Solve for n: Positive Integer Sequence with Greatest Divisor and LCM

[quddusaliquddus]

for what values of n (where n is >=2) is there a sequnce of integers-positive so that the greatest num in the set divides the LCM of all the numbers left?

 

[matt grime]

Suppose n is prime, is there anyway of finding numbers less than n whose product is divisible by n? If n is not prime could you find some constraint on its prime factors?

 

[quddusaliquddus]

questions...answers...and questions...

Question 1) I am not sure, but whatever number <= n e.g. M, multiplied by n e.g. M*n should give you a number. Any number divisable by n (prime) would have to have n as a factor wouldn't it? Does this mean you can't have numbers <=n when n is prime whose products are divisable by n?
Question 2) I'll have a look at that.

 

[matt grime]

Well, you initial question starts off by mentioning n, and then doesn't use n at all in the second part when you set up the problem, so it's a little vague. And what sequences are allowed?

I took it to mean: suppose S is some set of numbers between 1 and n (not inclusive of n) wthout repetition. When can the lcm of the elements of S be divisible by n.

If n is prime, there is no subset of 1...n whose lcm is divisible by n, as there is no subset whose product is divisible by n, and the lcm divides the product.

Try it for some examples of n, and please clear up what you mean.

Examples,
n=6 S={2,3} the lcm is divisible by 6
n=9 there is no subset whose lcm is divisible by 9.

 

[quddusaliquddus]

ok.......

 

[matt grime]

But do you admit that you say: let n be an integer greater than 2, and then go on to ask a question that is independent of n? Could you rewrite the question?

 

[quddusaliquddus]

there's nothing to admit other than i might've done mistakes

 

[matt grime]

Sorry, just trying to get you to say if n is tha largest number in the sequence or if it is the number of terms in the sequence. I couldn't decide what "ok..." meant.

 

[quddusaliquddus]

i tried to type "ok" but it didint lemme...i typed some fullstops to satisfy the 12-letter rule (spaces also don't work)

 

[quddusaliquddus]

Answer...

I don't know if you guys want to see the answer. But i think I've got it...bty...i didnt do it. Someone else helped me. if anyone's interested in seeing the answer...lemme know. I ask cos its a little long.

 

[matt grime]

You're stilll not to my mind cleared up the question - is n the number of terms in the sequence or the largest term in the sequence? The answer is relatively straight forward in either case.

 

[quddusaliquddus]

I wish i could say its easy...lol. n is the number of numbers.

 

[matt grime]

But then unless you define the concept of 'sequence' oddly the answer is all n>2:

let p_i i =1,...n-1 be a set of n-1 distinct primes (this forms a sequence), then p_1, p_2...P_{n-1},q will do where q is the product of p_1...p_{n-1}

 

[quddusaliquddus]

This is an answer i got somewhere else. Thought you might like to look at it:

" suppose n=p*q is composite, p,q>1, gcd(p,q)=1. then consider S={1,2,...,n}

then p,q belongs to S
hence p|LCM{1,2,...,n-1}
q|LCM{1,2,...,n-1}

since gcd(p,q)=1, then n=p*q|LCM{1,2,...,n-1}

suppose n=p^k, where p is odd prime
by Catalan's Theorem, then only two powers (greater than 1) differ by 1 is 2^3+1=3^2
and we have p^k+1 is divisible by 2 hence non prime
for the case p=2 k=3, we take {3,4,5,6,7,8,9,10}
otherwise
S={2,...,p^k,p^k+1} has p^(2k+1)+1 factors to u,v such that u*v=1 and we are done.

still think about the case where n=p... and n=2^k "

 

[matt grime]

So it looks as though you mean n conescutive numbers then.

 

[quddusaliquddus]

yeah.sorry.