But you found your own mistake, which is why maths is the loveliest of all the -athematics.

Evolving wrote:The trouble with Thommo's proof A and Scott's proof is that it doesn't work if the multiple of 3 is even (or, equivalently, if the two odd numbers are primes).

I look at is as factors. Basically, for a number to be divisible by 24, it must contain the factors 2, 3, and 4 (2, 2). For this to occur as the result of a product of 4 numbers, The set of factors for each number must contribute at least 2, 2, 2, and 3.

This will occur in any sequence 4 consecutive positive numbers. In every such set, there will be at least 2 even numbers, both divisible by 2 and thus providing at least one of the 2 factors each. Every other even number is a multiple of 4, and thus will provide the factor of 2 twice, thereby contributing all needed 2 factors. Every third number is a multiple of 3. It is possible that one of the even numbers may be that multiple of 3, in which case it will provide the factors 2, 3, or 2, 2, 3, and the other even number will provide the factors 2, 2, or 2. If that even number is not the number containing a factor of 3, another number in the set must be a multiple of 3 since every third number will be and thus provide that factor of 3.

Thommo wrote:

Calilasseia wrote:I'm tempted to suggest that the above problem is a special case for a more general theorem. Viz:

Every sequence of k consecutive integers, when multiplied together, produces another integer that is divisible by (k!).

How to prove this conjecture is currently exercising my mind somewhat.

Same ways as given before. The induction is easy, for example.

Base case: 1 x 2 x ... x k = k! by definition.

Induction step: Assume true for n:

n+1 x n+2 x ... x n+k = k! x c

Then n+2 x n+3 x ... x n+k x n+k+1 = k! x c x(n+k+1)/(n+1) = k(k! x c) + k! x c which is divisible by k!.

Which means that binomial coefficients are always integers; after all,



Edit: Thommo, I think there's a mistake in your proof:
k! x c x(n+k+1)/(n+1) = k! x c + (k! x c)k/(n+1)

Pulsar wrote:Edit: Thommo, I think there's a mistake in your proof:
k! x c x(n+k+1)/(n+1) = k! x c + (k! x c)k/(n+1)

Yes, there is. I spotted that just before I went out earlier.

I think you need instead to observe that for all factors i <= k if im divides (n+1) then im+1 divides (n+i+1) instead, which is considerably less elegant.

Thanks for pointing out the error.

Thommo wrote:

Pulsar wrote:Edit: Thommo, I think there's a mistake in your proof:
k! x c x(n+k+1)/(n+1) = k! x c + (k! x c)k/(n+1)

Yes, there is. I spotted that just before I went out earlier.

I think you need instead to observe that for all factors i <= k if im divides (n+1) then im+1 divides (n+i+1) instead, which is considerably less elegant.

Thanks for pointing out the error.

One could use Pascal's rule



combined with the boundary values



from which it follows that all binomial coefficients are integers.

I'm not sure how we'd use that? We know they are all integers as the numbers are products of integer factors, it's trivial that dividing by one of those factors, namely (n+1) produces an integer, no?

Thommo wrote:I'm not sure how we'd use that? We know they are all integers as the numbers are products of integer factors, it's trivial that dividing by one of those factors, namely (n+1) produces an integer, no?

I'm just pointing out that the fact that all binomial coefficients are integers immediately proves Cali's hypothesis. After all, a binomial coefficient C(n,k) is a product of k consecutive numbers divided by k!.

Ahhh, I get it now, thanks.

I'm not sure it does though - all binomial coefficients C(n,k) are products of k or more consecutive numbers divided by k!

Thommo wrote:Ahhh, I get it now, thanks.

I'm not sure it does though - all binomial coefficients C(n,k) are products of k or more consecutive numbers divided by k!

There are exactly k consecutive numbers in the numerator.

A lot of combinatorics questions are exceptionally hard in my opinion. This isn't one of them though.

Pulsar wrote:

Thommo wrote:Ahhh, I get it now, thanks.

I'm not sure it does though - all binomial coefficients C(n,k) are products of k or more consecutive numbers divided by k!

There are exactly k consecutive numbers in the numerator.

I'm sorry, I must be having a bad day. You're right.

newolder wrote:@spike, The question as asked has two solutions (roots) for n. 10 and -9 are those roots but since -9 is not a physical number of sweets that one can hold in a bag, the solution n=10 is taken as physical.

What about antimatter sweets?

Sadegh wrote:

newolder wrote:@spike, The question as asked has two solutions (roots) for n. 10 and -9 are those roots but since -9 is not a physical number of sweets that one can hold in a bag, the solution n=10 is taken as physical.

What about antimatter sweets?

No thanks. (They would annihilate the bag.)

Maybe it's a special magnetic bag.

How does -9 sweets correspond to a collection of magnetic antimatter sweets?

newolder wrote:How does -9 sweets correspond to a collection of magnetic antimatter sweets?

The bag would have the magnets in it.

If you "add" them to a collection of 9 ordinary matter sweets, you'll have 0.

Sounds a bit contrived and nothing to do with the original question. Whatever floats your biscuit.

newolder wrote:Sounds a bit contrived

It's humor.

lol