Mind Boggling Number

[mfgoode]

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

[cyrix]

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

[patdumpsite]

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.

[cyrix]

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" 387420489-250000000 iterations.

cyrix

[Gary Edstrom]

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.

[michael]

Try to prove that 9^(9[sup]9)[/sup]>((9!)!)!


-michael

[jinydu]

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).

[wblipp]

Originally Posted by jinydu

That reminds me. How do you obtain an approximation for the factorial of any natural number, n?

Stirling Series

[rogue]

Originally Posted by michael

Try to prove that 9^(9[sup]9)[/sup]>((9!)!)!


-michael

I can't prove that, but I could show the opposite is true. 9[sup](9⁹ = 9^387420489 while ((9!)!)! = (362880!)!

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.

[wblipp]

Originally Posted by Gary Edstrom

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.

I think the constraint "only three digits" should be interpretted to mean "and no other symbols, either." Then exponentiation can be shown by positioning as 9^9[sup]9[/sup]. If we allow non-digit symbols, then simple repetition of (x)! can turn a single 9 into an arbitrarily large number.

[Maybeso]

Originally Posted by wblipp
I think the constraint "only three digits" should be interpretted to mean "and no other symbols, either."

And to anticipate the obvious, let's restrict it further to
"The largest number that can be written using only 3 digits, base 10, and no other symbols, is 9^9[sup]9[/sup].