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Topic: 9^9^..ad infinitum (Read 1144 times) |
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TenaliRaman
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9^9^..ad infinitum
« on: Mar 1st, 2004, 6:25am » |
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Let x=9^9^9^...ad infinitum
Find the last ten digits of x.
Unfortunately,my elementary modular arithmetic skills haven't helped me here and this seems too daunting to be executed on a computer.
Seeing that this was on my College Puzzle Board, i am expecting it to be easy.
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| « Last Edit: Mar 1st, 2004, 6:25am by TenaliRaman » |
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Barukh
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Re: 9^9^..ad infinitum
« Reply #1 on: Mar 1st, 2004, 6:47am » |
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I'm confused... Is x a finite number? Or I misunderstand the meaning of '^' operation?
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #2 on: Mar 1st, 2004, 6:52am » |
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:: 0123456789 ::
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ThudnBlunder
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Re: 9^9^..ad infinitum
« Reply #3 on: Mar 1st, 2004, 7:01am » |
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That makes three of us.  (Evidently John is not.)
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towr
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Re: 9^9^..ad infinitum
« Reply #4 on: Mar 1st, 2004, 7:09am » |
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on Mar 1st, 2004, 6:25am, TenaliRaman wrote:Let x=9^9^9^...ad infinitum
Find the last ten digits of x. |
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I suppose you mean something like
f1 = 9
fn = fn-1^9
gn = fn % 10000000000
and we should find lim n [to] [infty] g(n)
Rather than trying to find (lim n [to] [infty] f(n)) % 10000000000 = [infty] % 10000000000 which is undefined
You can use
h1 = 9
hn = hn-1^9 % 10000000000
to find the answer, since the 11th (or higher) least significant digit won't affect the lesser significant digits in these kind of multiplications..
But it doesn't seem to converge..
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| « Last Edit: Mar 1st, 2004, 7:22am by towr » |
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rmsgrey
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Re: 9^9^..ad infinitum
« Reply #5 on: Mar 1st, 2004, 7:51am » |
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Does fn=9^fn-1 give convergence on gn?
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towr
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Re: 9^9^..ad infinitum
« Reply #6 on: Mar 1st, 2004, 7:59am » |
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I was just thinking about that..
I'm not sure how to get an answer (if there is one) in that case though..
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #7 on: Mar 1st, 2004, 8:13am » |
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on Mar 1st, 2004, 7:01am, THUDandBLUNDER wrote:
That makes three of us. (Evidently John is not.) |
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My answer is speculation based on inspection of the first few results. Obviously the number gets very large very quickly, so without some sort of proof, there is no way to know for sure.
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #8 on: Mar 1st, 2004, 8:19am » |
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How about representing 910 in base 3 (1003) or 9 (109)? Essentially the problem then becomes: given a number in base 3 or 9 that is a 1 with an infinite number of zeros after it, what are the last ten digits in base 10?
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Sameer
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Re: 9^9^..ad infinitum
« Reply #9 on: Mar 1st, 2004, 8:22am » |
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hmmm I think this one goes with that thread of .999... = 1 thing I guess..
::
Here's my approach Let's say the last ten digits were represented by number m.
Let m be broken up into 10 digits a1,a2,... a10.
Then a10*9 = a10 (since it converges?) => a10 = 0
Recursively all numbers come to 0.
So the answer is 0000000000
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towr
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Re: 9^9^..ad infinitum
« Reply #10 on: Mar 1st, 2004, 9:16am » |
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I very much doubt you can ever get 9^x to end in 0..
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Sameer
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Re: 9^9^..ad infinitum
« Reply #11 on: Mar 1st, 2004, 9:47am » |
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Crap you are right.. I just modified the problem from powers to multiplicated  .. back to drawing board... looks like any number raised to power 9 can have the last number as itself... so i guess u just work backwards...
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #12 on: Mar 1st, 2004, 11:04am » |
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on Mar 1st, 2004, 9:16am, towr wrote:| I very much doubt you can ever get 9^x to end in 0.. |
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9x must end in 1 or 9. Think. 92 = 81. Multiply 81 by 9, and the last digit clearly is 9. Again, 9 times 9 is 81. The last digit at least must equal 1, although carrying means the 10s digit may not equal 8.
Of course, since this is 99 and not 92, the answer will be more restrictive -- either one or nine. I suspect it will be nine. And I still think my first guess is correct.
The cycle goes on and on... where does it end? It does not!
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #13 on: Mar 1st, 2004, 11:16am » |
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I ran this simple little C++ program and it just keeps scrolling and scrolling, apparently disregarding the overflow bit, and never comes to any conclusion. Maybe that is the problem -- some inaccuracy introduced from overflowing and blundering forward anyway.
that integer type is G++'s 64 bit unsigned integer. It is not standard. I think MS VC++ calls it "int64" or "_int64".
