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
2·29 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
2×29 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
100011_{2} 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
1000001_{2} 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
940_{16} Posts 
Quote:


20040215, 16:36  #9  
"Mark"
Apr 2003
Between here and the
2^{5}·3·67 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
2^{6}·37 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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