Division Diversity

The process of division is a basic component of calculation, an essential part of arithmetic and algebra, and is used in practical ways for comparing quantities and sharing values...

With division so important, it should not be surprising that divisibility is such an issue. The number twelve is divisible by two, three, four and six, but thirteen has no such divisors. Calculations with composite numbers (those with divisors) are both easier and more plentiful than with primes.

We get so used to dealing with divisibility that we can instantly identify whether a particular number is divisible by two. Those numbers for which two is a divisor are, of course, called even while those prime to two are dubbed odd. We can instantly identify whether a number is divisible by five: its last decimal digit is either zero or five itself.

When we use decimal numbering there are some useful tips for testing the divisibility of a number: starting at the obvious point, no number is divisible by zero, and every number (whether prime or not) is divisible by one.

To test for other factors (the numbers by which a number can be divided without remainder), we have observed tricks: these are usually pointed out early in the study of arithmetic, so they will probably be obvious to you by now:

2 is a factor of all decimal numbers ending in 0, 2, 4, 6 or 8 (even).

If you add all the digits of a number, then repeat the process with the result until you have a single number, you have established its "digital root". For example if you start with 1967, 1+9+6+7 = 23, and 2+3 = 5, so five is the digital root of 1967. The digital root of a number is a useful aid to determining divisibility.

3 is a factor of all decimal numbers whose digital root is 3, 6 or 9.

4 is a factor of all decimal numbers whose tens digit is even and last digit is 0, 4 or 8, or whose tens digit is odd and last digit is 2 or 6.

5 is a factor of all decimal numbers whose last digit is 0 or 5.

6 is a factor of all decimal numbers which are even and possess a digital root of 3, 6 or 9.

9 is a factor of all decimal numbers whose digital root is 9.

Things get a bit more complicated at this point, and the job of identifying factors of a number needs a bit more than mechanically checking its individual digits. Let's look at 8 as a factor. If a number is divisible by 8 you can be sure it is even - so it goes without saying that the last digit will be even. The rest of the number (as if it ended at the tens column rather than the units column) must be a multiple of four. Say we are asking "Is 8 a factor of 7128 ?" Since we know 7128 is even, this is the equivalent of asking "Is 4 a factor of 712 ?" We can confirm that it is, just by looking at the last two digits. Put these two steps together and we see that we need only look at the last three digits of a number to identify whether 8 is a factor.

We can generalise on this principle: to find whether a number is divisible by 2^n, you need only confirm that the last n digits are divisible by 2^n.

For even numbers in general we need only combine these tips: note that to find whether 6 is a factor, you must identify both 2 and 3 as factors, because 2x3 = 6.

Turning to twelve, we can be sure it is a factor of any number which is divisible by both 3 and 4.

So far everything is very simple: we have useful tips requiring almost no calculation and can identify factors (divisors) of even quite large numbers all the way up to twelve, barring 7 and 11. If we had shortcuts for those two we can extend the list to 22, missing out only 13 and 17.

Check out multiples of seven and you will not be able to find a pattern easily. There is a pattern, but it is well hidden. Let's attack three factors all at once: seven, eleven and thirteen. Note that if we find a tip for seven, it will be useful for 14 and 21 as well, just by marrying the tip to those for 2 and 3. Likewise 11 yields a similar tip for the factor 22 (11x2). Multiply 7x11x13 to give 1001.

Now this shows some promise... for example, factors of 5005 must include not only 5 but also 7, 11 and 13. Can we extend this? You bet! 238238 must have 7, 11, 13 and 238 among its factors. 238 can itself be broken down further: one factor is 2 because it is even. We are now left to find any other possible factors in 119 (which is 238/2). Since its digital root is two (1+1+9=11, 1+1=2), we know that 3 or any of its multiples are definitely not factors. Putting a little work into it, we find that 119 is itself a prime, so we shall delve no further.

But there's even more use we can make of this fact (7x11x13=1001). Any number with digits in the pattern XYZXYZ has factors 7, 11, and 13 (also XYZ, of course). Any six digit number in the pattern ABCABC has 7, 11 and 13 as factors. What can we say about ABCDEF ? Well, (ABCDEF - ABCABC) will give us a result somewhere between -999 and +999 that shares the same factoring relationship with ABCDEF in regard to 7, 11 and 13. Now that last sentence may pack a wallop and seem difficult. So we will look at it step by step...

