Friday, July 16, 2010

Wednesday Math (two days late), Vol. 121: Divisibility by 11 revisited

Blogging is down to a minimum due to the summer session. It feels like I'm always writing assignments or grading them, especially Mondays through Thursdays. Only two weeks to go.

Since this in Volume 121 of the allegedly weekly math posts and 121 is 11 x 11, I'm going to write about the multiples of 11. This was discussed this before last October when I reviewed the 19th Century math text Robinson's New Higher Arithmetic.

It's easy to spot the multiples of 11 less than 100, 11, 22, 33, 44, etc., but not quite as obvious when the number has more than two digits. Let's take for example the number of feet in a mile, 5,280.

The trick to checking is known as an alternating sum, where we start with the digit in the ones place, subtract the digits in the tens place, add the digit in the hundreds place, subtract the digits in the thousands place, and continue this pattern of add and subtract for all the digits. For 5,280, the answer would be 0-8+2-5, which equals -11.

If the sum is a multiple of 11, the original number is a multiple of 11. We can state a rule that if the number is negative, add multiples of 11 until the result is zero or more, and if the total is more than 10, subtract 11 until the result is between 0 and 10 inclusive. Now we get an added bonus that the result is the remainder when you divide by 11.

Back a century ago, this trick was a popular way to check your work when doing any simple math. Let me do a sum and get it wrong to show how casting out elevens can show me that I made a mistake.

_3717 Casting out 11s: 7-1+7-3 = 6+4 = 10
+ 244 Casting out 11s: 4-4+2 = 2
_____
_3951 Casting out 11s: 1-5+9-3 = 2

10+2 = 12, so if we subtract 11 from 12 we get 1, which is what the remainder of the sum should be. Since we got a remainder of 2 instead, there must be a mistake. (I forgot to carry the 1 when I added 7+4 and the correct sum is 3,961.)

Besides mistakes that can be done with simple arithmetic errors, casting out 11s also catches simple transposition of digits mistakes. Let's say I write 3,691 instead of 3,961. Now we cast out elevens on both.

3,691 -> 1 - 9 + 6 - 3 = -5, so we add 11 and get 6
3,961 -> 1 - 6 + 9 - 3 = 1.

This is very handy, but not perfect. Most transposition errors humans will make are switching two digits that are adjacent, but if I wrote 3169 instead of 3961, both will give the answer 1 when casting out elevens.

This is known now as modular arithmetic, a topic I covered way back in Vol. 5 of Wednesday Math.

4 comments:

Lockwood said...

My very first post after the "hello, world" piece was on this topic, and I think you were my blog's first commenter, confirming that the alternating sum trick works in all base systems.

47th Problem of Euclid said...

Do you know the Trachtenberg System of Speed Mathematics? In his system, multiplying by 11 is the easiest type of multiplication that fully uses his method, and is the one he teaches first.

Karlacita! said...

You can also do this with 9s -- is it because it's close to the base?

Matty Boy said...

Hey, Lockwood! Thanks for reminding me.

Hey, 47th! I got the book when I was in grade school, but I didn't absorb everything.

Hey, Karlacita!! Yes, being one above the base and one below the base always works in a similar way, so if we used base 16 as our standard number system, the tricks would work with 17, which is written 11(base 16), and 15 which is F(base 16).