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Lewis Carroll's method for dividing by 9 (1897)

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Brock
1381009.  Thu May 06, 2021 3:02 pm Reply with quote

As discussed in post 1380996. (The Nature article cited explains it extremely poorly, and appears to confuse "under" and "over".) As pointed out by suze:

Quote:
While Dodgson thought he had discovered something new, he was soon advised by a correspondent in Denmark that he hadn't, and that the method had been described in a school math text book published in Copenhagen half a century earlier.

All the same, the method was new to him and had not been published in English before.


So I'm going to call it "Lewis Carroll's method" for sentimental reasons :-)

To calculate 753096 / 9.

Step 1. Calculate the remainder by repeatedly summing the digits of 753096:

7 + 5 + 3 + 0 + 9 + 6 = 30
3 + 0 = 3

Step 2. Write the remainder above the units digit of 753096:

Code:

     ¦3
-------
75309 6


Step 3. Subtract downwards in the units place. 3-6 is negative, so "borrow" 1 from the tens place and do 13-6=7. Put the result above the tens digit:

Code:

    7¦3
-------
75309 6


Step 4. Subtract downwards in the tens place. As you "borrowed" 1 in the last step, it's 6-9 rather than 7-9, but that's negative, so you "borrow" 1 from the hundreds and do 16-9=7. Put the result above the hundreds digit:

Code:

   77¦3
-------
75309 6



Step 5. Subtract downwards in the hundreds place. You "borrowed" 1 in the last step, so it's 6-0=6. Put the result above the thousands digit:

Code:

  677¦3
-------
75309 6


Step 6. Subtract downwards in the thousands place: 6-3=3. Put the result above the ten-thousands digit:

Code:

 3677¦3
-------
75309 6


Step 7. Subtract downwards in the ten-thousands place: 3-5 is negative, so it's 13-5=8. Put the result above the hundred-thousands digit:

Code:

83677¦3
-------
75309 6


So the answer is 83677, remainder 3.

Whether this is any easier than the usual method I shall leave others to decide!

 
Jenny
1381015.  Thu May 06, 2021 4:12 pm Reply with quote

I can't honestly see how this is easier than the usual method of division.

 
suze
1381017.  Thu May 06, 2021 4:48 pm Reply with quote

Thanks Brock, that is a much better explanation than the one that Dodgson gave in Nature. It's really not easier than the conventional method, though.

By the wonders of YouTube, I have discovered another method for dividing by 9. It is probably equivalent to Lewis Carroll's Method and possibly easier to follow, but I still can't see any real reason to use it.


The Dane who published Lewis Carroll's Method half a century before Dodgson was one Adolph Steen. Dodgson clearly wasn't familiar with Dr Steen, and neither was I until today, but he's important enough to have a page on Danish-language Wikipedia.

As well as being a mathematician, Dr Steen was also involved in politics. Denmark was an absolute monarchy until 1849, and when it elected its first meaningful Parliament he was a member of it.

The book referred to by the Danish correspondent seems likely to have been Elementær Arithmetik (1843, Kjøbenhavn: C A Reitzel). I shall leave the translation of the book title from the Danish as an exercise! In something which I learned today, the silent <j> was dropped from the name of Denmark's capital in 1851.

 
Brock
1381027.  Fri May 07, 2021 2:59 am Reply with quote

Jenny wrote:
I can't honestly see how this is easier than the usual method of division.


I think it has two clear advantages:

(1) There's no multiplication involved; you don't need to know the nine times table.
(2) It has a built-in check, because the final subtraction should always yield the answer zero. I didn't include it above, but the left-hand 8 in the quotient (83677) becomes a 7 (because you "borrowed" 1 in the previous step), and then 7-7=0. I think that's extremely useful.

It also arguably involves less writing than conventional long division.

Its main disadvantage is that it's not immediately generalizable to other common divisors (apart from 11, for which Dodgson gave a similar technique). In his later paper on "Abridged Long Division", he generalizes it to divisors like 6997, which are a few away from a multiple of a power of ten; but in practice such divisors are so rare that it hardly seems worth the effort of learning a whole new technique for them.

