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

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

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.

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

Of course, if you allow the use of [URL=http://mathworld.wolfram.com/ArrowNotation.html]Donald Knuth's Arrow Notation[/URL], 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.

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

-michael

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).

[QUOTE=jinydu]That reminds me. How do you obtain an approximation for the factorial of any natural number, n?[/QUOTE]

[URL=http://mathworld.wolfram.com/StirlingsSeries.html]Stirling Series[/URL]

[QUOTE=michael]Try to prove that 9[sup](9[sup]9[/sup])[/sup]>((9!)!)!
:whistle:

-michael[/QUOTE]

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

9[sup]387420489[/sup] 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.

[QUOTE=Gary Edstrom]Of course, if you allow the use of [URL=http://mathworld.wolfram.com/ArrowNotation.html]Donald Knuth's Arrow Notation[/URL], there is no limit to the size of number that can be represented with even just 2 digits.[/QUOTE]

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[sup]9[sup]9[/sup][/sup]. If we allow non-digit symbols, then simple repetition of (x)! can turn a single 9 into an arbitrarily large number.

[QUOTE]Originally Posted by [B]wblipp [/B]
I think the constraint "only three digits" should be interpretted to mean "and no other symbols, either." [/QUOTE]And to anticipate the obvious, let's restrict it further to
"The largest number that can be written using only 3 digits, [b]base 10,[/b] and no other symbols, is 9[sup]9[sup]9[/sup][/sup]. :smile: