A Clock in Three Pieces

A clock numbered 1 through 12 clockwise falls and breaks into three pieces. The sums of the numbers on the three pieces are all equal. What are the numbers on each piece, given that no strange-shaped piece is allowed, so every piece is a single unbroken arc of the dial?

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#Equal sums force 26

The numbers total $1 + 2 + \cdots + 12 = 78$ . Three equal pieces means each piece sums to $78 / 3 = 26$ . The divisibility check passes cleanly, which is exactly what makes the puzzle a trap.

#Only two arcs reach 26

An unbroken piece is a run of consecutive numbers around the rim. Scanning every such run, only two of them total 26,

11 + 12 + 1 + 2 = 26, \qquad 5 + 6 + 7 + 8 = 26. \tag{1}

Every other arc overshoots or undershoots, so these are the only legal pieces worth 26.

Equal sums force 26 on each piece, and the only unbroken arcs that reach 26 are eleven-twelve-one-two and five-six-seven-eight. Whatever is left, three-four and nine-ten, lands on opposite sides of the dial, so the third piece can never be a single unbroken arc. With no strange shapes allowed, no such break exists.

#The leftover cannot be whole

Those two arcs are disjoint, and removing both leaves $\{3, 4, 9, 10\}$ , which also sums to 26. But 3 and 4 sit on one side of the dial while 9 and 10 sit on the far side, so this set is two separate clumps rather than one arc. As a single piece it would be strange-shaped, which the rules forbid. Hence no break into three unbroken equal pieces exists. Divisibility is necessary but not sufficient, and here the geometry refuses what the arithmetic permits.