# Strange Proof Question

## Homework Statement

The integers 1 thru 40 are painted on the outside
rim of a circular disk, in “random” order. Prove that some consecutive
set of three integers must have a sum that is greater than 61. Hint: this

n/a

## The Attempt at a Solution

I have no idea where to even start on this one.

I'll get you on the way:

Let the elements on the disk be numbered $$\{x_1,...,x_{40}\}$$. For conveniece we also put $$x_{41}=x_1,~x_{42}=x_2,...$$. Assume that for every i:
$$x_i+x_{i+1}+x_{i+2}\leq 61$$

Now try to evaluate the following sum in two ways (one way is the inequality above):

$$\sum_{i=1}^{40}{x_i+x_{i+1}+x_{i+2}}$$

I don't know how I would calculate that sum though, considering we don't know which integers are next to each other... or would i take the case of the three integers are all at least about 20?... I don't really follow.

Ok, another hint:

$$\sum_{i=1}^{40}{(x_i+x_{i+1}+x_{i+2})}=\sum_{i=1}^{40}{x_i}+\sum_{i=1}^{40}{x_{i+1}}+\sum_{i=1}^{40}{x_{i+2}}$$

The evaluation of these three sums does not depend on the order of the 40 elements...

Thank you, completely missed that, so in this case the individual sums are all equal, and the sum is 820 for each individual sum, so 2460 in total.

and I also notice, when I divide this total sum by 40 (the number of possible "i" s, I get 61.5), which is greater than 61)

Yes. So now you need to use the inequality $$x_i+x_{i+1}+x_{i+2}\leq 61$$ on the sum. This will provide an upper bound of this sum and will hopefully yield a contradiction.

so I'm dividing the total by 40, because really I just added up the sum of 120 randomly selected numbers (3 of each number), so by dividing by 40, I get the sum of 3 such randomly selected numbers, and this yields 61.5, which is a contradiction because we assumed it was less than or equal to 61?

Yes, that's good to.

The way I saw it was:

$$2460=\sum_{i=1}^{40}{x_i+x_{i+1}+x_{i+2}}\leq \sum_{i=1}^{40}{61}=40.61=2400$$

Thank you very much for the help. The only thing is I don't see how you got the 40.61 value?
Thanks again.

The sum $$\sum_{i=1}^{40}{61}$$ means (by definition)

$$(61+61+61+...+61)$$

where you take the sum 40 times. Thus this equals 40.61.

In general (and I suggest you remember this, as it will come in handy), if you summate a constant, then

$$\sum_{i=1}^n{c}=c.n$$

Oh 40.61 = 40X61. Ok I followed that, I just didn't see how you got 40.61. That clears everything up. Thanks again.

Ow, Im sorry for the confusion. I should have used other notation