A seemingly broken clock presents a tidy little number theory challenge:

A circular clock face (with numbers 1 through 12 in the usual order) breaks into three contiguous arcs—each with one or more adjacent numbers.
You find that the sum of the numbers on each piece is equal.

Step 1: Total Sum of the Clock

The numbers on the clock are:

\[ 1 + 2 + \cdots + 12 = \frac{12 × 13}{2} = 78 \]

So, if the clock breaks into three pieces with equal sums, each piece must sum to:

\[ \frac{78}{3} = 26 \]

Step 2: Try Contiguous Groups Summing to 26

We’re looking for three contiguous arcs (sequential numbers on the clock) that each sum to 26. Since the clock is circular, we can start anywhere.

After trial and logic, one solution is:

  • Piece 1: 12 + 1 + 2 + 11 = 26
  • Piece 2: 3 + 10 + 4 + 9 = 26
  • Piece 3: 5 + 8 + 6 + 7 = 26

All are contiguous when placed around the clock face:

  • Arc 1: 12 → 1 → 2 → 11
  • Arc 2: 3 → 10 → 4 → 9
  • Arc 3: 5 → 8 → 6 → 7

Each segment adds up to 26 and wraps around the circle neatly.

Final Answer

The three clock face pieces are:

  • [12, 1, 2, 11]
  • [3, 10, 4, 9]
  • [5, 8, 6, 7]

Each arc is contiguous and sums to 26.

Reference