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amistre64
Group Title
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
amistre64 Group Title
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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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
thats not work, thats an answer .... teaches me nothing :/
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
I thought this was open study.
 one year ago

Malachy Group TitleBest 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
 one year ago

amistre64 Group TitleBest 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!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
you seem to have used something different than that i posted in the question
 one year ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
yup  I also get 21 using the same method as @amistre64
 one year ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
I have never heard of Malachys  oh well  another new topic to learn...
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
do you have a reference (a link) to this method?
 one year ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
my method? i thought it up
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
lol, we may never know unless they show up again ;)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
you jest with us mere mortals sir! :)
 one year ago

amistre64 Group TitleBest 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
 one year ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
can you give a generalized rule for your method?
 one year ago
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