If we place the numbers 1,2,3,...,10 in a circle (as in a clock) in any order, then at least three consecutive numbers in our circle will sum to 16 or higher. Present an argument on why this is true; don’t quote specific examples as proof, since they won’t cover all 9!/2 = 181400 possible configurations. (Hint: One direction to approach might be by contradiction. Assume there is a configuration where no three consecutive numbers sum to 16 or higher; e.g. The maximum any three consecutive numbers in this configuration could sum to would be 15. Hence the sum of all triples of consecutive numbers can’t exceed a certain number. Then reach a contradiction by arguing that the sum of all triples of consecutive numbers must be larger than what is allowed.)

I understand how a normal proof by contradiction works but this has made me very confused. Thank you!

  • $\begingroup$ Your title is misleading. You aren't trying to prove that "3 consecutive numbers from $1$ to $10$ can add up over 16"; that's obvious ($10+9+8$ is more than 16). The point is that the integers from $1$ to $10$ are arranged randomly, "in any order," in a circle, and yet still there must be three consecutive (in the sense of adjacent) numbers that sum to at least $16$. That is what you're trying to prove. $\endgroup$ – symplectomorphic Nov 14 '17 at 22:29
  • $\begingroup$ @symplectomorphic noted and edited title. Thank you $\endgroup$ – Mario Liden Nov 14 '17 at 22:34

Consider your arrangement. There are $10$ triples of consecutive numbers so if each sum were at most $16$ then the sum of all ten sums would be at most $160$. On the other hand, in that sum each of the ten digits is counted exactly three times. Thus the sum of the ten sums has to equal $$3\times (1+\cdots + 10)=3\times 55=165>160$$ A contradiction.

  • $\begingroup$ thank you, my issue was realizing that the numbers were in a random order. Originally I was looking at 10+9+8 which is obviously over 16 so I was having difficulty thinking about what to prove. $\endgroup$ – Mario Liden Nov 14 '17 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.