I appreciate all the enthusiastic comments, everyone. I can already tell this board will be a good place to get help. But please give me time to study your replies. For now, can we limit the focus to the "adding down" rather than the proof issue.
Flansburg really does give many examples, and they all do indeed add all the way down to 9, even large numbers:
2+3+4+3+7= 19 = 23,437 - 19 = 23,418 = 2+3+4+1+8 = 18 = 1+8=9
@JeffM, your answer is so kind and elegant I want to frame it and put it on the wall. I feel guilty to have to confess that, alas, it is way above my head. Is there an easier proof, I wonder?
I sense Hallsoflvy is right in saying it has something to do with the base ten system.
I wouldn't bother about this point if it weren't for the face that Flansburg at the outset of his book seems to stress that this is an essential basic principle: "If you understand this," he says on p. 3, "then you have just turned on the calculator in your brain."
But he seems to expect that readers will simple be able to intuit the underlying principle from the examples. I fear it's pointless to try to proceed in the book unless I understand this first.
I will appreciate and study any further comments, if you give me time.
Many thanks
Halls of Ivy is absolutely right that it has to do with decimal notation.
And elegance that is incomprehensible is worthless. So rip that post of mine off the wall.
Let's start by explaining this in terms of two digit numbers. What does 43 mean? It means (4 * 10) + 3. Now what is the sum of the digits?
It is 4 + 3 = 7. Now let's subtract the sum of the digits
in its expanded form from the actual number
in its expanded form.
\(\displaystyle (4 * 10) + 3 - (4 + 3) = (4 * 10) - 4 + 3 - 3 = 4 * 10 - 4 = 4 * (10 - 1) = 4 * 9, which/ is/ evenly/ divisible/ by/ 9.\)
Now this is not an accident. Consider the number \(\displaystyle ab = (a * 10) + b.\)
That is ab is not meant as multiplication but the first and second digits respectively of a two digit number. You with me?
The sum of the digits is \(\displaystyle (a + b).\)
So the number minus the sum of its digits \(\displaystyle (a * 10) + b - (a + b) = (a * 10) - (a * 1) + b - b = a * (10 - 1) = 9a.\)
The reason that this all works is that the unit digit subtracts out, and we get 10 times the tens-digit minus the tens-digit which is always 9 times the tens-digit, so the number is ALWAYS divisible by 9. 10 - 1 = 9, now and forever.
You solid with that?
Now here is where things get a little tricky. There is a form of proof (called weak mathematical induction) that says if I can prove that something is true for some particular integer and if I can
also prove that if that something is true for
any integer equal to greater than that particular integer, then it is also true for the
next higher integer as well, it is true for all integers equal to or greater than the particular integer. The intuition is this. In this case, I prove our adding up to a multiple of 9 idea is true for a two-digit number. Then I prove that
if the adding up idea is true for for an n-digit number, where n is not smaller than 2, then it is also true for an (n+1)-digit number. Now this means that because it is true for 2, it is true for 3, which means that it is true for 4, but then it must be true for 5, and that in turn must mean it is true for 6,etc, world without end. The whole line of dominoes falls.
So let's say our divisible by 9 idea works for any number with n digits. Let's consider the number x expressible in (n + 1) digits.
\(\displaystyle Then\ x = (10^n * y) + z, where\ z\ is\ a\ number\ expressible\ in\ n\ digits.\)
\(\displaystyle Let\ u = the\ sum\ of\ the\ digits\ of\ z.\) But we know that z - u = 9v.
Still with me?
OK then the sum of the digits of x is y + u. So what is the difference between x and the sum of its digits?
\(\displaystyle (10^n * y) + z - (y + u) = (10^n * y) - (y * 1) + z - u = y(10^n - 1) + 9v.\)
Now when I subtract 1 from a power of 10 greater than or equal to 2 I get a number that is a string of 9's
10 - 1 = 9
100 - 1 = 99
1000 - 1 = 999
10000 - 1 = 9999.
So that number is evenly divisible by 9. (If I was giving a super formal proof I'd need to prove that, but it is obvious.)
\(\displaystyle Let\ 9w = (10^n - 1) \implies y(10^n - 1) + 9v = 9wy + 9v = 9(wy + v).\) Evenly divisible by 9.
I've given it my best shot.