Suppose this is divisible by seven. We already know ABCABC is divisible by seven (it is ABCx7x11x13), so the number (ABCDEF - ABCABC) must also be divisible by 7. On the other hand, If ABCDEF is not divisible by 7, then neither will (ABCDEF - ABCABC) be divisible by 7. The same goes for 11 and 13. But (ABCDEF - ABCABC) is simply (DEF - ABC). Take 863429 as an example: the smaller number is -434 which is (429 - 863). You can safely forget about the minus sign: if p is divisible by q, then so is -p. Mini-calculations will show that 434 is divisible by 7, but not by 11 or 13. The same therefore applies to 863429: 7 is a factor, but neither 11 nor 13 are.

So to test for factors 7, 11 and 13 you can just chop down a six digit number to three digits first. It's actually better than that: you can cut down any length number the same way: alternately add and subtract groups of three: to find whether 7, 11 or 13 are factors of 3,487,262,386,876 simply calculate (3 - 487 + 262 - 386 + 876 = 268) and then test 268 instead. (It doesn't matter whether you end up with a negative or a positive number: if 268 is divisible by 7, so is -268.) Coincidentally we already group large numbers into packets of three digits by placing commas.

Divisors and Decimals

Take a few moments and calculate 1/7 as a decimal fraction: you will find it comes to 0.14285714.. with the digits from 1 to 7 repeated in the same sequence. Now try 2/7; maybe it surprises you that this is 0.28571428.. - again, repeating the six-digit string 142857 in the same sequence over and over. The only difference is the starting point. What do you think happens to 3/7 ? Its decimal incarnation is 0.42857142.. - again the same six digits in the same order, starting at a different point. So it goes with 4/7 (= 0.57142857..); 5/7 (= 0.71428571..) and 6/7 (= 0.85714285..). Why should this be? Why are the repeated parts of these decimals six digits long? Why are they the same six digits? Why does the sequence then repeat?

The answers are all there to behold; maybe you will marvel at them as I do. Converting a fraction to decimal form involves carrying out the division that is "delayed" in fractional form: Converting 1/7 we divide seven into ten and then note the remainder (in this case, 3). When we continue expanding the decimal, we end up dividing 30 by 7 to get 4 and remainder 2. Repeating over and over, there are only seven different remainders we can get when we divide by seven: 1 to 6 and zero. If we get a positive remainder, it will define the next step in the division process, and there are only six possible ways; if the remainder becomes zero, the expansion stops and the decimal fraction has a specific length. But as the remainders turn up, eventually it must end with a zero, or inevitably repeat one that has been used before. If you are dividing by n the remainders must either reach zero or begin repeating after (n-1) digits at most.

Evaluating 1/8, for example, you get remainders 1, 2, 5, then zero. With 1/7 you get remainders 1, 4, 2, 8, 5, 7, then 1. When you get the remainder 1 the second time, the division is exactly as it was the first time, so your are doomed to repeating with remainder 4, then 2 and so on. Try another fraction and study what happens: 1/11 produces remainders 10, 1, 10 and so on, so you should not be surprised that its decimal expression has a two-digit repeated string: 0.0909.. The fraction 1/13 has remainders 10, 9, 12, 3, 4, 1 and then 10 again: the decimal expression likewise has six digits repeated over and over: 0.07692307.. Study a few more fraction-to-decimal conversions and see if you can spot any relation between the denominator of the fraction and the number of digits in the repeated part of the decimal string. Note that the repeated part of the string can never be longer that the value of the denominator (divisor) itself.

There's something more that I find outstanding about 1/7: it is easy to memorise the six digit expansion because it goes 14, then 28, then 57. 14 is itself twice 7, 28 is twice fourteen, and 57 is nearly twice twenty-eight. Is this just coincidence? When I first noted this pattern, I surmised that maybe the extra one added to make 57 comes because of overlap: what if 1/7 is really 14/100 + 28/10000 + 56/1000000 + 112/100000000 +.. ? the 1 from 112 would "drip over" into the units column of the 56, making it 57.

