Question

Find the last 10 digit's of \( 999^{999} \) ?

Answer 1

9*111

9*3*37

\[(9*3*37)^9)^3)^{37}\]

not sure if something like that would be a good idea

Answer 2

well .. i was thinking like factoring it. it's a programming problem
http://projecteuler.net/problem=48
if you do this directly on any language ... returns Not an integer

Answer 3

i wouldnt know right off hand what that is :)

but, maybe
\[(1000-1)^3)^9)^{37}\]

Answer 4

this looks promising
\[ (1000 - 1)^{999} = \sum_{i=0}^{999} \binom{999}{i}1000^i\]

Answer 5

yes it does :)

Answer 6

and maybe a (-1)^(999-i)

perhaps

Answer 7

yeah ... i can calculate it programatically (for this particular problem). I still need to generalize it to every terms. this won't be nice.
this isn't nice. I hope there is nice algebra.

Answer 8

gotta sleep .. we'll see this tomorrow.

Answer 9

good luck ;)

Answer 10

\[999^{999}\mod 10^{10} = (1000-1)^{999}\mod 10^{10}=(10^3-1)^{999}\mod 10^{10}\]\[=\sum_{k=0}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[=\sum_{k=4}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10} \] \[=0+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\Big)\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2\Big)\mod 10^{10}+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\]\[\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)*10^6\mod 10^{10}+999,000-1\]\[\Big(\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)\mod 10^{4}\Big)*10^6+999,000-1\]

Answer 11

wow i cant understand this

Answer 12

how did you get zero for