experimentX
  • experimentX
Find the last 10 digit's of \( 999^{999} \) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
9*111 9*3*37 \[(9*3*37)^9)^3)^{37}\] not sure if something like that would be a good idea
experimentX
  • experimentX
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
amistre64
  • amistre64
i wouldnt know right off hand what that is :) but, maybe \[(1000-1)^3)^9)^{37}\]

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More answers

experimentX
  • experimentX
this looks promising \[ (1000 - 1)^{999} = \sum_{i=0}^{999} \binom{999}{i}1000^i\]
amistre64
  • amistre64
yes it does :)
amistre64
  • amistre64
and maybe a (-1)^(999-i) perhaps
experimentX
  • experimentX
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.
experimentX
  • experimentX
gotta sleep .. we'll see this tomorrow.
amistre64
  • amistre64
good luck ;)
Valpey
  • Valpey
\[999^{999}\mod 10^{10} = (1000-1)^{999}\mod 10^{10}=(10^3-1)^{999}\mod 10^{10}\]\[=\sum_{k=0}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[=\sum_{k=4}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10} \] \[=0+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\Big)\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2\Big)\mod 10^{10}+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\]\[\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)*10^6\mod 10^{10}+999,000-1\]\[\Big(\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)\mod 10^{4}\Big)*10^6+999,000-1\]
perl
  • perl
wow i cant understand this
perl
  • perl
how did you get zero for |dw:1348816150400:dw|
perl
  • perl
oh I see, because (10^3)^4 mod 10^10= 0
experimentX
  • experimentX
this ain't easy ... you can write a short program to determine what are the first ten terms. still this is quite long.
perl
  • perl
well ive never seen mod used in binomial problems this is cool
experimentX
  • experimentX
well ... I'm still looking for some better method. this might blow up my processing chips
perl
  • perl
valpeys solution is fine, why dont you use that?
experimentX
  • experimentX
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 \[ 1^1 + 2^2 + 3^3 + ... + 1000^{1000}\]
perl
  • perl
oh
experimentX
  • experimentX
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
anonymous
  • anonymous
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
anonymous
  • anonymous
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
experimentX
  • experimentX
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/

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