 
Brock
1381029.  Fri May 07, 2021 3:31 am Reply with quote

suze wrote:

By the wonders of YouTube, I have discovered another method for dividing by 9. It is probably equivalent to Lewis Carroll's Method and possibly easier to follow, but I still can't see any real reason to use it.


Ooh, I like that! Hadn't seen it before, and I agree that it's easier to follow than Carroll's method. Whether it's "equivalent" in any formal sense, I wouldn't like to say. (Of course all methods that give the right answer are equivalent, in one sense.)

Regarding the expression of the remainder as a decimal fraction, isn't it simpler to remember that n/9 = 0.nnnnnnn... for all single-digit values of n?

 
tetsabb
1381033.  Fri May 07, 2021 3:59 am Reply with quote

I seem to remember shortcuts that helps one decide whether numbers are divisible by 7 and 13. But I can not dredge either from my memory.

 
dr.bob
1381034.  Fri May 07, 2021 4:02 am Reply with quote

Brock wrote:
(1) There's no multiplication involved; you don't need to know the nine times table.


This is certainly true. However, as noted above, this method only works for dividing by 9. If you want to divide by any other number, you'll need to rely on long division and for that you'll need to learn your times tables. Given that you'll have learned the times tables for 1-8, you're pretty likely to have learned the times table for 9 as well, so I think it'd be a very niche market indeed of people who need to divide something by 9 but haven't learned the 9 times table :)

Not to mention the fact that the 9 times table is possibly only second to the 1 times table in its easiness to learn.

Brock wrote:
(2) It has a built-in check, because the final subtraction should always yield the answer zero.


Having a built-in checksum is very elegant indeed and would definitely avoid mistakes. I do like that though, whether it's enough to encourage me to use a separate system only when dividing by 9, I'm not entirely sure.

 
suze
1381048.  Fri May 07, 2021 11:56 am Reply with quote

Brock wrote:
Whether it's "equivalent" in any formal sense, I wouldn't like to say. (Of course all methods that give the right answer are equivalent, in one sense.)


The good husband reckons that the two methods are conceptually the same, but presented "the other way up". The Lewis Carroll Method hinges on 10-1 being 9, while the one in that video hinges on 9+1 being 10. But as you say, any method which arrives at the correct answer is ultimately equivalent to any other.


tetsabb wrote:
I seem to remember shortcuts that helps one decide whether numbers are divisible by 7 and 13. But I can not dredge either from my memory.


The same good fellow tells me that there are in fact two ways to test for divisibility by 7.

I hope I don't garble this, but the simpler one is as follows. Remove the last digit from the number, double it, and subtract this from the number with the last digit removed. Repeat the process until you are left with a single digit. If the single digit is 0 or ±7 then the original number is divisible by 7. If not, it isn't.

As an example, start with 999,999.

The last digit is 9, double this is 18, and 99,999-18 = 99,981.
The last digit is 1, double this is 2, and 9,998-2 = 9,996.
The last digit is 6, double this is 12, and 999-12 = 987.
The last digit is 7, double this is 14, and 98-14 = 84.
The last digit is 4, double this is 8, and 8-8 = 0.
Therefore, 999,999 is divisible by 7.

While he was doing that, I found this page, which presents similar rules for every prime number up to 50.

 
Brock
1381049.  Fri May 07, 2021 12:13 pm Reply with quote

suze wrote:

I hope I don't garble this, but the simpler one is as follows. Remove the last digit from the number, double it, and subtract this from the number with the last digit removed. Repeat the process until you are left with a single digit. If the single digit is 0 or ±7 then the original number is divisible by 7. If not, it isn't.


Hadn't come across that one before - it follows from 21 being a multiple of 7. Very neat!

Quote:
As an example, start with 999,999.


Of course, if you happen to know that 7 x 11 x 13 = 1001, you will be aware that any number of the form abc,abc (e.g. 265,265) is divisible not only by 7 but by 11 and 13 as well. There's an old party trick that relies on this fact.

 
Jenny
1381107.  Sat May 08, 2021 10:31 am Reply with quote

suze wrote:
As an example, start with 999,999.

