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	Choose the correct option:
A) 32
B) 28
C) 29
D) 30
E) None of these
Explanation :   At first, let's find the number fo integers in the given series.   =&gt; Tn = a + (n - 1) d.   =&gt; 467 = 3 + (n - 1) 8.   =&gt; 464/8 = n - 1.   =&gt; 58 + 1 = n. i.e n = 59.   If we closely observe the given set 3 + 467 = 11 + 459 = .... = 470.   The total number of elements in the given set = 59.   This means that we have 59/2 i.e 29 pairs of values that add to 470, and one lone number (235) that has no other number to pair up with.   So, We have 29 prohibited number pairs: (3 and 467), (11 and 459), (19 and 451), . . .   We can take 1 number from each of these pairs, so we can have 29 numbers in set S.   PLUS we can have the lone number 235.   So, set S can have a maximum of 29+1 i.e 30 numbers.   Answer : 30
<result>D</result>
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