Where a class is only recently new to being allowed to use a calculator, in my experience year 8, there is a good deal of fun to be had in exploring things you can do with these things that you couldn’t do before.

I recommend exploring the R-P and P-R functions when exploring Pythagoras - R-P gives the hypotenuse and one included angle when supplied with the two shorter sides. For low-ability classes, this gives a short-cut to trig that works well enough for many. It certainly provides techniques for checking or for searching for answers.

However, this article is about producing numbers longer than the display on the calculator. Some have eight digits, many have 11, a few have 12.

Consider the reciprocal , 1/n where n is an integer.

1/2 = 0.5

1/3 is 0.33333333 etc

1/4 = 0.25

1/5 = 0.2

1/6 = 0.1666666666 etc

1/7 = 0.142857142857142857

1/8 = 0.125

1/9 = 0.111111111 etc

What we can see? Some of these stop, the others repeat and 1/7 is more interesting, repeating in a cycle. What is special about the ones that stop? Could it be that the reciprocal denominator (the number on the bottom) has factors that are also factors of ten? Is it true, then, that if the n in 1/n has factors of only two and five, its decimal will come to a finish (have finite length)? Does that mean that all the other such n have infinitely long decimals?

Further, if a seventh (1/7) repeats in a cycle of six, then does a third repeat in a cycle of one or two (yes, one of those)? What about nine? let’s then look at n being only prime, since they look more interesting...

[side issue to explore more thoroughly what I said about n’s factors being two and five]

I write the decimal form of 1/7 without needing a calculator. I can write 2/7 = 0.285714

1/7 = 0.142857

2/7 = 0.285714

3/7 = 0.428571

4/7 = 0.571428

5/7 = 0.714285

6/7 = 0.857142

(repeat) without effort, partly because i can double the other number as fast as I write it and mostly because it is the same set of digits in the same order, just starting in a different place. Of course it is. Look at the hand division (side issue for the class that can’t do this). The point is that there are only six possible remainders, visible if you divide 2 by seven by hand. All six sevenths are the same 142857 digits, starting in different places, as in the insert. I’d get the class to find this and to explain what they can see, so it is their discovery, not mine. They might come up with ideas about why this is so, seeing the cycle length of six is one less than seven, possibly seeing why – because there are only six possible remainders within the division process.

So explore larger n (maybe in a later lesson):

1/11 = 0.090909090 etc is that a cycle of two? [Might it be ten?, n-1?]

1/13 = 0.076923 076923 etc a cycle of six. Why six?

1/17 = 0.05882352941..... but this doesn’t repeat. I say it must repeat.

Why? Because , as I wrote earlier, I can do long division, and there are only 16 possible remainders (zero and 17 can’t happen, the reciprocal would have an ending). Therefore the longest possible repeat is 16 digits long.

1/13 = 0.076923 2/13 = 0.153846

3/13 = 0.230769 5/13 = 0.384615

4/13 = 0.307692 6/13 = 0.461538

9/13 = 0.692307 7/13 = 0.538461

10/13 = 0.769230 8/13 = 0.615384

12/13 = 0.923076 11/13 = 0.846153

This now gives me an argument that says 1/13 is weird - it only has six of the twelve possible digits. That means I’ve missed a trick somewhere. Does 2/13 look like 1/13? A little research (class to do this) produces a table (on the board, teacher collecting the data from the class, being a secretary to their direction – this works really well if they don’t feel steered).

So 13 gives us two sets, that’s all twelve possible remainders.

So what happens to 1/17? Can anyone be persuaded to look at 10/17? Obviously it starts 0.5882352941 but does anyone’s calculator give any more digits? If you were to put 1/17 in memory (so as to retrieve it more easily) then what happens to 5/17? 0.2941176471 Dare i suggest that 1/17 now starts 1/17 = 0.0588 2352 9411 7647 1 ? How many digits is that? 17? {that’s why I grouped the digits in sets of four, not three] But shouldn’t it have repeated by now? What is 2/17? [0.117647059). Does this fit onto (I’d write it underneath and label it appropriately) what we have? Then I suggest that 1/17 is 0.0588 2352 9411 7647 059 and repeats forever. Pause to let the class catch up, talk about this, write some notes. Try to make this their discovery, not your explanation. Perhaps the collective discovery; if you’ve never done this before, get excited.

