20040214, 16:03  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Mind Boggling Number
Mind Boggling Number.
The largest number that can be written using only 3 digits is 9^9^9. Mathematician and editor Joseph S. Madachy asserts that 1)With a knowledge of the elementary properties of numbers 2) a simple desk calculator The last 10 digits of this fantastic number (and other bigger nos.) have been calculated. For the last 10 digits of 9^9^9 these have been calculated and are 2,627,177,289. Can any one give me a method with the above conditions? Note 9^9^9 is not equal to 9^81 
20040214, 16:25  #2 
Jul 2003
Thuringia; Germany
58_{10} Posts 
begin with 9
then multply with 9 (so you get max. 1 digit more) if the result has more than 10 digits, remove the first (highest) and iterate this 9^9 times (this will take a while, but it works) When your calculator has more than 11 digits to work (normaly 13) you could "optimize" this by taking a few iteration at once (multiplying with 9^3). So you have to do only 9^9/3 steps. Are there better possibilities to solve the problem? Cyrix 
20040214, 22:34  #3 
Dec 2003
Albany, NY
2·3 Posts 
9^(9^9) = 9^387420489
Thus you need only multiply 9 by itself 387420489 times. To make your calculations easier, I suppose you could keep multiplying 9 by itself until the last 10 digits started repeating themselves (which is bound to happen). I haven't given it any thought, but will this repetiton begin before we are done computing the actual value? I suspect that it might. 
20040215, 00:07  #4 
Jul 2003
Thuringia; Germany
72_{8} Posts 
The order of 9 in the multiplicative group Z(10^10)* (the group of all integers relativly prime to 10^10), which means the lowest integer p>0, for which 9^p == 1 mod (10^10), is 250,000,000 (calculated with Maple).
With this knowledge you have to do "only" 387420489250000000 iterations. cyrix 
20040215, 08:40  #5 
Oct 2002
5·7 Posts 
Of course, if you allow the use of Donald Knuth's Arrow Notation, there is no limit to the size of number that can be represented with even just 2 digits. Since there is no up arrow on the standard keyboard, let's use "^" instead. Now, you can write the number 9^^9 in arrow noation. This can be written out as 9^(9^(9^(9^(9^(9^(9^(9^9))))))). If that isn't big enough, you could write 9^^^9. You couldn't even begin to expand it, much less comprehend its value.

20040215, 08:53  #6 
Dec 2003
Belgium
5×13 Posts 
Try to prove that 9^{(9[sup]9})[/sup]>((9!)!)!
michael 
20040215, 11:16  #7 
Dec 2003
Hopefully Near M48
3336_{8} Posts 
That reminds me. How do you obtain an approximation for the factorial of any natural number, n? I want at least the first few digits to be accurate, but avoid overflowing my calculator (which is limited to numbers < 10^100).

20040215, 15:33  #8  
"William"
May 2003
New Haven
23×103 Posts 
Quote:


20040215, 16:36  #9  
"Mark"
Apr 2003
Between here and the
14502_{8} Posts 
Quote:
9^{387420489} has fewer than 387420489 digits 362880! itself has well over a million digits. That means that (326880!)! will easily exceed 387420489 digits. I doubt I need to do any math in order for that to be obvious. 

20040215, 18:40  #10  
"William"
May 2003
New Haven
23·103 Posts 
Quote:


20040215, 23:10  #11  
Aug 2002
Portland, OR USA
2×137 Posts 
Quote:
"The largest number that can be written using only 3 digits, base 10, and no other symbols, is 9^{9[sup]9}[/sup]. Last fiddled with by Maybeso on 20040215 at 23:10 

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