# Numbers on a circle

• Apr 25th 2011, 07:18 PM
boldbrandywine
Numbers on a circle
Let S = {1,2,...12,}. Suppose the elements of S are scattered at random around a circle. Show that there exists some string of three consecutive numbers whose sum is at least 20.

My attempt: I know that 1+...+12 = 78, and there are 12 ways to add three consecutive numbers around the circle, the smallest three adding to 6 (1+2+3), and the highest adding to 33 (10+11+12). Note however the circle doesn't have to be "nice" in the sense of a clock. So we could have 7+12+2 = 21 and so on. I'm just stumped on where to continue from here.
• Apr 25th 2011, 08:06 PM
TKHunny
Seems ideal for a pigeon hole problem. There just aren't that many buckets to put things in.

First idea. Start with 12. There aren't too many ways to keep 12 from going over.

1, 12, 2
1, 12, 3
1, 12, 4
1, 12, 5
1, 12, 6
2, 12, 3
2, 12, 4
2, 12, 5
3, 12, 4

That's it. Any other arrangement and we just have to look at the 12 and we're done.

For each of those cases, what can you put next? It's already at least 15, so there are not too many choices.
• Apr 25th 2011, 08:14 PM
boldbrandywine
Why are you only looking at the 12 though? I'm not understanding why that becomes important. And all these sums are less than 20. Why wouldn't we look at the ones that are 20?
• Apr 25th 2011, 08:14 PM
abhishekkgp
Quote:

Originally Posted by boldbrandywine
Let S = {1,2,...12,}. Suppose the elements of S are scattered at random around a circle. Show that there exists some string of three consecutive numbers whose sum is at least 20.

My attempt: I know that 1+...+12 = 78, and there are 12 ways to add three consecutive numbers around the circle, the smallest three adding to 6 (1+2+3), and the highest adding to 33 (10+11+12). Note however the circle doesn't have to be "nice" in the sense of a clock. So we could have 7+12+2 = 21 and so on. I'm just stumped on where to continue from here.

let a_1, a_2, ..., a_{12} be the numbers.
there are 12 possible strings of 3 consecutive numbers, viz, s_1=a_1+a_2+a_3, s_2=a_2+a_3+a_4, s_3= ...
assume each of these have magnitude less than 20. sum all these 12 numbers up.
then, s_1+s_2+ ... + s_12 <=12x19=228
note that s_1+s_2+... + s_12 = 3(a_1+a_2+...+ a_12)= 3x6x13=234
228<234 and hence....
• Apr 25th 2011, 08:20 PM
TKHunny
I'm not ONLY looking at 12. That's just where I started.

In this case, I just felt like starting with 12. It seemed like a good place to start. No matter where you start, eventually you'll have to place the 12. Why not look there first?

Certainly, there are other approaches.
• Apr 26th 2011, 12:35 AM
haedious
I would try to find how many combinations can satisfy the opposite of this question: sum of three consecutive numbers does not exceed 20.