Can you solve this math problem???

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Looks much bigger than a googolplex to me but so what?

(9^62773 + 2)^83721

I think = ((9^63774)^83721) + 83271 + 83271

In base 9 , it would look like a 1 with 63774^83721 zeroes
plus whatever (2 x 83271) is in base 9.

100000000.....0000000273406 in base 9 , with a gazillion oodles of zeroes.

I can't remember if there's a neat trick to convert base 9 to base 10 that would
quickly make an imaginable number like this, but unless I goofed, that's it in base 9.

edit: I think I goofed a little, but the number may be imaginable because this particular
math problem might involve patterns that involve powers of 2,9,10, and 11. My mistake
involves using 2 x 83271 instead of 2 ^ 83271 , or possibly broke another rule about
simplifying exponents. I'm not fast enough at doing math in my head to be confident
about the numbers above, especially since they are huge.

But integer arithmetic, no matter how large the numbers get, they are counting numbers,
very much rational and knowable, especially with computers. In fact, a DVD burned with
all ones on it contains only the exact number of all possible DVD's minus one!

Just to be clear: (9^62773 + 2)^83721

The "+ 2" is essentially meaningless, by comparison to the thing it is added to. Even when raised to a moderately larger power (83,721), its contribution to the result is tiny.

So, the essential part is the (9^62773)^83721, where you just multiply the exponents. Hence 9^5,339,139,333.

To express that as a tens exponent, just multiply the exponent by log10(9), and get a little more than 10^5,094,833,715.

Now, a googol is 10^100, so it's clearly quite a bit larger than a googol.

However, a googolplex is 10^googol. That's 10^10,000,000,000,000, ... ,000,000,000 (where the exponent, fully written out, would have 333 groups of three zeroes. Hence it is hugely less than a googolplex.

Not me...I can't even do the ones that start "If a train left Chicago...";)

My answer to that is that as the number is way, way bigger than the number of objects in the entire observable universe, the question itself is meaningless and invalid; the answer you get if you grind it to its logical mathematical conclusion is not correct.

(I'd say that about mathematics and logic as a whole, it's a nice approximation, but it isn't real, so don't be surprised when it gives us absurdities. But that's another topic.)

The binary number that a common data file is made of is often an exact count of
units such that the number if it counted atoms couldn't fit in the universe, and
almost always evaluates to over a googol unless it's got much less than 100 bytes.
You can't use scientific notation to compress a data file. And the math question's answer
for all we know could be a file, and it could be seriously corrupted if you don't add the 2,
especially because it is in parenthesis which is raised to a power.

Arithmetic CAN count things that can't fit in the universe, like the exact number of
possible unique DVD's, which is equal to a DVD full of 1's plus 1, and is somewhere
around 2 to the power of 32,000,000,000. It can therefore generate any possible
DVD, by calculating any number from zero to the highest one that can be burned.
We can no longer say that the number is too big to care about all the digits if we
want to burn a DVD, or we will get a coaster.