Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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!! :)
 2 years ago
 2 years 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!! :)
 2 years ago
 2 years ago

This Question is Closed

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

hba Group TitleBest ResponseYou've already chosen the best response.0
I thought this was open study.
 2 years 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
 2 years 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!
 2 years 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
 2 years 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?
 2 years ago

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
my method? i thought it up
 2 years 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 :)
 2 years ago

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
can you give a generalized rule for your method?
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.