Prove how many of these numbers are divisible by 3? 5 integers written around circle

Five integers are written around a circle in such a way that no two or three consecutive numbers give a sum divisible by 3. Among those 5 numbers, how many are divisible by 3?

I don't know how to prove the answer for a general case

Ilikebugs said:

Five integers are written around a circle in such a way that no two or three consecutive numbers give a sum divisible by 3. Among those 5 numbers, how many are divisible by 3?

I don't know how to prove the answer for a general case
3?Five integers are written around a circle in such a way that no two or three consecutive numbers give a sum divisible by [FONT=MathJax_Main]33[/FONT]. Among those [FONT=MathJax_Main]55[/FONT] numbers, how many are divisible by [FONT=MathJax_Main]33[/FONT]?

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Nice problem. I would let the 1st integer be x and decide what the 2nd can (or can't ) be. The 2nd number must be in the form of 3y+1 or 3y + 2 for some integer y. Play with this for a bit and see where it gets you. Please reply back.

If x =0 mod 3, the next number and previous number has to be either 1 or 2 mod 3. Both have to be the same mod 3 as otherwise their sum will be a multiple of 3. If they are 1, the 3rd and 4th numbers have to be either 0 or 1 mod 3. Since 3 and 4 are next to eachother, they both cant be 0, and since they are in a triplet with a number thats 1 mod 3, they both cant be 1 so one is 0 and one is 1. So 2 numbers are 0 mod 3

Answer

Every number divided by 3 leaves a remainer of 1 or 2 or is actually dividable by 3. Let's call these 1, 2 and 3.

Because no 2 or 3 consecutive number together can have a sum of a multiple of 3, no threes can be next to eachother as it will lead to being a multiple. As there are 5 integers in a circle, a maximum of 2 threes can be achieved.

A 1 and 2 may never be together because they will add to 3 and three 1s will also add to 3. The same with 2 as they add to 6. This means that atleast 2 threes need to be used

3
1 1
1 3
Or

3
2 2
2 3

These are the only 2 combinations available both containing 2 multiples of three.

Hope this helped!