Size of a Special Subset

The set $\{ 50, 51, 52, \ldots, 100 \}$ satisfies the condition, and has $51$ elements.

Let $A = \{ a_1, a_2, \ldots, a_n \}$ , where $a_i$ are arranged in increasing order and $A$ satisfies the given conditions. Consider the set $B$ consisting of the $n-1$ differences $a_2 - a_1, a_3 - a_1, \ldots a_n - a_1$ . If $a_j-a_1$ is in $A$ , then $a_j - a_1 = a_k$ for some $k< j$ (since the $a_i$ are arranged in increasing order). If $k\neq 1$ , then $a_j= a_k + a_1$ , which would contradict the assumption. Clearly, at most one of the elements of $B$ can be equal to $a_1$ , since they are all distinct. Hence, at most one of these differences is in A.

Since all the numbers that are in $B$ are also in $T$ , we have $100 =|T| \geq | A \cup B| \geq |A| + (n - 1) - 1 = 2n-2$ . Hence $n \leq 51$ , so $A$ can contain at most $51$ elements.

Let $T_n = \{1, 2, \dots, n\}$ . $A_n$ is any subset of $T_n$ subject to the condition as in the problem. The inductive hypothesis is

$\max |A_n| \leq \begin{cases} (n+1)/2 & \mbox{ if n is odd} \\ n/2+1 & \mbox{ if n is even}\\ \end{cases}$

The base case when $n=1$ is trivial. Now we prove the inductive hypothesis. It is easy to see that $A_n- \{n\}$ is a possible $A_{n-1}$ . Hence if $n$ is even, $|A_n| \leq |A_{n-1}|+1 \leq n/2 + 1$ . If $n$ is odd, if $A_n$ does not contain $n$ , then $A_n$ is a kind of $A_{n-1}$ and its cardinality $<= (n+1)/2$ . If $A_n$ contains $n$ , then $A_n-\{n\}$ does not have 2 distinct elements with sum $= n$ . Hence it contain at most one element from each of the set {1,n-1}, {2, n-2}, \dots, {(n-1)/2,(n+1)/2}. Hence $A_n$ contains at most $(n+1)/2$ elements.

Logically, the subset of T, A, should consist of large number in such a way that a+b is more than 100, which is the largest number in set T

So the maximum number of |A| is when A={50,51,52,...,100} which means that the sum of any two numbers from A will result more than 100. so maximum number of |A| is 51.

Looks like 51 to me. If 49 is in A, 50,51 cannot be in A If 48 is in A, 50,51,52 cannot be in A. if n<50 is in A, we must exclude 51-n elements from A.

So, including 50,51,...100 in A, we get 51 elements.

Looking for the subset of T where all values of a+b are not in A. Specifically, a+b < 0 or a+b > 100. Since T is all positive integers 1-100, a+b can never be less than 0. Hence, a+b > 100, and the average value of (a+b) > 50. The minimum average must be 50.5, so the lowest two values of a and b must be 50 and 51 respectively. All values of T between 50 and 100 satisfy the condition that a+b > 100.

Since elements 1-49 are NOT in A, the number of elements in A is: 100-49 = 51.

If (a+b) is not to be a part of set A then the sum should be greater than the greatest number possible in A. And if A is required to have maximum no. elements then its elements should be larger so that their sum would exceed the greatest no.. keeping all this in mind largest possible set A will have elements 50 to 100.

sum of any two numbers from {50,51,52........,100 } is more than 100 does not belong to the set.so that set is the maximum set containing 51 elements.

I claim the answer is $51$ . Here is a construction that validates this. If $49\in A$ we must have $50, 51\notin A$ . If $48\in A$ we must have $50, 51, 52\notin A$ . If for any $X<50$ we have $X\in A$ , there must be $51-X$ elements $a_1, a_2,\ldots, a_{51-X}\notin A$ . Thus, if we have $50, 51,\ldots,100\in A$ this gives a total of $51$ elements.

If A ={50, 51, 52..., 100}, then the sum of any 2 distinct elements a and b cannot equal to another element of A as the minimum value of a + b =50+51=101. Here, | A |=51 and the first element of A is 50. If the first element of A $leq$ 49,