Difference between revisions of "1986 AIME Problems/Problem 12"
m (→Solution) |
m (→Solution) |
||
| (2 intermediate revisions by 2 users not shown) | |||
| Line 4: | Line 4: | ||
By using the greedy algorithm, we obtain <math>\boxed{061}</math>, with <math>S=\{ 15,14,13,11,8\}</math>. We must now prove that no such set has sum greater than 61. Suppose such a set <math>S</math> existed. Then <math>S</math> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>. | By using the greedy algorithm, we obtain <math>\boxed{061}</math>, with <math>S=\{ 15,14,13,11,8\}</math>. We must now prove that no such set has sum greater than 61. Suppose such a set <math>S</math> existed. Then <math>S</math> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>. | ||
| − | <math>S</math> can't have more than 5 elements. To see why, note that at least <math>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element and so on), and each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum. | + | <math>S</math> can't have more than 5 elements. To see why, note that at least <math>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element and so on), and each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum. If those subsets were disjoint, we would directly arrive at a contradiction; if not, we could remove the common elements to get two disjoint subsets. |
| − | |||
Thus, <math>S</math> would have to have 5 elements. <math>S</math> contains both 15 and 14 (otherwise its sum is at most <math>10+11+12+13+15=61</math>). It follows that <math>S</math> cannot contain both <math>a</math> and <math>a-1</math> for any <math>a\leq 13</math>, or the subsets <math>\{a,14\}</math> and <math>\{a-1,15\}</math> would have the same sum. So now <math>S</math> must contain 13 (otherwise its sum is at most <math>15+14+12+10+8=59</math>), and <math>S</math> cannot contain 12, or the subsets <math>\{12,15\}</math> and <math>\{13,14\}</math> would have the same sum. | Thus, <math>S</math> would have to have 5 elements. <math>S</math> contains both 15 and 14 (otherwise its sum is at most <math>10+11+12+13+15=61</math>). It follows that <math>S</math> cannot contain both <math>a</math> and <math>a-1</math> for any <math>a\leq 13</math>, or the subsets <math>\{a,14\}</math> and <math>\{a-1,15\}</math> would have the same sum. So now <math>S</math> must contain 13 (otherwise its sum is at most <math>15+14+12+10+8=59</math>), and <math>S</math> cannot contain 12, or the subsets <math>\{12,15\}</math> and <math>\{13,14\}</math> would have the same sum. | ||
Latest revision as of 18:41, 1 September 2020
Problem
Let the sum of a set of numbers be the sum of its elements. Let 
be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of have the same sum. What is the largest sum a set with these properties can have?
Solution
By using the greedy algorithm, we obtain 
, with 
. We must now prove that no such set has sum greater than 61. Suppose such a set existed. Then must have more than 4 elements, otherwise its sum would be at most 
.
can't have more than 5 elements. To see why, note that at least 
of its subsets have at most four elements (the number of subsets with no elements plus the number of subsets with one element and so on), and each of them have sum at most 54. By the Pigeonhole Principle, two of these subsets would have the same sum. If those subsets were disjoint, we would directly arrive at a contradiction; if not, we could remove the common elements to get two disjoint subsets.
Thus, would have to have 5 elements. contains both 15 and 14 (otherwise its sum is at most 
). It follows that cannot contain both 
and 
for any 
, or the subsets 
and 
would have the same sum. So now must contain 13 (otherwise its sum is at most 
), and cannot contain 12, or the subsets 
and 
would have the same sum.
Now the only way could have sum at least 
would be if 
. But the subsets 
and 
have the same sum, so this set does not work. Therefore no with sum greater than 61 is possible and 61 is indeed the maximum.
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
