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amistre64
 3 years ago
Let n=\(2^3*3^2*5*7^3*11\)
Find the smallest factorial: k! that is divisible by n
Show your work!! :)
amistre64
 3 years ago
Let n=\(2^3*3^2*5*7^3*11\) Find the smallest factorial: k! that is divisible by n Show your work!! :)

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1thats not work, thats an answer .... teaches me nothing :/

hba
 3 years ago
Best ResponseYou've already chosen the best response.0I thought this was open study.

Malachy
 3 years ago
Best ResponseYou've already chosen the best response.1=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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1 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!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you seem to have used something different than that i posted in the question

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1does your method, using my posted information, equate to 27 as well? or is it 21?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0yup  I also get 21 using the same method as @amistre64

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im pretty adept at brute mathing; i see some potential in Malachys if i can get it verified ;)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0I have never heard of Malachys  oh well  another new topic to learn...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0do you have a reference (a link) to this method?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i see in mine the 7^3 produces the biggest parts; and 7*3 = 21

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1my method? i thought it up

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0true  but still have to show that all numbers before this are factored out as well  which you did using brute force :)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0you said " i see some potential in Malachys" <<< is this made up? :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1lol, we may never know unless they show up again ;)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0you jest with us mere mortals sir! :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1=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

Malachy
 3 years ago
Best ResponseYou've already chosen the best response.1Actually i was missed as 2^3*3^2*5*7*3^11 thats why that ans came

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1can you give a generalized rule for your method?
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