Really big number

[Chasney913]

So, a couple years ago, a magazine had a contest as to the largest number created with 3 digits. Most people say 999. However, using simple operators (no factorials, etc.), the answer was 9^(9^9). This is different than 9^9^9, interpreted as (9^9)^9, which is a number about 2E77, unless my calculator is not working correctly. However, 9^(9^9) is quite a bit over what any calculator can handle. I know it is less than a googolplex, but I was wondering, around how large is it? Just a number of zeros would be fine, because an exact number would be difficult to do. If anyone knows, or can explain how I should go about finding out, it would be appreciated.

[DarkerLine]

Well, if you take the base 10 logarithm of a number, and round down, you get one less than the number of digits. In this case, log(9^(9^9)) is (9^9)log 9, which your calculator will be able to calculate - and give you the answer 369693099.6. So the number of digits is 369693100.

Weregoose's routines page gives you a routine for computing such high powers (approximately). In this case, the output is {4.281244195 369693099} which represents the number 4.281244195E369693099 and is actually accurate to about 6 decimal places.

Edit: now that I think about it, perhaps that routine should be modified so that it doesn't give misleadingly accurate answers in the mantissa? In this case, we only have 5 digits of the fractional part of the log: 0.63157 - when we raise 10 to this power to get the mantissa, we can only guarantee that 4.2812 is accurate - including the digits makes it seem like they're all correct, which is false.

[Chasney913]

Thanks, it's a bit easier to visualize the number if I can get a sense of how large it is. I can see why this number is the biggest; it beats 999 by a long shot.

I agree that the fractional part should take into account accuracy. The 4.281244195 makes it seem like it has a extremely accurate answer, whereas it is only accurate to 5 places.

I suppose you could make the number bigger if you used factorials or other higher operators. Contest idea! Mind you, you could just keep adding on factorials, 9!!!!!!!!^(9!!!!!!!!!^9!!!!!!!!!), so you'd need to limit the quantity of each operator.

Anyways, thanks again.

[alexrudd]

[Harrierfalcon]

"Digit's" is a fairly general term. If it's limited to base 10, don't bother me.

However, if you count all bases, then technically it could be infinity, as "999" could be base 2, base 16, or as high of a base as you want .

[Chasney913]

I believe it was supposed to be limited to base 10. It didn't explicitly say, but I probably mentioned something along that line, like it's supposed to be in the number system most people are familiar with. Also, I've never known base 2 to use the digit "9".

[simplethinker]

There's infinity in a (used-to-be) common numerical system, the babylonian sexagesimal system. This uses symbols with an actual value so this still counts as a real-valued number. They had no placeholder like 0, so it can be interpreted as any number of 'digits' as you want.