The last digit is 9, double this is 18, and 99,999-18 = 99,981.
The last digit is 1, double this is 2, and 9,998-2 = 9,996.
The last digit is 6, double this is 12, and 999-12 = 987.
The last digit is 7, double this is 14, and 98-14 = 84.
The last digit is 4, double this is 8, and 8-8 = 0.
Therefore, 999,999 is divisible by 7.


As an exercise, though, I mentally divided 999,999 by 7 and it didn't take me any longer than it would have done to go through that exercise, assuming one knows one's 7 times table. Something like that would be helpful for numbers higher than 10 though, if you didn't have a calculator handy.

 
Numerophile
1381123.  Sat May 08, 2021 3:25 pm Reply with quote

Going back to division by 9, the MooMooMath technique in the YouTube video suze linked to is essentially using the fact that 10 = 9 + 1: it is very easy to divide by 10 because that is the base of the number system, and each group of 10 is a group of 9 plus one left over - we just need to count up these 'left-over' elements and see how many more groups of 9 they give. It has one advantage over the Lewis Carroll method, in that it only uses addition rather than subtraction - you are less likely to make errors when carrying digits than when borrowing them. But unless you totally failed to learn the nine times table, I can't see that it is really any quicker than simply doing the division.

However, it is easy to generalise this technique to divide by 10-1 for some n greater than 1, and here it does have a distinct advantage over long division. I will illustrate this (non-mutant) algorithm by dividing 85,374,611 by 99 and by 999:

1. Write out the number you want to divide in groups of digits of the same number as in the divisor (counting from the right); leave a larger gap before the last group:
Code:
(85,374,611 / 99)                  (85,374,611 / 999)

85 37 46    11                      85 374    611


2. Now do the same again, moving each group one place to the right and dropping the last group; repeat this until you run out of groups:
Code:
(85,374,611 / 99)                  (85,374,611 / 999)

85 37 46    11                      85 374    611
   85 37    46                          85    374
      85    37                                 85
            85


3. Now add up the columns on each side of the larger gap (you can ignore the rest of the grouping):
Code:
(85,374,611 / 99)                  (85,374,611 / 999)

  853746    11                       85374    611
    8537    46                          85    374
      85    37                                 85
            85
--------   ---                      ------   ----
  862368   179                       85459   1070


4. At this point, if the total on the right is less than your divisor, you are finished: the left hand total is the quotient, and the right hand total is the remainder. If the right hand total is greater than or equal to the divisor, you will need to divide that by the divisor, and add the result to the left hand column. The chances are that this will be trivial, unless you were dividing a very large number!
Code:
(85,374,611 / 99)                  (85,374,611 / 999)

  853746    11                       85374    611
    8537    46                          85    374
      85    37                                 85
            85
--------   ---                      ------   ----
  862368   179                       85459   1070
       1    80                           1     71
--------                            ------
  862369 remainder 80                85460 remainder 71



I will leave it as an exercise for the reader to carry out the same calculations by long division, then compare and contrast...

 
Numerophile
1381142.  Sun May 09, 2021 3:52 am Reply with quote

And here is a slight variation in presentation, which facilitates division by 10-y, where y is any number (preferably small, or a power of 10!) less than 10.

You will use four columns A, B, C, D.
1. Write the number to be divided at the head of column A.
2. Divide the number in column A by 10 and write the quotient in B and the remainder in C.
3. Multiply the number in B by y and write the answer in D.
4. Add the numbers in C and D and write the answer in A on the next line.
5. Repeat steps 2-4 until the figure in B is 0.
6. Add up the figures in column B; this will give the quotient, and the final figure in column C will give the remainder. (Actually there is a possibility that this figure will lie between 10-y and 10, in which case you will need to add 1 to the quotient and subtract 10-y from the remainder.)

Code:
Example: 85,374,611 divided by 997 (i.e. 1000 - 3)

       A       B       C       D
85374611   85374     611  256122
  256733     256     733     768
    1501       1     501       3
     504       0     504
           -----
           85631 rem 504

Note that if y is 1, as in my previous post, you can dispense with column D; in this case just add the figures in B and C to get the next number for A.

 

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