[QUOTE=Captain Carrot]
One simple proof (well, not really a rigorous proof in any sense of the word):
You have a number with digits a,b,c,d.
Multiplied by 11, it becomes a,a+b,b+c,c+d,d.
The odd digits sum to a+(b+c)+(d), and the even digits sum to (a+b)+(c+d).
This holds for any number of digits, not just four.
[/QUOTE]
No, that’s not rigorous, but I think it can be made rigorous, and the result is a proof that I think could be followed even by non number-theoreticians. (Though the proof must necessarily be a bit longer for that reason–baby steps must be made explicit. And this is irrelevant to the OP’s purpose, since he’s teaching a number theory class anyway. But still, I present this in the interest of math fun.)
The following is still not fully rigorous, but I think it’s clear it can be made so.
Say you have a number …dcba (The … means there may be more digits than just the four I’ve mentioned.)
Thats … 1000d + 100c + 10b + a.
Multiplying by 11 you get
… 11000d + 1100c +110b +11a
which could be written as:
… dd000 + cc00 + bb0 + aa
What happens when you put these all in columns to add them up?
Unfortunately, the columns in the font this board uses don’t line up, but try to pretend:
…
dd000
cc00
bb0
aa
Say instead of adding these columns up in the usual fashion, you simply added each column up and wrote down the answer beneath in (i.e, without carrying.) You will get a list of five numbers, to wit, the numbers
d, d + c, c + b, b + a, a
(its clear that if our original number had had more than four digits, then the list I just typed out would not have an element reading “d” but rather “e + d” with another element previous to it with either an “e” (and that would begin the list) or else an “f + e” with yet another element previous to it, and so on. The comments I am going to make here generalize to all those cases.)
Now, say you determined yourself to add up the odd numbered numbers from this list (the 1st, 3rd, and 5th in this case) and subtract the even numbered ones from that total. You get
d - d - c + c + b - b - a + a.
In other words, zero.
So the list of numbers we got from the unorthodox column addition will always come to zero when adding the odd column results and subtracting the even column results.
But what about orthodox addition? In this case, those numbers on the list we got from unorthodox addition which have two digits will cause problems, because of carrying. If one of the numbers on the list had been 15, then if we had added in the orthodox fashion, we would have written down, not 15, but rather, 5 and carried the one over to the next column.
Do address that possibility, say we had, after unorthodox addition, come up with the following list of numbers:
z, y, x, w, u.
If any of these had two digits, then in orthodox addition the tens digit would have been carried. Notice that if any had two digits, the tens digit had to be a ‘1’. This is because, in each of the columns above, at most two non zero numerals appeared. The largest two digit number resulting from adding two one digit numbers is 18—the digit to be carried, therefore, is 1 at maximum.
So, say w had been a two digit number. Then the numeral for the answer to the addition problem (the numeral standing for the number we got by multiplying our original number by 11 in the first place) would consist in the following digits:
z, y, x + 1, w – 10, u.
Now, recall that z – y + x – w + u equaled 0. What does this new list add up to, then? Well, the x digit is an odd one, so it is added. The w digit is subtracted. So the new list adds up to z – y + (x + 1) – (w – 10) + u,
Or,
z – y + x + 1 – w + 10 + u
or
z – y + x – w + u + 10 + 1
or
0 + 11 == 11.
Had the carrying been done from x (an odd digit) to y (an even digit) instead, then the total would have been -11 rather than 11.
Either way, its clear a single carry changes the total by either -11 or 11. If there are multiple carries, then, they must cause the total to change by some number you can get from adding and subtracting 11’s. But, since we are adding and subtracting all these 11s from an “initial” total of zero, this just means the total must be some multiple of 11.
And that’s the proof.
-FrL-
