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amistre64Best ResponseYou've already chosen the best response.2
9*111 9*3*37 \[(9*3*37)^9)^3)^{37}\] not sure if something like that would be a good idea
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
i wouldnt know right off hand what that is :) but, maybe \[(10001)^3)^9)^{37}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
this looks promising \[ (1000  1)^{999} = \sum_{i=0}^{999} \binom{999}{i}1000^i\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
and maybe a (1)^(999i) perhaps
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
gotta sleep .. we'll see this tomorrow.
 one year ago

ValpeyBest ResponseYou've already chosen the best response.0
\[999^{999}\mod 10^{10} = (10001)^{999}\mod 10^{10}=(10^31)^{999}\mod 10^{10}\]\[=\sum_{k=0}^{999}\binom{999}{k}(10^3)^k*(1)^{999k}\mod 10^{10}\]\[=\sum_{k=4}^{999}\binom{999}{k}(10^3)^k*(1)^{999k}\mod 10^{10}+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(1)^{999k}\mod 10^{10} \] \[=0+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(1)^{999k}\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3\binom{999}{2}(10^3)^2+\binom{999}{1}(10^3)^1\binom{999}{0}(10^3)^0\Big)\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3\binom{999}{2}(10^3)^2\Big)\mod 10^{10}+\binom{999}{1}(10^3)^1\binom{999}{0}(10^3)^0\]\[\Big(\binom{999}{3}10^3\binom{999}{2}\Big)*10^6\mod 10^{10}+999,0001\]\[\Big(\Big(\binom{999}{3}10^3\binom{999}{2}\Big)\mod 10^{4}\Big)*10^6+999,0001\]
 one year ago

perlBest ResponseYou've already chosen the best response.0
wow i cant understand this
 one year ago

perlBest ResponseYou've already chosen the best response.0
how did you get zero for dw:1348816150400:dw
 one year ago

perlBest ResponseYou've already chosen the best response.0
oh I see, because (10^3)^4 mod 10^10= 0
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
this ain't easy ... you can write a short program to determine what are the first ten terms. still this is quite long.
 one year ago

perlBest ResponseYou've already chosen the best response.0
well ive never seen mod used in binomial problems this is cool
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
well ... I'm still looking for some better method. this might blow up my processing chips
 one year ago

perlBest ResponseYou've already chosen the best response.0
valpeys solution is fine, why dont you use that?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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 \[ 1^1 + 2^2 + 3^3 + ... + 1000^{1000}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
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
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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 http://projecteuler.net/
 one year ago
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