3.2 Problem 2: The Clock Puzzle

Loud thinking: “If one side must have the same sum than the other, then it is unlikely that all large numbers are on one side and all small numbers are on the other side. A mixture is needed. What about putting the largest and the smallest together? 12 + 1 = 13. The same sum is given by 11+2 = 13. Well, all 12 numbers can be paired to give 13. There are 6 pairings. Put three of them on one side and the other three on the other side!”

   

PIC

To prove that this is the unique solution, we must show that there exists no other consecutive sequence of numbers on the clock face that sums to the half of all numbers. Let’s choose two unknown numbers x and y such that 1 x < y 11. The sum of all numbers on the face from y + 1 to x in clockwise direction is the sum from y + 1 to 12 plus the sum from 1 to x. This is: the sum of all numbers from 1 to 12 minus the sum of all numbers from 1 to y plus the sum of all numbers from 1 to x:

(                  )
  12 ⋅-13-- y(y-+-1)  + x(x-+-1)-
    2          2           2

This quantity must be half of all numbers from 1 to 12, that is, it must be 12⋅13
  4. Simplifying the expression leads to:

(y - x) ⋅ (y + x + 1) = 6 ⋅ 13

As y - x < y + x + 1, we conclude that y + x + 1 = 13 and y - x = 6. Resolving for x and y gives: y = 9 and x = 3. Hence the unique solution is a consecutive sequence from 10 to 3 in clockwise direction.