## Wednesday, October 28, 2009

### Wednesday Math, Vol. 94: Robinson's New Higher Arithmetic, Part 4

In this final installment of my review of Robinson's New Higher Arithmetic, a math text published 114 years ago, I'd like to bring up two techniques from the book that could still have value today. Robinson called them "casting out nines" and "casting out elevens". The modern names are "mod 9" and "mod 11".

All calculations back in the day were done by hand. How could you be sure you didn't make a mistake? Robinson recommended methods he would call "proof of addition" or "proof of subtraction" or "proof of multiplication" that you could use to check your work.

The first one is casting out nines. You may have learned this at some point in your education, but that isn't guaranteed since it isn't required knowledge, and you might have forgotten it as well. To cast out nines, add all the digits together of a number. If the sum of the digits is more than ten, add up those digits, and repeat the process until you get a number less than ten.

Example: Casting out nines in 2,598,960. 2+5+9+8+9+6+0 = 39, 3+9 = 12, 1+2 = 3.

What this tells us is that if we divide 2,598,960 by 9, we get a remainder of 3. If after casting out nines we get 9 as the final answer, that means the number is divisible by 9.

The idea is that if we add a column of numbers, the remainder of the total when divided by nine should be the total of the remainders of the original numbers when divided by nine. Likewise, if I multiply two numbers together, the remainder of the product will be the product of the remainder. If this didn't work, you messed something up. Since casting out nines is fairly easy, it's more likely you messed up the addition of the column or the multiplication.

There's also casting out elevens, which is only slightly trickier. Instead of adding all the digits, we start at the ones place, add that digit, then subtract the tens place digit, then add the hundreds place, then subtract the thousands place, alternating back and forth until all the digits are used. Let's do this with 2,598,960. I'll alternate the colors of the digits, where black are the ones to be added and red are the ones to be subtracted.

2 - 5 + 9 - 8 + 9 - 6 + 0 = 1.

Because we are adding and subtracting, we might get a negative number, though we did not here. When that happens, add 11 until we get a positive number.

You might think there's not much point to this in a world full of spreadsheets and calculators, since the machines aren't going to make addition or multiplication mistakes, but people still make transcription errors, and casting out nines or casting out elevens can catch these. Here are two common types of errors we humans make.

Duplicating the wrong digit:
We were supposed to write 377 and we wrote 337 instead. Casting out nines, the simpler of the two methods, will catch this if we do the process to both the original list and the list we typed into calculator or computer.

Transposing digits: The true number was 275 and we wrote 257 instead. Casting out nines won't help here.

2+5+7 = 2+7+5 = 14 in either case.

This is why Robinson's included the slightly more difficult back-up plan of casting out elevens.

2
- 5 + 7 = 4, while 2 - 7 + 5 = 0.

In a spreadsheet like Excel, there is the mod function so =mod(2598960,9) will give you 3 and =mod(2598960,11) is 1, but the whole idea is that you might have mistyped something from another source, so it makes sense to do the casting out of nines and elevens on the original source by hand.

The curriculum is changing constantly, but the amount of time students are in class isn't increasing that much. This means that when new material is added, usually something has to go by the wayside. It might be time to double back and put casting out nines and casting out elevens back in the mix.

Emphyrio said...

It's not obvious to me that adding up the digits successively is equivalent to dividing by 9 and taking the remainder. I'll believe it if you say so, but I'd be interested to see a proof, as long as it doesn't involve group theory. :-)

Matty Boy said...

It's not really group theory, but every power of ten is a mulitple of 9 + 1.

1 = 0x9 + 1
10 = 1x9 + 1
100 = 11x9 + 1
1000 = 111x9 + 1

That means every digit adds a bunch of multiples of 9 and some remainder, which is the digit itself. Add all the digits and you get the remainder for the entire number. If the sum is more than 9, you can keep adding the digits until it's less than 9.

Emphyrio said...

Ah, now it makes sense -- thanks!

Matty Boy said...

Now I can tell the truth. It really is group theory, only cleverly disguised. ;^}

Dr. Zaius said...

I get it! No I don't.

art bartok said...

i was pleased to see a transcription error in the paragraph preceding the one about transcription errors.

Matty Boy said...

Thanks, Art. It's fixed.