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not enough :)

Okay, let us do this in steps. Who can find me the prime-power factorization of that number?

including 34 and 1? (with the respective number of zeroes after their numbers)

17*2*10^(let me count..)=-1*17*2*5^(...)*2(...)

there is a really easy way to do this but i forgot, finding primes of really really large numbers

\(17\cdot2\cdot 5^n\cdot 2^n\) where \(n\) is the number of 0s that I'm too lazy to count? :p

10^36 * 34 * (-10)
17*2*10^(36) * (-10)

152 factors?

wait, itd be greater than that.

17*2^(37)*5^(36)
If I trust that Ishaan counted correctly

^ Ruby counts better. It's 37 0s, so should be 38 and 37, if I'm not mistaken?

I used python interpreter to count
17*2^38* 5^37*(-1)

(-1)*17*2^(38)*5^(37)
keep forgetting the minus
now what?

Combinations of the 2s, 5s, and -1?

Why do I feel like Pascal's triangle is about to pop out of my screen and bite me?

oh I always forget that theorem...
probably because I never use it

what's it called again?

how does it work?

Interesting.

http://en.wikipedia.org/wiki/Divisor_function

ahhhh

Or, as across updated the question to read, the τ function.

@funinabox that's not how the theorem works...

instead of multiply* dunno why i said subtract

ok, who else started seeing "math processing error" all of a sudden?

it was just me I guess
it's gone now

I forgot to close the parenthesis up there.

oh wow, and now I think I actually understand the theorem
thanks again across

I noticed anoter typo. Oh well, you guys get the idea. ;P

♥ elegant proofs, elegant theorems.