Please, don’t ask me why - it’s a long story that’ll bore you to tears (but makes me and my co-workers laugh) - but I need to calculate the following, and FAST!

2 ^ 2 ^ 100 (using ^ to mean “to the exponent of”)

If it helps 2 ^ 100 is 1,267,650,600,228,229,401,496,703,205,376

The calculator that comes with Windows can’t do it. Microsoft Excel can’t do it. I tried this web site and still no luck.

The answer will have 381,600,854,690,147,056,241,701,702,033 digits in it, which is far more than the number of atoms in the universe. No one has the ability to write that number down, much less calculate it.

I should also mention I just cut and pasted that from the Windows calculator. It should be taken as a ballpark figure; not all of the given digits can be expected to be accurate due to rounding errors.

Punoqllads:The answer will have 381,600,854,690,147,056,241,701,702,033 digits in it, which is far more than the number of atoms in the universe.

After I posted we did our own calculations as to the size of the answer: It would take 147,573,952,589,676,412,928 GB to store the answer. That’s 147,573 YB (yotta bytes) I couldn’t find the right prefix to bytes to go any smaller. Hmmmm, if it’s not already defined I propose 1.476 ZB, or Zotti Bytes!

Cagey Drifter *Blast it! Can’t you see that they need the solution NOW?! We need less excuses and more answers!! *

What’s at stake is a box of Timbits! You’ve only got a 200 GB hard drive? Swap it out a few times!

The earlier posts have shown you how to get as many of the initial digits as you like. You can also find the last digits easily; for example, notice that successive powers of 2 have last digits 2, 4, 8, 6, 2, 4, 8, 6, … . So the last digit in 2[sup]2[sup]100[/sup][/sup] is 6. There are similar repeating patterns in the final n digits, though the sequences run longer before repeating.

The number of digits is actually a lot fewer than the number of atoms in the universe (though of course the number represented by those digits is vastly larger). The number of digits is about 3.816x10[sup]29[/sup]; the number of atoms in the observable universe is estimated at something like 10[sup]79[/sup]. 3x10[sup]29[/sup] atoms is only 633900 moles; if they’re all [sup]12[/sup]C atoms this weighs only 7600kg. So if you could write one base-10 digit per atom you could write the number in a solid block of carbon the size of a small car. (Be sure to get the clear-coat finish and undercoating so the answer doesn’t oxidize away!)

**Omphaloskeptic: ** So if you could write one base-10 digit per atom you could write the number in a solid block of carbon the size of a small car. (Be sure to get the clear-coat finish and undercoating so the answer doesn’t oxidize away!)
What a relief: I thought I’d have to up the prize to two boxes of Timbits!

I don’t think the problem is calculating the result, the problem is printing it out or saving that many digits on a disk.

What we need is that gentleman in Japan that memorized and then repeated PI to 80,000 or so places. Get him started on the project and just before the heat death of the universe, we would hear the entire number…

No need for the clear coat, just get your carbon in crystallized form, i.e. diamonds. The usual measure for diamonds is the Carat, of course. One Carat is equal 0.2 gram, so 7600 kg is 38,000,000 Carats.

Right. When you need to take a whole page to invent a new notation, and then another page to use it, then you’ve got something. But even then, it’s still far smaller than the vast overwhelming majority of integers. No, give me a transfinite cardinal of transfinite degree, at least. None of this piddly finite stuff.

What’s the order of operations here? If you were to write it out algebraically without parentheses, it would be right-to-left. However, the carot is a binary operator used since it’s usually impossible to write exponents of exponents on a computer in algebraic notation, and identical binary operators are usually, if not always, calculated from left to right. I think 2 ^ 2 ^ 100 is the same as (2 ^ 2) ^ 100, which can be calculated far more easily than 2 ^ (2 ^ 100). The original poster mentions 2 ^ 100, which suggests he wants it calculated from right to left, but it seems more likely to be asked to solve it from left to right.