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amistre64
 2 years ago
Best ResponseYou've already chosen the best response.29*111 9*3*37 \[(9*3*37)^9)^3)^{37}\] not sure if something like that would be a good idea

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0well .. 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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2i wouldnt know right off hand what that is :) but, maybe \[(10001)^3)^9)^{37}\]

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

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2and maybe a (1)^(999i) perhaps

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0yeah ... 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.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0gotta sleep .. we'll see this tomorrow.

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

perl
 2 years ago
Best ResponseYou've already chosen the best response.0wow i cant understand this

perl
 2 years ago
Best ResponseYou've already chosen the best response.0how did you get zero for dw:1348816150400:dw

perl
 2 years ago
Best ResponseYou've already chosen the best response.0oh I see, because (10^3)^4 mod 10^10= 0

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

perl
 2 years ago
Best ResponseYou've already chosen the best response.0well ive never seen mod used in binomial problems this is cool

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0well ... I'm still looking for some better method. this might blow up my processing chips

perl
 2 years ago
Best ResponseYou've already chosen the best response.0valpeys solution is fine, why dont you use that?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0because that is only for particular 999 which I posed for this particular problem. My original problem is to find the first ten digits of \[ 1^1 + 2^2 + 3^3 + ... + 1000^{1000}\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0for 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.0The 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.0And first ten digits of 1000^1000 is 1000000000 Thus, the first ten digit of 1^1+2^2+3^3+...+1000^1000 is 1000000000

experimentX
 one year ago
Best ResponseYou've already chosen the best response.0i 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/
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