Difference between revisions of "2002 AIME I Problems/Problem 9"
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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. | ||
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| + | 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>. | ||
{{incomplete|solution}} | {{incomplete|solution}} | ||
Revision as of 13:04, 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
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.
Since answers on the AIME can't be greater that 
and the sum of the paintable integers found so far is 
, any remaining paintable integer must be less than or equal to 
. However, this implies that 
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 
.
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 | ||