So could 1/19 be homework, or do ‘we’ discover that one and let 1/23 be homework? (or leave 1/19 as homework and do 1/23 ‘now’ in class). What is amazing is that the kids are producing very long numbers from a ten or eleven - digit display. If they move to a computer they’ll get 15+ digits, but still need to employ these techniques by the time they explore 1/43, 1/47, 1/101, 1/103.

There are some special cases, which can be discovered by wondering again about 1/13 - why was it a two-setter, not one like 17.

I set the kids to explore (next lesson, if they are interested) in groups, checking each other and generating material that is then shared. I found it valuable to credit results on the board (e.g. Jamie somehow ‘owns’ the result for 1/31)

1/19 = 0.052 631 578 947 368 421 05 263 18 digits long

1/23 = 0.043 478 260 869 565 217 391 304 348 22 digits long

1/29 = 0.034 482 758 620 689 655 172 413 793 103 28 digits long

1/31 = 0.032 258 064 516 129 032258 3/31 = 0.096 774 193 518 387 096 two sets of 15 digits

1/37 is a special, being only three digits long, 0.027027027 . There is a related result for 1/27 (27 is not prime), 0.037037037. Why is that?

Could it be that we might multiply 27 by 37, or 1/27 by 1/37 ? What happens?

27*37=999 and 1/999 = 0.001001001

Side track here: 1/9 = 0.1111, 1/99 =0.01010101 1/999 = 0.001001001001, 1/9999 = 0.000100010001

Does this tell us something significant? Well, actually it points to a general understanding to much of what is going on.

If we know what 1/999 is, then surely 2/999 is obvious, 14/999 is predictable and so is 27/999 and 88/999. Of all of these, 27/999 is the only one that cancels down, to 1/37, simply because 999 has so few factors;

999 = 3.3.3.37 = 27.37. Thus we can predict all decimals of the form N/999.

So what we then need to look at (and the class needs time for them to come up with this as an idea) are the factors of numbers of the for 99...99. Presumably six nines, 999 999, has factors of 7 and 13, since we have decimal repeats for those numbers? By now some of the class are very excited, some are losing it, so what has to happen is team work, and recording. Some will only ‘get’ this by being the writers, some will only ever ‘get’ the repetitive elements, finding long decimals and hunting for factors, some ‘get’ the whole thing and will be really quite excited that maths is doing something they were not expecting – being interesting.

99 = 3.3.11

999 = 3.3.3.37

9999 = 3.3.11.101 So predict 1/101 without the calculator (009900990099).

99999 = 3.3.11111 = 3.3.41.271 so 1/41 should have a five figure repeat, 0.02439 02439

999999 = 96 = 3.3.3.7.11.13.37 this has to include the factors of 99 and 999; also, 1001=7.11.13

999999999 = 99 = 34.37.333667

999999999999 = 912 = 33.7.11.13.37.101.9901

999999999999999 = 915 = 33.31.37.41.271.2906161

I’d use a spreadsheet to find factors beyond this point, especially for prime numbers of nines (7,11,13). We know some factors from the work already done – in a sense, now we have an explanation, we can use the previous results to help.

For example, fifteen nines includes all the factors of 999 and 99999 but also a factor of 31 (since 1/31 had a repeat of 15 digits), ie. 915 = 999 999 999 999 999 = 999.1001001001001 = 33.31.37.41.271.2906161 and I think 2906161 is prime. 2906161*344096559 = 999999999999999. Confirmed on the ‘net.

Similarly, we know that 922 has a factor 23 (1/23 has 22 digits) and 928 has a factor of 29 for a similar reason.

DJS 20160912

I wrote this first and then went to research in the modern way, on the internet.

