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

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9*111 9*3*37 \[(9*3*37)^9)^3)^{37}\] not sure if something like that would be a good idea
well .. i was thinking like factoring it. it's a programming problem if you do this directly on any language ... returns Not an integer
i wouldnt know right off hand what that is :) but, maybe \[(1000-1)^3)^9)^{37}\]

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Other answers:

this looks promising \[ (1000 - 1)^{999} = \sum_{i=0}^{999} \binom{999}{i}1000^i\]
yes it does :)
and maybe a (-1)^(999-i) perhaps
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.
gotta sleep .. we'll see this tomorrow.
good luck ;)
\[999^{999}\mod 10^{10} = (1000-1)^{999}\mod 10^{10}=(10^3-1)^{999}\mod 10^{10}\]\[=\sum_{k=0}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[=\sum_{k=4}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10} \] \[=0+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\Big)\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2\Big)\mod 10^{10}+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\]\[\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)*10^6\mod 10^{10}+999,000-1\]\[\Big(\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)\mod 10^{4}\Big)*10^6+999,000-1\]
wow i cant understand this
how did you get zero for |dw:1348816150400:dw|
oh I see, because (10^3)^4 mod 10^10= 0
this ain't easy ... you can write a short program to determine what are the first ten terms. still this is quite long.
well ive never seen mod used in binomial problems this is cool
well ... I'm still looking for some better method. this might blow up my processing chips
valpeys solution is fine, why dont you use that?
because that is only for particular 999 which I posed for this particular problem. My original problem is to find the first ten digits of \[ 1^1 + 2^2 + 3^3 + ... + 1000^{1000}\]
for 999, the one is parts 1000 and 1 <--- this one factor makes things easy. for 998 the other part is 2 .... I can't ignore powers of 2
The first ten digits of 1^1 +2^2 +3^3 +...+1000^1000 is the first ten digits of 1000^1000 As 1000^1000 > 10^10 *999^999
And first ten digits of 1000^1000 is 1000000000 Thus, the first ten digit of 1^1+2^2+3^3+...+1000^1000 is 1000000000
i think the problem is to find the last ten digit of it. You don't have to find out the solution using purely maths only, you can use computer to calculate. the problem is taken from here

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