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experimentX

  • 2 years ago

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

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  1. amistre64
    • 2 years ago
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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

  2. experimentX
    • 2 years ago
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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

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

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

  5. amistre64
    • 2 years ago
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    yes it does :)

  6. amistre64
    • 2 years ago
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    and maybe a (-1)^(999-i) perhaps

  7. experimentX
    • 2 years ago
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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.

  8. experimentX
    • 2 years ago
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    gotta sleep .. we'll see this tomorrow.

  9. amistre64
    • 2 years ago
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    good luck ;)

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

  11. perl
    • 2 years ago
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    wow i cant understand this

  12. perl
    • 2 years ago
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    how did you get zero for |dw:1348816150400:dw|

  13. perl
    • 2 years ago
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    oh I see, because (10^3)^4 mod 10^10= 0

  14. experimentX
    • 2 years ago
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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.

  15. perl
    • 2 years ago
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    well ive never seen mod used in binomial problems this is cool

  16. experimentX
    • 2 years ago
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    well ... I'm still looking for some better method. this might blow up my processing chips

  17. perl
    • 2 years ago
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    valpeys solution is fine, why dont you use that?

  18. experimentX
    • 2 years ago
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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 \[ 1^1 + 2^2 + 3^3 + ... + 1000^{1000}\]

  19. perl
    • 2 years ago
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    oh

  20. experimentX
    • 2 years ago
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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

  21. sauravshakya
    • one year ago
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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

  22. sauravshakya
    • one year ago
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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

  23. experimentX
    • one year ago
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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/

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