Mathematics *** Numbers Theory?

Question

Hi There, Can anyone solve this problem? you can also email me your answers. [e-mail]
Any comments is extremely appreciated.
Regards,
Qaviliom.


Why for m greater than 7 we have:

2^m=< LCM(1,2,3,4,...,m)=< e^m

Please prove that for m>=7 we have:

2^m=< LCM(1,2,3,...,m) =< e^m

Answer 1

Let p_n be the nth prime number. Let a_n(m) be the largest power of p_n which is less than or equal to m. Then lcm(1,2,3,...,m)
=2^a_1(m)*3^a_2(m)*5^a_3(m)*
...*b(m)^1
where b(m) is the largest prime not greater than m, so
lcm(1,2,3,...,m)<=m^c(m)
where c(m) is the number of primes not greater than m.
ln lcm(1,2,3,...,m)<=c(m)*ln(m)
(ln lcm(1,2,3,...,m))/m<=
c(m)/(m/ln(m))
By the Prime Number Theorem,
lim(m->oo)((ln lcm(1,2,3,...,m))/m)<=1
hence
lim(m->oo)
lcm(1,2,3,...,m)^(1/m)
<=e
thus for any small positive delta, for sufficiently large m depending on delta we must have
lcm(1,2,3,...,m)*(1/m) and thus
lcm(1,2,3,...,m)<=(e+delta)^m for all sufficiently large m.
This is not exactly the result you asked for, but I hope it helps.

Answer 2

good proof ksoileau
Remark that the right inequality is stronger then the problem asked some days ago :
http://answers.yahoo.com/question/index;_ylt=Aspr1sgRNFZ2Pe9BDozFRm_ty6IX;_ylv=3?qid=20071006175914AA4pVHE&show=7#profile-info-9ea2e1cd35ec4eb78e3ab1b7124c047caa

That is because LCM(1,2,3,...,m) is bigger than the product of primes less than m and of course e^m<4^m. So it was to be expected that a strong theorem like Prime Number Theorem to be applied.

As for left hand, that is rather simple:
LCM(1,2,..,m) = LCM(LCM(1,2,...,m-1),m) =
= m*LCM(1,2,...,m-1)/GCD(m,LCM(1,2,...,m-1))

Now we prove by induction: for m=7 verify directly
Assume m-1 is true.
Then m/GCD(m,LCM(1,2,...m-1)) is at least 2. Because it cannot be 1, m should have one prime divisor in common with LCM(1,2,..,m-1)).
By induction hypothesis:m*LCM(1,2,...,m-1)/GCD(m,LCM(1,2,...,m-1)) >=2^(m-1)*2 =2^m

edit: actually i did an error:m/GCD(m,LCM(1,2,...m-1)) can be 1, it is not one exactly when m is prime

Answer 3

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Answer 4

because m to 7 is smaller but greater to lcm in brackets 2-4 so there for e is not equal to m so m is overall calculated to 7