Let T be the set of integers 3, 11, 19, 27, ... 451, 459, 467 and S be a subset of T such that the sum of no two elements of S is
470. In this case, the maximum possible number of elements in S will be how much?
Show Hint
When the condition is about sums being a constant, pair up elements that sum to that constant. The maximum subset is obtained by taking one from each pair, and including any element that pairs with itself.
Step 1: Understanding the Concept:
Identify the number of terms and identify pairs that sum to 470. From each pair, at most one can be in S. Step 2: Key Formula or Approach:
AP Term: \( T_n = a + (n-1)d \). Max subset size \( = \text{Lone elements} + \text{Count of pairs} \). Step 3: Detailed Explanation:
Set T is an AP: \( a=3, d=8 \).
Total terms: \( 467 = 3 + (n-1)8 \Rightarrow 464 = 8(n-1) \Rightarrow n-1 = 58 \Rightarrow n = 59 \).
Pairs summing to 470: \( 3+467=470, 11+459=470 \), etc.
There are \( 58/2 = 29 \) such pairs.
The middle term is \( (59+1)/2 = 30 \)th term. Value \( = 3 + 29 \cdot 8 = 235 \).
\( 235 + 235 = 470 \). To follow "sum of no two elements", we can pick one from each of the 29 pairs, plus the lone middle term 235.
Total \( = 29 + 1 = 30 \). Step 4: Final Answer:
Maximum elements = 30.