Code:#include <iostream>
using namespace std;
int main (void)
{
unsigned long long int n;
unsigned long long int mask = 10000000000;
for (n = 9; n < ULONG_LONG_MAX; n = n * n * n * n * n * n * n * n * n)
{
cout << (n % mask) << endl;
}
return 0;
} |
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ThudnBlunder
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Re: 9^9^..ad infinitum
« Reply #14 on: Mar 1st, 2004, 11:18am » |
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John, the last digit of 92n+1 is indeed 9 for any positive integer n.
Quote:Let x=9^9^9^...ad infinitum
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But 9[smiley=infty.gif] is not an integer and therefore has no final digit(s). 
We can't even claim that the final digit is 9.
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| « Last Edit: Mar 1st, 2004, 11:37am by ThudnBlunder » |
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towr
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Re: 9^9^..ad infinitum
« Reply #15 on: Mar 1st, 2004, 11:18am » |
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The last digit will allways be 9, since 9 is odd, 9^9 is odd and ends in a 9 etc.. As long as fn-1 is odd, 9^fn-1 = fn must end with a 9 (and is thus odd)
But how do we get the other numbers?
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| « Last Edit: Mar 1st, 2004, 11:20am by towr » |
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #16 on: Mar 1st, 2004, 11:41am » |
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on Mar 1st, 2004, 11:18am, towr wrote:The last digit will allways be 9, since 9 is odd, 9^9 is odd and ends in a 9 etc.. As long as fn-1 is odd, 9^fn-1 = fn must end with a 9 (and is thus odd)
But how do we get the other numbers? |
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We have the initial case, now all we need is the inductive step 
I figured that the last digit will always be 9, I was talking about other powers though, in response to another post.
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #17 on: Mar 1st, 2004, 11:43am » |
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on Mar 1st, 2004, 11:18am, THUDandBLUNDER wrote:| But 9[smiley=infty.gif] is not an integer and therefore has no final digit(s). |
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Just like 0.9999999... is not 1, right? It has no final digits, at least not toward the unit's place, right?
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kellys
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Re: 9^9^..ad infinitum
« Reply #18 on: Mar 1st, 2004, 11:45am » |
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9-adically, x=0
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ThudnBlunder
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Re: 9^9^..ad infinitum
« Reply #19 on: Mar 1st, 2004, 12:09pm » |
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on Mar 1st, 2004, 11:43am, John_Gaughan wrote:
Just like 0.9999999... is not 1, right? It has no final digits, at least not toward the unit's place, right? |
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A different kettle of fish altogether.
0.9999... represents a convergent series which sums to the finite number of 1. No problem there.
Whereas we have a tower of exponents whose value diverges to an infinite number as it gets taller.
Are you still claiming that the final digit of ((((((9^9)^9)^9)^9)^9)^)^..... is 9?
Or do you mean, to paraphrase your sig, that the last digit of ((((((9^9)^9)^9)^9)^9)^)^..... is 9
for large values of ((((((9^9)^9)^9)^9)^9)^)^.....?
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| « Last Edit: Mar 1st, 2004, 12:26pm by ThudnBlunder » |
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #20 on: Mar 1st, 2004, 12:19pm » |
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The last three digits are 489. Let's see if you can figure out the rest 
edit: ah crap, that is not it... but I am close, I swear!
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| « Last Edit: Mar 1st, 2004, 12:23pm by John_Gaughan » |
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ThudnBlunder
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Re: 9^9^..ad infinitum
« Reply #21 on: Mar 1st, 2004, 12:21pm » |
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OK, the answer is ......489
What is the question?
If the question is 'What are the last three digits of (((((((9^9)^9)^9)^9)^9)^9)^9)^9
then the answer is ...089
And the last ten digits are ...9107814089
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| « Last Edit: Mar 1st, 2004, 12:54pm by ThudnBlunder » |
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Sameer
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Re: 9^9^..ad infinitum
« Reply #22 on: Mar 1st, 2004, 1:10pm » |
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how did you arrive at this answer?
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"Obvious" is the most dangerous word in mathematics.
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ThudnBlunder
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Re: 9^9^..ad infinitum
« Reply #23 on: Mar 1st, 2004, 1:20pm » |
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Quote:| how did you arrive at this answer? |
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I had a peek at John's screen just as his numbers stopped 'scrolling and scrolling' and then quickly checked the answer with Mathematica.
What are the first few digits of 9^(9^(9^9))?
How feasible would it be to calculate them?
What would be your best guess for the 1st digit?
What else can be said about this number? (Yes, I know it's big.)
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| « Last Edit: Mar 1st, 2004, 1:49pm by ThudnBlunder » |
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John_Gaughan
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Re: 9^9^..ad infinitum
« Reply #24 on: Mar 1st, 2004, 1:32pm » |
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on Mar 1st, 2004, 1:20pm, THUDandBLUNDER wrote:| I had a peek at John's screen just as his numbers stopped scrolling and scrolling and then I quickly checked the answer with Mathematica. |
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I answered several questions that this thread did not ask in the process of writing that little program. It still is not correct. I really need to pay attention either to riddles or to my job, but not both at the same time
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