide by 7. What is the remainder? Wait, divide it by 13.... NOw what is the remainder? I just discovered this morning that this works.. and proved a general version of it which I will include next time with a conjecture about a couple of extensions to all primes... not sure if or who might have done it before... and I can add that if you write the two digit number out eight times, it will be divisible by 17. It will also work if you write down any four digit number four times, or any eight digit number twice (wow, what a clue)

If you have trouble dividing big long numbers and finding the remainder, try using the Google search engine. Just enter something like "4512451245124512 mod 17 =" in the search box and the Google calculator takes over and gives you the remainder (modulus)

Ok, IF you get stuck on that one, you can try this, write down any THREE digit number, and copy it again...xyzxyz to make a number... now divide by seven again, and by 13???? The reason this one works may be a little easier(I've know it has been around in calculator books before).... see if you can figure it out. I'll be back in a couple of days with the additional material, and if you KNOW this has been done already, especially if you have a citation, Please advise.

Repeating Decimal Periods and Patterns

In my last post I asked, "IF you try to find the decimal number for 1/17 on the TI-84 it says 1/17 = .0588235294... and that's all you get. Now the challenge.... can you write out the complete decimal expansion (one complete cycle of the repeating expansion) of 1/17 using only what the calculator gave you and simple mental arithmetic...?"

The repeating portion of a decimal fraction is called the repetend, and the number of digits in the repeating cycle is called the period or length of the repetend. For 2/7, as shown above, the repetend is 285714 and has a length of 6. To figure out the original question I asked, it helps to know a rule about the length of repetends; for any prime number p > 5 the length of the repetend of 1/p is a factor of p-1. This seems first to have been stated by Johann H. Lambert in 1769. I find it odd that we had discovered so much about Primes so early, and this little tidbit came so late. For seven above, the length of 1/7 (or n/7) is six digits long. In looking at 1/17, the rule says that the period of the repetend must be a factor of 16, so the length will be 2, 4, 8, or 16 digits. Looking at what is given, we realize that the period can not be any of the smaller numbers, so it must have a 16 digit cycle, but how do we find the other digits?

Look back at 2/7 and another interesting property of repeating decimals may pop out at you. This one was published by Henry Goodwyn in 1802. He noted that if you divide the period into a first half and a second half, the sum of the digits in corresponding positions add up to nine.. For 2/7 above the repeat cycle is 285714 so the halfs are 285 and 714. Note that 2+7 = 8+1= 5+4 = 9, and Goodwyn showed this is true for ALL repeating decimals..... so it must be true for 1/17. If we know the period is 16 digits, then the first eight are 05882352 and to find the next eight, we just use the nines compliment of each number in the first eight, so they are 94117647, and we see that 1/17 = .0588235294117647...Now suppose we suddenly needed 2/17, we could double the string above, or just double the first half, and use the rule to complete the cycle.

In the age of computing when we can get out to a couple of hundred digits with almost any software, it might not seem like much of a tool....but sitting around drinking beers at the lake, and someone asks, "by the way, what is the repetend of 1/23?", you can whip out your pencil and do the first 11 digits and then coast the rest of the way adding nines-compliments.

I feel like I should have picked that one up somewhere in the seventh grade, but stumbled across it a few days ago reading an old book on Primes by the late George Loweke, who taught at Wayne State Univ. before retiring a few miles up the coast of Lake Michigan from me. Thanks George, but you should have mentioned this one to me sooner.

A Problem About Elevens, and Some Methods of "Casting Out Sevens"

Expanding an Archive blog from 2008 with some new insights: 


Sometimes I enter the math contest they host at the Wild About Math blog site, and usually am among the people who get a correct answer, but haven't been the lucky name pulled from the hat yet. A recent problem about divisibility got me thinking about testing by divisibility by sevens again.
The problem was actually about divisibility by eleven, and asked
Consider all of the 6-digit numbers that one can construct using each of the digits between 1 and 6 inclusively exactly one time each. 123456 is such a number as is 346125. 112345 is not such a number since 1 is repeated and 6 is not used.

So.... How many of these 6-digit numbers are divisible by 11?

The answer, of course, is none.
Of course, is a danger word, like obviously, or trivially, in which we dismiss the idea that there is thinking involved.  If I were presenting this question to a class, and I have, I would say it differently. 

"None of these numbers are divisible by eleven, can you figure out why without test dividing any of them? "
 If you don't see it, don't worry, I'll spill the beans on how I would prove it down the page.  Just take a moment and try it yourself. 

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The divisibility rule for eleven most commonly known is to take the digits abcdef and add every other one from the back, and then subtract the alternate ones.  So f-e+d-c+b-a  For example, 123456 would be  6-5+4-3+2-1=3.  Now they can only be divisible by eleven if the result is 0, or a multiple of 11. 

