Why does 11111111^2=12345678987654321
s...@radix.net
interesting facts. I'm sure you've seen something like this: "A duck's
quack does not echo - nobody knows why.", that sort of thing.
One of the items in the list is the fact that 11,111,111 x 11,111,111 =
12,345,678,987,654,321.
I wondered why.
After scratching my head for a while, I came up with a place to start. A
number in decimal notation is nothing more than a polynomial, where x,
instead of being unknown, is ten. 5,924 = (5*10^3) + (9*10^2) + (2*10^1) +
(4*10^0) = 5x^3 + 9x^2 + 2x + 4.
The problem in question is "just" asking us to square a 8th order
polynomial where the coefficients are all one. I thought I would call a
polynomial with all coefficients = +1 a "Unit polynomial". I doubt it's a
useful enough concept to deserve its own name, but why not. For the sake
of compact notation, call the nth order Unit polynomial Un(x).
To start at the most basic case, U0(x) = 1, and 1^2 = 1. Nothing
interesting there.
U1(x) = 1*x + 1, and [U1(x)]^2 = x^2 + 2x + 1. Hopping back to base ten
for a moment, U1(10) = 11 and [U1(10)]^2 = (10)^2+ 2*10 + 1 = 121. Since
we already know that 11^2 = 121, this confirms what we've done so far.
I kept working along these lines, and tried getting "recurrence relations"
for [Un+1(x)]^2 given [Un(x)]^2. I was getting the feeling that squaring
the nth order Unit Polynomial would give you coefficients 1 for the
highest order term, then 2, then 3, etc., up to a maximum, and then back
down to 1 for the constant term. But there was nothing simple that seemed
to come out of it to explain why and how.
Then I saw an episode of the _Project Mathematics!_ TV show. It's a show
produced by Caltech, the same group that produced the _Mechanical
Universe_ TV series for introudctory physics. I like both series, in part
because they use excellent computer graphics to explain concepts, and
because they provide historical information that is usually lacking on
most textbooks.
In this particular episode of _Project Mathematics!_ they showed an
example of squaring a polynomial by contructing an actual square matrix,
where the top row and the left column contain the terms of the polynomial
in question, and you fill each cell with the product of the top row and
left column cells. For example, to square 5x^2 - 4x + 7 you would do the
following:
5x^2 | -4x | +7
============================
5x^2 || 25x^4 | -20x^3 | +35x^2
----------------------------
-4x || -20x^3 | +16x^2 | -28x
----------------------------
+7 || +35x^2 | -28x | +49
----------------------------
Add up all the terms of the same power, which are on the same diagonal
from lower-left to upper-right, and you get: 25x^4 - 40x^3 + 86x^2 - 56x +
49.
So to square, for example, U2(x) = x^2+x+1:
x^2 | +x | +1
======================
x^2 || x^4 | +x^3 | +x^2
----------------------
+x || x^3 | +x^2 | +x
----------------------
+1 || x^2 | +x | +1
----------------------
Gathering like terms gives [x^2+x+1]^2 = x^4 + 2x^3 + 3x^2 + 2x + 1. Let
x=10 and pull out your calculator, and you'll see that 111^2 = 12,321.
This visual trick to square polynomials is actually the key to the
problem. When we go from squaring Un(x) to squaring Un+1(x) we're adding a
row and a column to the matrix. It's pretty obvious looking at the matrix
that there is one x^2n term and one con stant term, two x^(2n-1) terms and
two linear terms, etc. And there are (n+1) terms of x^n along the
diagonal. So in general [Un(x)]^2 = x^2n + 2x^(2n-1) + 3x^(2n-2) + ... +
(n+1)x^n + ... + 2x + 1.
Or going back to base ten (x=10):
n Un(10) [Un(10)]^2
- ----------- ------------
1 11 121
2 111 12,321
3 1,111 1,234,321
4 11,111 123,454,321
5 111,111 12,345,654,321
6 1,111,111 1,234,567,654,321
7 11,111,111 123,456,787,654,321
8 111,111,111 12,345,678,987,654,321
And that's the original problem!
To ponder further: now that we've run out of base ten digits with squaring
U8(10), what would [U9(10)]^2 look like?
[U9(x)]^2 = x^18 + 2x^17 + 3x^16 + 4x^15 + 5x^14 + 6^13 + 7x^12 + 8x^11 +
9x^10 + 10x^9 + 9x^8 + 8x^7 + 7x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1.
(Now that we know the system, generating this long 18th order polynomial
is just a matter of typing.) Plug i n x=10 and you'll get the 10x^9 term
becoming 1*10^10, which then adds to the 9*10^10 term to become 10*10^10 =
1*10^11. That adds to the 8*10^11 term to become 9*10^11. Therefore
[U9(10)]^2 = 1,234,567,900,987,654,321 - so much for patterns! I'm sure
you 'd have the same problem with [U7(8)]^2 expressed in base 8. But
that's another day's problem!
