harsh.champ wrote: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
(A)32
(B)28
(C)29
(D)30
(E)27
The ans. is D.
Nice question - where did you get it?
Notice that the first term (3) and the last term (467) have a sum of 470. So, set S can contain only 1 number from this pair.
Also notice that the second term (11) and the second to last term (459) have a sum of 470. So, set S can contain only 1 number from this pair.
And so the pattern continues.
So, how many pairs of numbers do we have? To answer this, we need to know how many numbers are in the sequence.
There is a formula for this kind of sequence where each term is derived by adding (or subtracting) a constant value to the previous term. The formula for finding the nth term is tn = a + d(n-1) where a is the first term and d is the constant value (difference).
In the sequence 3, 11, 19, 27, . . . ,451, 459, 467 the first term (a) is 3 and the constant value (d) is 8
So, the nth term will be: tn = 3 + 8(n-1)
We want to know the value of n such that 3 + 8(n-1) = 467 (the last term of the sequence).
Solving 3 + 8(n-1) = 467 for n, we get n=59. So, there are 59 terms in our sequence.
This means that we have 29 pairs of values that add to 470, and one lone number (235) that has no other number to pair up with.
We're now ready to count.
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 (30) numbers.
