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Lesson Five: Recovering a missing number from a sequence - Arithmancy: HP math geeks unite [entries|archive|friends|userinfo]
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Lesson Five: Recovering a missing number from a sequence [Aug. 1st, 2006|02:46 pm]
Arithmancy: HP math geeks unite



Number sequences are quite important in Arithmancy as well. A single digit of a particular number sequence is often critical for an Arithmantic spell to work. However, you may not have access to the particular sequence you need, such as a witch’s identification number, or a wizard’s mobile number, as people are understandably suspicious of those who come around asking for credit card numbers and such. However, there are useful spells for determining a chosen number in a sequence without having to be given the sequence at all. This lesson demonstrates one such powerful spell.

The Spell

  1. Have the individual write down the number sequence in question (for example, a credit card number).

  2. Then ask them to scramble the digits and write down the new number under the original one.

  3. Ask them to subtract the smaller number from the larger. (They may need to use a Muggle calculator for this step.)

  4. Now ask the subject to cross out any nonzero digit in the result and then read the remaining digits to you in any order. (They should keep track of this by marking off each digit as they say it..)

You can now determine the digit your subject chose.

Using the technique of casting out nines, add up the digits as they read.

8, 7, 6, 5, 4
8+7 = 15 → 1+5 = 6
6+6 = 12 &rarr 1+2 = 3
3+5 = 8
8+4=12 &rarr 1+2 = 3

The digit they crossed off was 9 – [the number you got when summing the digits].

Why does this work? The magic comes from the structure of our base ten number system. When you cast out nines, what you are actually doing is summing a sequence of digits and keeping the remainder after division by 9. So it doesn't matter what order you add the numbers in -- 1+2+3 is the same as 3+1+2. In both cases, the remainder upon division by 9 in the same.

Furthermore, if you drop one number from the sequence, it changes the remainder by that amount. In our example above, the sum of the digits (without casting out nines) is 30. No matter how the subject rearranges the digits (see step 2), both sequences of digits will have the same remainder when divided by 9. When the two numbers are subtracted, the result will have the property that the sum of it's digits will be divisible by 9. That is, the remainder upon division by 9 will be 0.

Consider our example above: 87654. The sum of the digits is 30, which has a remainder of 3 when divided by 9. If we rearrange the digits to get 67845, the sum of the digits remains 30. 87654 - 67845 = 19809, and the sum of those digits is 27 -- which is divisible by 9.

So essentially, you have forced your subject to construct a sequence of digits number that you know add up to a multiple of 9. That means that whatever digit they cross off and leave out, you will be able to find it by adding up to the next multiple of nine. For example, if the person crossed off one of the 9s in 19809, the total you would get from the remaining digits is 18. Since the total must add up to a multiple of 9, and since the next one is 27, the number they crossed off had to have been 27-18 = 9. The quick way to find out the missing number is to cast out nines as you add and then subtract the final total you got from 9. You're just skipping a few steps along the way when you do that.

[User Picture]From: emmagrant01
2006-08-04 10:50 pm (UTC)

Re: Lesson 5

Another explanation of it requires modular arithmetic. If your students have a background in that at all, it might be a good way to go. (But you could still talk about it in terms of remainders, actually.)

The basic idea is to notice that if you add up the digits of a number (like 12345 will give 15), and then if you scramble the digits (to say 43521) and add those up, you get the same sum (4+3+5+2+1 = 15, by commutativity). When you take these two numbers and subtract them, you get a new number, 31166, the sum of whose digits is divisible by 9.

It turns out that the result will always have the property that the sum of its digits (and hence the number itself, actually) is divisible by nine. (There are lots of good and fairly understandable proofs of that fact online.) One way to think about why this is true is through modular airhtmetic. Modular arithmetic is basically doing arithmetic with remainders. In this case, we're looking at remainders upon division by (or "modulo") nine, because it turns out that the sum of the digits of a number has the same remainder modulo 9 as does the number itself. Nine is special that way.

"Casting out nines" was an old pre-calculator trick for checking your arithmetic in a problem, and works the same way. When you subtract a number with a remainder (mod 9) of 6 from one with a remainder (mod 9) of 8, your answer should have a remainder (mod 9) of 2. (If it doesn't, you have an arithmetic mistake somewhere.) To see how this works, think about the fact that mulitples of 9 are nine apart on the number line. Similarly, numbers of the form 9k+2 (a mulitple of 9 plus 2, like 20 or 65) are also separated by nines. In other words, if you take two numbers of the form 9k+2 and subtract them, their difference will always be a multiple of 9.

So what this means is that whatever answer they get (after taking a number, scrambling the digits, and then subtracting) will have the property that the sum of its digits will add up to 9. So when they call out all but one of the digits, you just have to add them up and then subtract from the next highest mulitple of nine to see what was missing.

I hope that made more sense! Thanks for asking a question, and for checking out the class notes! :-)
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