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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. The maximum possible number of elements in S is

Explanation:

The sum of the first and last terms in T is 470.

Likewise, the sum of the second and second-last terms is also 470.

In general the sum of the nth term from the beginning and the nth term from the end is 470.

∴ Only one of each of these pairs of terms will be in S. (For instance only one of 3 and 467 can be in S)

∴ The set S can have a maximum of half of the terms in T.

The terms in T are in A.P. with a common difference of 8.

Last Term = 467 = 3 + (n − 1) × 8

∴ n = 59                     

∴ Total number of terms in the set T = 59

∴ there are 29 pairs of numbers in T that add up to 470 and the 59th number is 235, which occurs in the middle of the series.

∴ S will be a set with 30 terms, with 29 terms which are from the pairs adding up to 470, and 235.

Hence, option (d).