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amistre64

Let n=\(2^3*3^2*5*7^3*11\) Find the smallest factorial: k! that is divisible by n Show your work!! :)

  • one year ago
  • one year ago

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  1. Malachy
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    27!

    • one year ago
  2. amistre64
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    thats not work, thats an answer .... teaches me nothing :/

    • one year ago
  3. hba
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    I thought this was open study.

    • one year ago
  4. Malachy
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    =2^3*3^13*5*7 here 13 is the maximum power raised on any number. thus we have to take till the number untill 3 comes 13 times 3=1 6=1 9=2 12=1 15=1 18=2 21=2 24=2 27=3 Total of rhs=15>13 thus 27! will be the ans

    • one year ago
  5. amistre64
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    ------------------ 2 2 2 3 3 5 7 7 7 11 2 ------------------ 1 2 2 3 3 5 7 7 7 11 2 3 ------------------ 2 2 3 5 7 7 7 11 2 3 4 ------------------ 3 5 7 7 7 11 2 3 4 5 ------------------ 3 7 7 7 11 2 3 4 5 6 ------------------ 7 7 7 11 2 3 4 5 6 7 ------------------ 7 7 11 2 3 4 5 6 7 8 9 10 11 ------------------ 7 7 2 3 4 5 6 7 8 9 10 11 12 13 14 ---------------------------- 7 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 ----------------------------------------------- i think its 21!

    • one year ago
  6. amistre64
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    you seem to have used something different than that i posted in the question

    • one year ago
  7. amistre64
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    does your method, using my posted information, equate to 27 as well? or is it 21?

    • one year ago
  8. asnaseer
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    yup - I also get 21 using the same method as @amistre64

    • one year ago
  9. amistre64
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    im pretty adept at brute mathing; i see some potential in Malachys if i can get it verified ;)

    • one year ago
  10. asnaseer
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    I have never heard of Malachys - oh well - another new topic to learn...

    • one year ago
  11. asnaseer
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    do you have a reference (a link) to this method?

    • one year ago
  12. amistre64
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    i see in mine the 7^3 produces the biggest parts; and 7*3 = 21

    • one year ago
  13. amistre64
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    my method? i thought it up

    • one year ago
  14. asnaseer
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    true - but still have to show that all numbers before this are factored out as well - which you did using brute force :)

    • one year ago
  15. asnaseer
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    you said " i see some potential in Malachys" <<<--- is this made up? :)

    • one year ago
  16. amistre64
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    lol, we may never know unless they show up again ;)

    • one year ago
  17. asnaseer
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    you jest with us mere mortals sir! :)

    • one year ago
  18. amistre64
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    =2^3*3^2*5*7^3*11 here 3 is the maximum power raised on any number. thus we have to take till the number untill (2 or 7 ???) comes 3 times ???? Total of rhs=??? thus 21! will be the ans seems simple enough to me

    • one year ago
  19. Malachy
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    Actually i was missed as 2^3*3^2*5*7*3^11 thats why that ans came

    • one year ago
  20. amistre64
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    can you give a generalized rule for your method?

    • one year ago
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