So if we think about the numbers involved in all these numbers, they all have three odd numbers, and three even, so no matter how you split them, you'll have one group is odd, the other is even, so the difference is an odd number.  You can't get zero.  But can we get eleven?  Ummm, NO, because if we put all the big ones in one clump (6+5+4), and all the small ones in another (3+2+1), the difference is less than eleven.... so NONE of them are divisible by 11. 

 My interest was piqued by a response by Jonathan, of the jd2718 blog made these observations about the 6! or 720 possible numbers formed with the six digits as described:

As a consolation, all 720 of them are multiples of 3.

Half of them (360) are even (multiples of 2)

One in 6 (120) are multiples of 5.

Eight in thirty (192) are multiples of 4.

None are multiples of 9.

Fourteen of 120 (84) are multiples of 8.

Multiples of 7? New puzzle, good place to stop.


It is Not surprising that Jonathon covered all the other one digit divisors but did NOT test seven. Seven is the one that books on mental math simply say "divide by seven"; but thinking about it, it shouldn’t be too hard
I wrote recently about a mental test for divisibility by seven but it did have one flaw. It worked fine to tell you a number was divisible by seven, but if the number was not divisible by seven it did not give the correct modulus. Casting out nines will always give you the correct modulus. The sum of the digits of 2134 is 10 which has a digit root of 1, so if you divide 2134 by 9 you get a remainder of one. My rule for seven didn't give the true remainder.

The remainder when you divide a number by seven (or any other number) is frequently called the modulus. I've mentioned this before, it is just a way to divide integers into sets. Odd numbers are all equivalent to 1 mod 2 (they all leave a remainder of one when you divide by 2) and even numbers are all equivalent to zero. If a number is equivalent to zero mod (something) that means it is divisible by that (something). One of the things that makes them effective is the simple rule that if a+b=c then for any modular index n(the number you are dividing by) c mod n = a mod n + b mod n. For example 25 mod 7 = 4.  We could have found that by breaking it into parts.  Since 25 = 20 + 5 we find 20 mod 7 = 6, and 5 mod 7 = 5... and the  6+5 = 11 which is equivalent to 4 mod 7. So now we have the basics.

Every increase of one in the units increases the modulus 1, and every increase in the tens increases the modulus three, (for example 21 mod 7 = 0; 31 mod 7 = 3,and 41 mod 7 = 6). I figured out that the rest would be an increase of two in the hundreds, six in the thousands (think minus one) four in the ten-thousands (minus three?) and five in the hundred-thousands (minus two….) so the sequence applied to our test number, 546231… and what a coincidence, if you multiply 5(-2)+4(-3)+6(-1)+2(2) +3(3) + 1(1) …YOU GET -14, which is zero mod 7, the sequence of moduli used in each place makes it a multiple of seven…(546231 = 0 [mod 7]…

 but wait, if you used the A-2B method in the blog  above, we would notice that 231 by itself is a multiple of seven, since 23 - 2(1) =21… and 546 is also. [54 - 2(6)=42]..   You could also apply that approach step by step, 54623-2*1=54621; and 5462-2(1) = 5460.  The zero we can throw away because 546 tens is not divisible by 7 if 546 is not, so we proceed to 54-2(6) = 42 and hey, we have a divisor by 7.  In each case we just combined two moduli to reduce the size of the number. 

There is even a neat graphic approach that works the same way.  I came across a nice video of that at a site by Presh Talwalker called Mind Your Decisions.  Each step uses the same modulus increase approach as the method above, but challenge your students to figure out why it works (not an easy task). 

To check other numbers, you can just write them out multiplied times the correct modulus for that place value and probably check it in your head, but you would have to check each one… so making up a sequence, 234561(this covers everything up to six digits, but if you have a super memory, figure out another period), we would think 2×5=10 (the first digit times the first modulus) drops to 3, plus 3(second digit)x4(second modulus) is 15 which drops to modulus of one, now add 4(6) to get 25 and drop to mod 4, then add 2×5 to get 14, and we are at mod 0 [so 234500 is divisible by seven] and we know that 61 is NOT divisible by seven, so we can be assured that 234561 == 61 == 5 mod 7… ok, not EASY, but certainly could be done sans calculator…. [footnote, for those of you who have just gone through a pre-calc class and somewhere along the way they taught you about vectors, you probably imagined that you would never run across another dot product in your life, but you just did. IF you think of the sequence of modular values as one vector (5,4,6,2,3,1) and the six-digits of the number as a second vector, then the dot product is an integer that has the same modulus as the original six digit number....come on... that's pretty cool use of vectors!!!]
It is easy enough to write/remember the modulus up to ten-thousands (6231) to make this pretty useful as a factoring tool if you really needed the correct modulus. If the number is 4723 we just think 4(6)=24--==; 3 + 7x2---==; still 3, + 2(3)=9 --==;2 and then + 3(1)= 5 so 6231 divided by seven leaves a remainder of five.