Difference between revisions of "2002 AIME I Problems/Problem 9"

(cheap proof of no remaining paintable integers)
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If <math>h</math> is 4, then <math>t</math> can't be 1 or 3 mod 4, but can be 2 or 0 mod 4. The same for <math>u</math>, 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.
 
If <math>h</math> is 4, then <math>t</math> can't be 1 or 3 mod 4, but can be 2 or 0 mod 4. The same for <math>u</math>, 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.
 
  
 
Since answers on the [[AIME]] can't be greater that <math>999</math> and the sum of the paintable integers found so far is <math>333+424=757</math>, any remaining paintable integer must be less than or equal to <math>999-757=242</math>. However, this implies that <math>h \le 2</math> in any remaining paintable integers that exist, which results in the third picket being painted twice. Thus, there are no other paintable numbers, so the sum of all paintable numbers is <math>\boxed{757}</math>.
 
Since answers on the [[AIME]] can't be greater that <math>999</math> and the sum of the paintable integers found so far is <math>333+424=757</math>, any remaining paintable integer must be less than or equal to <math>999-757=242</math>. However, this implies that <math>h \le 2</math> in any remaining paintable integers that exist, which results in the third picket being painted twice. Thus, there are no other paintable numbers, so the sum of all paintable numbers is <math>\boxed{757}</math>.

Revision as of 14:05, 10 July 2008

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.

Since answers on the AIME can't be greater that $999$ and the sum of the paintable integers found so far is $333+424=757$, any remaining paintable integer must be less than or equal to $999-757=242$. However, this implies that $h \le 2$ in any remaining paintable integers that exist, which results in the third picket being painted twice. Thus, there are no other paintable numbers, so the sum of all paintable numbers is $\boxed{757}$.

Template:Incomplete

See also

2002 AIME I (ProblemsAnswer KeyResources)
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