You will find a table of factors for multiple nines up to 30 of them within the wikipedia note on Repunit, quoted below, where Rn means the number, made only of ones, with n digits, so R6 = 111111 and hence 96 has factors, from the table, of (3.3).(3.7.11.13.37), entirely in agreement with ‘our’ earlier result in blue above. I note that R19 and R23 are prime.

(Prime factors coloured red means "new factors", i. e. the prime factor divides Rn but not divides Rk for all k < n) (sequence A102380 in the OEIS)[2]

R1 = 1

R2 =11

R3 =3 · 37

R4 =11 · 101

R5 =41 · 271

R6 =3 · 7 · 11 · 13 · 37

R7 =239 · 4649

R8 =11 · 73 · 101 · 137

R9 =32 · 37 · 333667

R10 =11 · 41 · 271 · 9091

R11 =21649 · 513239

R12 =3 · 7 · 11 · 13 · 37 · 101 · 9901

R13 =53 · 79 · 265371653

R14 =11 · 239 · 4649 · 909091

R15 = 3 · 31 · 37 · 41 · 271 · 2906161

R16 =11 · 17 · 73 · 101 · 137 · 5882353

R17 =2071723 · 5363222357

R18 =32 · 7 · 11 · 13 · 19 · 37 · 52579 · 333667

R19 =1111111111111111111

R20 =11 · 41 · 101 · 271 · 3541 · 9091 · 27961

R21 =3 · 37 · 43 · 239 · 1933 · 4649 · 10838689

R22 =112 · 23 · 4093 · 8779 · 21649 · 513239

R23 =11111111111111111111111

R24 =3 · 7 · 11 · 13 · 37 · 73 · 101 · 137 · 9901 · 99990001

R25 =41 · 271 · 21401 · 25601 · 182521213001

R26 =11 · 53 · 79 · 859 · 265371653 · 1058313049

R27 =33 · 37 · 757 · 333667 · 440334654777631

R28 =11 · 29 · 101 · 239 · 281 · 4649 · 909091 · 121499449

R29 =3191 · 16763 · 43037 · 62003 · 77843839397

R30 =3 · 7 · 11 · 13 · 31 · 37 · 41 · 211 · 241 · 271 · 2161 · 9091 · 2906161

Some of the brighter children will observe putative connections between factors and their related 9n, where the observed factor is n plus and minus one. R10 has a factor 11, R12 has 11 and 13, R18 has 19, R22 has 23 and R28 has 29. That is, factors for Rn often include n+1. But quite often too they include 2n+1, 4n+1, 6n+1 and 8n+1.

When setting homework on topics such as this it is difficult to reward the student who has worked without internet assistance and perhaps who has made errors, over one who simply went to find answers without doing any of the work – sadly, these are the same students who remain unenlightened, who have missed out on the benefits of the discovery and for whom all the work done by others remains irrelevant. I say their future lives are going to be similarly lacking in the excitement of new discovery. I see that as very sad.

Length of repeat: [value of n from 1/n; digits in repeat length; visible rule]

13 6 (n-1)/2

17 16 n-1

19 18 n-1

23 22 n-1

27 3

29 28 n-1

31 15 (n-1)/2

37 3 (n-1)/12

41 5 (n-1)/8

43 21 (n-1)/2

47 46 n-1

51 16 51=3x17, 1/17 has an n-1 repeat cycle

53 13 (n-1)/4

59 58 n-1

61 60 n-1

71 35 (n-1)/2

73 8 (n-1)/9

79 13 (n-1)/6

83 41 (n-1)/2

89 88 n-1

93 15 93=3x31, 31 has an (n-1)/2 cycle length

101 4

103 51 (n-1)/2

107 53 (n-1)/2

109 108 n-1

1/109 = 0.0091743 119266 055045871559633 0275229357798 1651376 14678899 0825688 0733944954 12844 036697247 06422 0183486 23853211 0091...

1/9 = 0.11111 1/99 = 0.0101010 1/999 = 0.001001001 1/9999 = 0.000100010001 etc

...and then you might take this further, into primes in other bases...... but that is more difficult, since your calculator probably won't do many bases.