Question

Harold, Tanya, and Ulysses paint a very long picket fence.

• Harold starts with the first picket and paints every $h$ th picket;

• Tanya starts with the second picket and paints every $t$ th picket; and

• Ulysses starts with the third picket and paints every $u$ th picket.

Call the positive integer $100h + 10t + u$ paintable when the triple $(h, t, u)$ of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.

Answer

Solution:

The three progressions $\{1, 1 + h, \ldots\},$ $\{2, 2 + t, \ldots\},$ $\{3, 3 + u, \ldots\}$ must partition the positive integers. If $h = 2,$ Harold paints picket $3,$ which Ulysses also paints, so $h \ge 3.$ If $h \ge 5,$ consider picket $4:$ Harold's next picket is $1 + h \ge 6,$ and Ulysses cannot paint it (that would need $u = 1,$ repainting everything from $3$ on), so Tanya must, forcing $t = 2.$ Then picket $5$ is unpainted unless $u = 2,$ but then Tanya and Ulysses together cover every picket from $2$ on, and Harold's picket $1 + h$ is painted twice. So $h = 3$ or $h = 4.$

If $h = 3,$ Harold paints $1, 4, 7, \ldots.$ Ulysses cannot paint picket $5$ (then $u = 2$ and he would repaint $7$ ), so Tanya does: $t = 3,$ covering $2, 5, 8, \ldots.$ What remains is exactly $3, 6, 9, \ldots,$ so $u = 3,$ giving $333.$ If $h = 4,$ Harold paints $1, 5, 9, \ldots;$ picket $4$ again forces $t = 2,$ and the leftover pickets $3, 7, 11, \ldots$ force $u = 4,$ giving $424.$

The sum of the paintable integers is $333 + 424 = 757.$

2002 AIME I Problem 9

Solution:

Problem 9 in Other Years