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

Solution 1

Note that it is impossible for any of $h,t,u$ to be $1$ , since then each picket will have been painted one time, and then some will be painted more than once.

$h$ cannot be $2$ , 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 fourth and fifth pickets to be painted, $t$ and $u$ must be $3$ as well. This configuration works, so $333$ is paintable.

If $h$ is $4$ , then $t$ must be even. 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 $4$ , in order for $5,6$ to be paint-able. Thus $424$ is paintable.

$h$ cannot be greater than $4$ , since if that were the case then the answer would be greater than $999$ , which would be impossible for the AIME.

Thus the sum of all paintable numbers is $\boxed{757}$ .

Solution 2

Again, note that $h,t,u \neq 1$ . The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . By the Chinese Remainder Theorem, the greatest common divisor of any pair of the three numbers $\{h,t,u\}$ cannot be $1$ (since otherwise without loss of generality consider $\text{gcd}\,(h,t) = 1$ ; then there will be a common solution $\pmod{h \times t}$ ).

Now for $4$ to be paint-able, we require either $h = 3$ or $t=2$ , but not both.

In the former condition, since $\text{gcd}\,(h,t),\ \text{gcd}\,(h,u) \neq 1$ , it follows that $3|t,u$ . For $5$ and $6$ to be paint-able, we require $t = u = 3$ , and it is easy to see that $333$ works.
In the latter condition, similarly we require that $2|h,u$ . All even numbers are painted. We now renumber the remaining odd pickets to become the set of all positive integers ( $1,3,5, \ldots \rightarrow 1',2',3', \ldots$ , where $n' = \frac{n+1}{2}$ ), which requires the transformation $h' = h/2,\ u' = u/2$ , and with the painting starting respectively at $1',2'$ . Our new number system retains the same conditions as above, except without $t$ . We thus need $\text{gcd}\, (h',u') \neq 1, h',u' \neq 1$ . Then for $3',4'$ to be painted, we require $h' = u' = 2$ . This translates to $424$ , which we see works.

Thus the answer is $333+424 = \boxed{757}$ .

Solution 3

The three conditions state that no picket number $n$ may satisfy any two of the conditions: $n \equiv 1 \pmod{h},\ n \equiv 2 \pmod{t},\ n \equiv 3 \pmod{u}$ . Note that the smallest number, $min \{ h,t,u \},$ divides the other $2$ , and the next smallest divide the largest number, otherwise there is a common solution by the Chinese Remainder Theorem. It is also a necessary condition so that it paints exactly once. Note that the smallest number can't be at least $5$ , otherwise not all picket will be painted. We are left with few cases (we could also exclude $1$ as the possibility) which could be done quickly. Thus the answer is $333+424 = \boxed{757}$ .

Solution 4

The wording of this problem is a bit dissatisfying and could have been improved by stating that there are at least h + t + u + 6 pickets instead of "very long". For example, a very long fence could have 5 pickets, h = 78, t = 47, u = 1. To compensate for this we should lean on the fact that the answer cannot exceed 999.

Clearly we have to accept that there are 4 pickets or more, and the 4th picket is not reached from setting u = 1. If the 4th picket is reached from h = 3, then the 5th picket will not be reached via u = 2, or else we'd be unable to have 7 pickets. So h = 3 => u = 3 => t = 3, and 333 is a solution. We now have h > 3 for any remaining cases, and thus we only have room for 1 more case.

Since the 4th picket must henceforth be reached via t = 2, we either have the 5th picket reached via u = 2 (requiring h to be any number exceeding the number of pickets, and we're stipulating that this is not allowed), or we have the 5th picket reached via h = 4, from which it follows that u = 4. So we have accumulated the solutions 333 and 424 and there really is not room for further investigation.

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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