Difference between revisions of "2002 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | == Problem == |
| + | Harold, Tanya, and Ulysses paint a very long picket fence. | ||
| + | |||
| + | Harold starts with the first picket and paints every <math>h</math> th picket; | ||
| + | |||
| + | Tanya starts with the second picket and paints every <math>t</math> th picket; and | ||
| + | |||
| + | Ulysses starts with the third picket and paints every <math>u</math> th picket. | ||
| + | |||
| + | Call the positive integer <math>100h+10t+u</math> paintable when the triple <math>(h,t,u)</math> of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers. | ||
| + | |||
| + | == Solution == | ||
| + | <math>h</math> cannot be 1 or 2, or that will result in painting the third picket twice. If <math>h=3</math>, then <math>t</math> may not equal anything not divisible by 3, and the same for <math>u</math>. Now for every picket to be painted, <math>t</math> and <math>u</math> must be 3 as well. So <math>333</math> 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. | ||
| + | |||
| + | There are no other paintable numbers (proof required), so the sum of all paintable numbers is 757. | ||
| + | |||
| + | {incomplete|solution} | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2002|n=I|num-b=8|num-a=10}} | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=8|num-a=10}} | {{AIME box|year=2002|n=I|num-b=8|num-a=10}} | ||
Revision as of 11:37, 10 July 2008
Problem
Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every 
th picket;
Tanya starts with the second picket and paints every 
th picket; and
Ulysses starts with the third picket and paints every 
th picket.
Call the positive integer 
paintable when the triple 
of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
Solution
Problem
Harold, Tanya, and Ulysses paint a very long picket fence.
Harold starts with the first picket and paints every th picket;
Tanya starts with the second picket and paints every th picket; and
Ulysses starts with the third picket and paints every th picket.
Call the positive integer paintable when the triple of positive integers results in every picket being painted exactly once. Find the sum of all the paintable integers.
Solution
cannot be 1 or 2, or that will result in painting the third picket twice. If 
, then may not equal anything not divisible by 3, and the same for . Now for every picket to be painted, and must be 3 as well. So 
is paintable.
If is 4, then can't be 1 or 3 mod 4, but can be 2 or 0 mod 4. The same for , 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.
{incomplete|solution}
See also
| 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 | ||
See also
| 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 | ||
