Lesson 3: Another way to pass a secret number to a group 
[Aug. 1st, 200607:14 pm]
Arithmancy: HP math geeks unite

This lesson demonstrates a spell that is similar to the previous one, but takes a slightly different path. It uses the Arithmantic technique of digit reversal, which is both simple and powerful.
The Spell
 Ask the subject to choose three numbers from 1 through 9, with all three digits being different. They should write these digits down to form a threedigit number, in any order they choose.
 Have the subject reverse the digits to create a second threedigit number.
 Tell them to subtract the smaller of these two numbers from the largest.
 Now, have them take the resulting number and reverse it, and add the result and its reverse together.
The result will be 1089.
How does this work? Let's look an an example:
Suppose our number is 836. Reversing that gives 638. We subtract: 836  638 = 198. Notice that when the number is reversed, the middle digit remains the same. Since all digits are different, we are also guaranteed by the reversal that the ones digit of the larger number will be smaller than the ones digit of the smaller number. This means that we will have to "borrow" from the second column when we subtract, and that the middle digit of the result will always be 9. Now, we reverse the answer to get 891 and add: 198 + 891 = 1089. At this point, we have our result  but why?
Bear with me while I introduce some algebra.
Back in primary school, you learned that a number like 836 is composed of 8 hundreds, 3 tens, and 6 units. So we can write that as 8×100 + 3×10 + 6. Similarly, any threedigit number could be written as a×100 + b×10 + c. The reverse of that number would be c×100 + b×10 + a. Just assume that the original number is larger, (and understand that we can make the same argument if it isn't.) Now, subtract:
(a×100 + b×10 + c)  (c×100 + b×10 + a)
Rearrange the terms to group 100s and 10s together:
(a×100  c×100) + (b×10  b×10) + (c  a)
This can be simplified to get:
(ac)×100 + (bb)×10 + (ca)
The result of the middle column will always be 9 (and not 0) when we actually do the subtraction, since c is smaller than a and we will have to "borrow" to subtract. We will also have to borrow from the hundreds place to subtract the two b digits. This means that the above number will look like:
(a1  c)×100 + 9×10 + (10×c  a)
At this point, notice that if we add the first and last digits of our result, (ca) + (a1c) = 1. In mathematical terms, this implies that the sum of the first and last digits will always be 9 (since 1 and 9 are equivalent modulo 10, if that makes any sense to you). But if you do a few examples, you'll probably be convinced why this is true. So no matter what we started with, we now have a number of the form d×100 + 9×10 + (9d).
Okay, now you may be able to see what will happen when we reverse this last number and add the two together.
[d×100 + 9×10 + (9d)] + [(9d)×100 + 9×10 + d]
Add the ones digits: (9d) + d = 9.
Add the tens digits: 9 + 9 = 18. (Place the 8 and carry the 1.)
Add the hundreds: d + (9d) = 9, and with the extra 1 we carried, we get 10.
So the final result, no matter what digits a, b, and c we started with, will always be 1089.
Edit: Thanks to krissielee for finding a counterexample for this! It looks like whenever the first and last digits are only one apart, you run into a problem. In that case, the difference between the two numbers will be 99, and the trick requires the difference to be a threedigit number to work.
This is a list of the ones that won't work:
192, 182, 172, 162, 152, 142, 132 293, 283, 273, 263, 253, 243, 213 394, 384, 374, 364, 354, 324, 314 495, 485, 475, 465, 435, 425, 415 596, 586, 576, 546, 536, 526, 516 697, 687, 657, 647, 637, 627, 617 798, 768, 758, 748, 738, 728, 718 879, 869, 859, 849, 839, 829, 819
And then the reverse of each, for a total of 112 numbers that won't work. There are 729 different threedigit numbers with all digits different and nonzero, so the chance of someone picking one of those is about 15%.
Interesting! 

