2002 AIME I Problems/Problem 9

Problem

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.

Solution

$h$ cannot be 1 or 2, or that will result in painting the third picket twice. If $h=3$ , then $t$ may not equal anything not divisible by 3, and the same for $u$ . Now for every picket to be painted, $t$ and $u$ must be 3 as well. So $333$ is paintable.

If $h$ is 4, then $t$ can't be 1 or 3 mod 4, but can be 2 or 0 mod 4. The same for $u$ , except that it can't be 2 mod 4. Thus u is 0 mod 4 and t is 2 mod 4. Since this is all mod 4, t must be 2 and u must be 4. Thus 424 is paintable.

There are no other paintable numbers (proof required), so the sum of all paintable numbers is 757.

Template:Incomplete

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2002 AIME I Problems/Problem 9

Problem

Solution

See also