2002 AIME I Problems/Problem 9


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 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 $333+424 = \boxed{757}$.

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AIME Problems and Solutions

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