I don't know if this is good for anything, but it was a fun distraction
for me. It does underline the equivalence of polynomials and decimal
notation, and how you can use one to help with problems in the other. And
I learned a new trick in the process! You can't have enough tricks in
your bag, you never know which one is going to be useful!
-Simon
--
___________________________________________________________________________
Simon S. Goldenberg, Ph.D. s...@radix.net http://www.goldenberg.org
Opinions are my own - who else in their right mind would want them?
---------------------------------------------------------------------------
"A Ph.D. is not an innoculation against stupidity." - Robert L. Park, Ph.D.
___________________________________________________________________________
Nathaniel Silver
Isn't it: 111,111,111 x 111,111,111 = 12345678987654321?
Chris Ferrante
Mr. Silver that a longhand multiplication is sufficient for proving this,
although what you had was kindof neat!! I'd like to know more about this
matrix thing to multiply binomials and such, I've never heard of that!!
Oh, wait, I just read the part about 111111111^2, it seems the pattern
develops from 11^2 = 121, 111^2 = 12321, etc... to 12345678987654321 at
111111111^2. That's interesting to note, but I think that was in your other
post too.
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<s...@radix.net> wrote in message news:82bvql$oo9$1...@news1.Radix.Net...
Kaimbridge
s...@radix.net wrote:
> There's an email going around the net with a list of amusing or
> interesting facts. I'm sure you've seen something like this: "A duck's
> quack does not echo - nobody knows why.", that sort of thing.
>
> One of the items in the list is the fact that
> 11,111,111 x 11,111,111 = 12,345,678,987,654,321.
First of all, because it don't: 11,111,111^2 = 123,456,787,654,321
P=) P=) P=) 111,111,111^2 = 12,345,678,987,654,321
Just do it manually and you can see why:
111 111 111
11 111 111 x111 111 111
x11 111 111 ============
=========== 111 111 111
11 111 111 1 111 111 110
111 111 110 11 111 111 100
1 111 111 100 111 111 111 000
11 111 111 000 1 111 111 110 000
111 111 110 000 11 111 111 100 000
1 111 111 100 000 111 111 111 000 000
11 111 111 000 000 1 111 111 110 000 000
111 111 110 000 000 11 111 111 100 000 000
=================== ======================
123 456 787 654 321 12,345,678,987,654,321
It's a pyramid with the high middle number equalling the number of 1s:
1^2 = 1
11^2 = 1 2 1
111^2 = 12 3 21
1111^2 = 123 4 321
11111^2 = 1234 5 4321
111111^2 = 12345 6 54321
1111111^2 = 123456 7 654321
11111111^2 = 1234567 8 7654321
111111111^2 = 12345678 9 87654321
~Kaimbridge~
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UBasic Programming Forum: http://www.InsideTheWeb.com/mbs.cgi/mb426556
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Before you buy.
Leslie Wright
able to derive, using your labelling and that the number is in base 10,
that:
Un(10) = [ 10^(2n+2) - 2*10^(n+1) + 1] / 81
The 81 in the divisor fascinates me. I recommend that you take a peek at
the quantities 81*U(n) for all n, not just up to 8. There you will see a
neat pattern, namely an integer of 2n+2 digits consisting of n 9s, then
one 8, then n 0s, then 1, viz:
81*U5(10) = 999998000001
81*U12(10) = 99999999999980000000000001
81*U2(10) = 998001
81*U3(10) = 99980001
81*U20(10) =999999999999999999998000000000000000000001
Cool, eh?
Les
Leslie Wright
you will readily convince yourself that:
Un(x) = [x^(n+1) - 1] / ( x - 1 ) = SIGMA( x^j, j=0 to n )
(This is a basic property of binomials of the form x^(n+1) - 1, and can
be readily proven by induction, and other means I am sure!)
If you let x = 10 and Un(10), you see where we get the funky 9^2 = 81 in
my previous post.
kewl....
Les
Leslie Wright
This part should read "If you let x = 10 and YOU SQUARE Un(10)...."
Les
Chris Pando
>
> There's an email going around the net with a list of amusing or
> interesting facts. I'm sure you've seen something like this: "A duck's
> quack does not echo - nobody knows why.", that sort of thing.
>
> One of the items in the list is the fact that 11,111,111 x 11,111,111 =
> 12,345,678,987,654,321.
>
> I wondered why.
>
... snippage ...
Leading me to wonder why
8,589,934,592 x 116,415,321,826,934,814,453,125
= 1,000,000,000,000,000,000,000,000,000,000,000
Chris
---
God is real, unless declared integer.