Question

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?

• A. 32
• B. 34
• C. 29
• D. 28
• E. 30

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.

Answer 1

Answer: (E)

Step 1: Identify the Arithmetic Progression:
T = 3, 11, 19, .... This is an AP with first term $a=3$ and common difference $d=8$. The last term is 467. Find the number of terms: $a_n = a + (n-1)d$. $467 = 3 + (n-1)8 \implies 464 = (n-1)8 \implies n-1 = 58 \implies n = 59$. So, T has 59 terms.
Step 2: Pair Elements that Sum to 470:
We need to avoid picking two numbers whose sum is 470. Let $x$ be an element. Its complement is $470 - x$. If $470-x$ is also in T, they form a forbidden pair. Check if $470-x$ is in T: $470-x$ must be of the form $3+8k$. $x = 3+8m$. Then $470 - (3+8m) = 467 - 8m = 3 + 8(58-m)$? Since $467 = 3 + 8 \times 58$. $467 - 8m = 3 + 8(58-m)$. Yes, it is of the form $3+8k$, so it is in T as long as $k$ is between 0 and 58. Thus, every element in T has a unique partner that sums to 470. The pairs are (3,467), (11,459), ..., up to the middle.
Step 3: Find the Middle:
The sum of the two numbers is 470. Their average is 235. Since the terms are spaced by 8, the two middle terms would be around 235. Find the term closest to 235: $3+8m \approx 235 \implies 8m=232 \implies m=29$. So term $x=3+8*29=3+232=235$. Its partner is also 235. So 235 is the only term that is exactly half of 470. Thus, 235 is the midpoint. The pair for 235 is itself.
Step 4: Form Pairs:
The 59 terms can be grouped into 29 pairs (each pair summing to 470) and one unpaired middle term (235). The pairs are: (3,467), (11,459), ..., (227,243) (since 3+8*28=227, 467-8*28=467-224=243), and the middle term is 235. So there are 29 pairs + 1 middle.
Step 5: Choose Maximum Subset:
From each pair, we can pick at most 1 element (to avoid sum 470). From the middle term (235), we can pick it because its complement is itself, so picking it doesn't create a sum of 470 with another distinct element (but if we pick it, we can't pick another 235, which is fin(e). Thus, maximum elements = 1 from each of the 29 pairs + the middle term = 29 + 1 = 30.
Step 6: Final Answer:
The maximum possible number of elements in S is 30.