The maximum value of A \left| A \right|

Let A A be a set of positive integers not exceeding 24. Find the maximum value of A \left| A \right| so that A A have sums of all subsets different.

Example: If A = { 2 , 3 , 4 } A=\{2,3,4\} , then the possible sums of all subsets (including empty set) are 0 , 2 , 3 , 4 , 5 , 6 , 7 , 9 0,2,3,4,5,6,7,9 , which are all different. In this case A = 3 \left| A \right| =3 .

3 5 4 8 7 6

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1 solution

Chan Lye Lee
Jul 24, 2018

Suppose the maximum value of A = n 7 \left| A \right|=n \ge 7 . From these n n elements, we consider the sums of subsets at most 4 elements. There are at least ( n 1 ) + ( n 2 ) + ( n 3 ) + ( n 4 ) ( 7 1 ) + ( 7 2 ) + ( 7 3 ) + ( 7 4 ) = 98 {n \choose 1}+{n \choose 2}+{n \choose 3}+{n \choose 4} \ge {7 \choose 1}+{7 \choose 2}+{7 \choose 3}+{7 \choose 4}=98 such subsets. On the other hand, the maximum sum of the four distinct integers not exceeding 24 is 24 + 23 + 22 + 21 = 90 < 98 24+23+22+21=90<98 . Thus there are at least two of the subsets have the same sum. Hence the maximum value of A = n 6 \left| A \right|=n \le 6 .

The answer is 6 and one such possible set A A is { 11 , 17 , 20 , 22 , 23 , 24 } \{11,17,20,22,23,24\} .

There is a result in combinatorial number theory. I did not remember the result properly.i was in a hurry. It has nothing to with this problem. The result is if we have a set of n elements. Then it has a sum free subset of size at least n/3.

Srikanth Tupurani - 2 years, 8 months ago

Nice solution.

Srikanth Tupurani - 2 years, 8 months ago

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