Product of r consecutive numbers divisible by r!

roger

1*2*3...r is congruent to 0modr!

2*3*4...(r+1) is congruent to 0modr!

3*4*5...(r+1)(r+2) is congruent to 0modr! because the last 2 factors must contain 2.

So if I carry on in this fashion is this sufficient to prove that the product of r consecutive numbers is divisible by r!?

It seems like circular argument to me and so I'm not sure if it's justified.

morphism

There's a slick combinatorial proof. How many subsets of size r does a set with r+k elements have?

d_leet

Ummm... Why must the last 2 factors contain 2? If you mean the last two factors are divisible by 2, you do realize that not both of r+1, and r+2 can be divisible by 2 right?

What exactly is this fashion? What happens when your set of r consecutive integers no longer contains r?

matt grime

When you've finished verifying all of the infinitely many cases you will have proved the result. It could be a while before we hear from you again, though.

roger

could someone clarify what's wrong with it?

matt grime

1. You should have said 'one of the last two factors r+1 and r+2 is divisible by 2'

2. It is not clear that your idea actually works for 4*5*....*r+3, or any other example - you will need to make more of an argument when you start wishing to distribute lots of prime factors, and explain why there are enough powers of 2, say, later on.

3. Even if it did, you are attempting to check infinitely many things in turn. That is not a proof - you will never be able to verify all of the cases by hand as you assert you wish to.

cliowa

Personally, I would start off in a slightly different direction: Let's say you have your set of r consecutive numbers, [itex]a_1[/itex] to [itex]a_r[/itex]. Have a look at the first one, [itex]a_1[/itex].
1st case: Let's say [itex]a_1[/itex] is divisible by r. Then you're done.
2nd case: Let's say [itex]a_1[/itex] is not divisible by r. Then integer division will leave you with a rest between 1 and r-1, true? Now, what implication does this have?

cubzar

If the first of the r numbers is n, and n[itex]\equiv[/itex]m(mod(x)) where 0[itex]\leq[/itex]x[itex]\leq[/itex]r, then n+(x-m) is a multiple of x. Because x-m[itex]\leq[/itex]r at least one of the r numbers will be a multiple of x. r! is the product of all such x.