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experimentX Group Title

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

  • one year ago
  • one year ago

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  1. amistre64 Group Title
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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

    • one year ago
  2. experimentX Group Title
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    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
  3. amistre64 Group Title
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    i wouldnt know right off hand what that is :) but, maybe \[(1000-1)^3)^9)^{37}\]

    • one year ago
  4. experimentX Group Title
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    this looks promising \[ (1000 - 1)^{999} = \sum_{i=0}^{999} \binom{999}{i}1000^i\]

    • one year ago
  5. amistre64 Group Title
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    yes it does :)

    • one year ago
  6. amistre64 Group Title
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    and maybe a (-1)^(999-i) perhaps

    • one year ago
  7. experimentX Group Title
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    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
  8. experimentX Group Title
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    gotta sleep .. we'll see this tomorrow.

    • one year ago
  9. amistre64 Group Title
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    good luck ;)

    • one year ago
  10. Valpey Group Title
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    \[999^{999}\mod 10^{10} = (1000-1)^{999}\mod 10^{10}=(10^3-1)^{999}\mod 10^{10}\]\[=\sum_{k=0}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[=\sum_{k=4}^{999}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10} \] \[=0+\sum_{k=0}^{3}\binom{999}{k}(10^3)^k*(-1)^{999-k}\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\Big)\mod 10^{10}\]\[\Big(\binom{999}{3}(10^3)^3-\binom{999}{2}(10^3)^2\Big)\mod 10^{10}+\binom{999}{1}(10^3)^1-\binom{999}{0}(10^3)^0\]\[\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)*10^6\mod 10^{10}+999,000-1\]\[\Big(\Big(\binom{999}{3}10^3-\binom{999}{2}\Big)\mod 10^{4}\Big)*10^6+999,000-1\]

    • one year ago
  11. perl Group Title
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    wow i cant understand this

    • one year ago
  12. perl Group Title
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    how did you get zero for |dw:1348816150400:dw|

    • one year ago
  13. perl Group Title
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    oh I see, because (10^3)^4 mod 10^10= 0

    • one year ago
  14. experimentX Group Title
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    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
  15. perl Group Title
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    well ive never seen mod used in binomial problems this is cool

    • one year ago
  16. experimentX Group Title
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    well ... I'm still looking for some better method. this might blow up my processing chips

    • one year ago
  17. perl Group Title
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    valpeys solution is fine, why dont you use that?

    • one year ago
  18. experimentX Group Title
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    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
  19. perl Group Title
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    oh

    • one year ago
  20. experimentX Group Title
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    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
  21. sauravshakya Group Title
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    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
  22. sauravshakya Group Title
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    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
  23. experimentX Group Title
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    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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