Of course, that means it would expand to 0.14285712, but maybe this last digit would be 4 (as we expect) from 224 "dripping over" and so on. Sure enough, this works, and continues as far as you would like to test it. The first pair of decimal points comes from 14/100, and the second pair comes from (14x2)/(100^2). Now this calls for a slight surgery so that the first term is (7x2^1)/(100^1), and term "n" is (7x2^n)/(100^n). If you are familiar with the rules of algebra, you will see that this can be simplified to 7x50^(-n). Is that neat, or is that neat ? I wonder, does the same thing happen with any other numbers? We are looking at seven, which is prime, so it may be interesting to look at another prime when its reciprocal (1/n) is expressed as a decimal: say 13. Is there an interesting pattern - infinite but similar terms adding together? Equipped with an inquisitive mind, the knowledge of a few rules of arithmetic, and a supply of paper, you have everything you need to assist you on such a quest.

Depending on your age and where you grew up, you may not be used to calling numbers like .893 "decimal fractions", but this is a usage approaching universal acceptance today. To call .893 just a decimal is confusing, because it sounds as if it is talking about the number base (ten, rather than two for binary, or sixteen for hexadecimal). Anything that can be expressed as a decimal fraction can also be expressed as a common fraction; in this case the fraction is 893/1000. This is easy - the denominator consists of the number one followed by as many zeros as the length of the decimal fraction.

There are two complications, though. The first is that the resulting fraction may need to be "reduced". You reduce it by dividing both the top (numerator) and the bottom (denominator) by all the divisors they share. The denominator will always have some number of factors two and five so those are the only ones that concern you. For example .8125 is (8125/10000), but 8125 has at least one factor of five. In fact 8125 = 5x5x5x5x13, while 10000 = 5x5x5x5x2x2x2x2. This second step, removing the common factors, is necessary to get the simple fraction answer: .8125 = (13/16).

I mentioned there were two complications: the other is that the decimal fraction may end with a repeating string of digits. When this is the case, the denominator has a different form altogether: it will be as long as the decimal fraction up to the point it repeats. For example the decimal fraction .69222.. will have a three-number denominator. Its form will be some number of nines, the length of the repeating part, followed by zeros to make up the length. Only the two repeats here, so the denominator will be 900.

The numerator is just a bit more complicated: you multiply the non-repeating part with the number made up of nines to the length of the repeating part, and then you add the repeating part. The 69 is non-repeating, so we multiply that by 9 (since only one digit repeats) and then add on that repeating part. 69 x 9 + 2 = 623. So we have constructed (623/900).

Let's take another case: .1234567 where the "4567" repeats. The denominator will be seven long, starting with four nines: 9999000. The non-repeating part of the decimal fraction is 12 so the numerator is 123 x 9999 + 4567 = 1234444. So the fraction is (1234444/9999000), which can be reduced to (308611/2499750) by dividing top and bottom by four. You can be sure that this is lowest terms because 9999000 is made up of 2x2x2x5x5x5x3x3x1111. You will be able to confirm quickly that 1111 is not a factor of 308611, and 308611 is not a multiple of 3 (since 3+0+8+6+1+1 = 19, itself not a multiple of 3).

You can test this out on numbers mentioned earlier: .142857142857.. (repeating all six digits) must be (142857/999999), and sure enough, reducing this to lowest terms, (1/7).

I will leave you with an interesting little party piece you can try on friends who are able to do a little arithmetic accurately: ask them to choose a three-digit number (1 to 999) secretly. Then ask them to divide it by 7 and tell you the remainder. Then ask them to divide the original number by 11 and tell you the remainder. Then repeat for a final time with 13. You now have three small numbers, and from these you quickly identify the original three-digit number.

Here's how it's done:

1. Call the remainder from division by 7 "x", the remainder from division by 11 "y" and the remainder from division by 13 "z".
2. Calculate 715 times x, 364 times y and 924 times z.
3. Sum these three results together.
4. If the result is less than 1000, you have finished. If the result has four digits, take away the first from the last three; if the result has five digits, take away the first two from the last three.

Say the person chose 762: divided by 7, the remainder is 6. Divided by 11, the remainder is 3. Divided by 13 the remainder is 8.

With the numbers 6, 3 and 8 you calculate:

        715 x 6 (= 4290)
        364 x 3 (= 1092)
        924 x 8 (= 7392)

These sum to 12774, and 774 - 12 = 762

The trick works because 715 has 11 and 13 as factors, 364 has 7 and 13 as factors, and 924 has 7 and 11 as factors. We also have to choose multipliers which add to (1001 x n + 1) to make our equations balance on both sides. Try it out and see whether you can trace the workings.

The subject of division has taken some time, yet we have hardly scratched the surface. Consider this: a town nearby Divisibility is Prime Numbers. As a matter of fact, Prime Numbers is where Divisibility cannot be seen.

May 2006