Let n=$2^3*3^2*5*7^3*11$

Find the smallest factorial: k! that is divisible by n

Show your work!! :)

Question

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

anonymous · Answer

27!

amistre64 · Answer

thats not work, thats an answer .... teaches me nothing :/

hba · Answer

I thought this was open study.

anonymous · Answer

=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 · Answer

------------------
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 · Answer

you seem to have used something different than that i posted in the question

amistre64 · Answer

does your method, using my posted information, equate to 27 as well?  or is it 21?

asnaseer · Answer

yup - I also get 21 using the same method as @amistre64

amistre64 · Answer

im pretty adept at brute mathing; i see some potential in Malachys if i can get it verified ;)

asnaseer · Answer

I have never heard of Malachys - oh well - another new topic to learn...

asnaseer · Answer

do you have a reference (a link) to this method?

amistre64 · Answer

i see in mine the 7^3 produces the biggest parts; and 7*3 = 21

amistre64 · Answer

my method?  i thought it up

asnaseer · Answer

true - but still have to show that all numbers before this are factored out as well - which you did using brute force :)

asnaseer · Answer

you said " i see some potential in Malachys" <<<--- is this made up? :)

amistre64 · Answer

lol, we may never know unless they show up again ;)

asnaseer · Answer

you jest with us mere mortals sir! :)

amistre64 · Answer

=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

anonymous · Answer

Actually i was missed as 2^3*3^2*5*7*3^11
thats why that ans came

amistre64 · Answer

can you give a generalized rule for your method?