The number of terms of the series forms the sum of first n natural numbers i.e. n(n + 1)/2. Thus the first 23 letters will account for the first (23 x 24)/2 = 276 terms of the series. The 288th term will be the 24th letter viz. x.

Consider the square root of 1000. The largest prime less than this is 31. So the set of all prime numbers less than or equal to 31 i.e., { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31} must be represented in set A. The above set itself cannot be set A, as the members of set A are non-prime. Therefore, each member of set A must represent exactly one of these prime numbers as, set A is such that maximum number of elements. If, any member of the set is a combination of 2 or more primes from the above set, then the number of elements in set A will get reduced as the members of this set should be co-prime to each other. Set A could be {22, 32, 52, 112, 132, 172, 192, 232, 292, 312,} It can be noted that there are more possibilities for set A. A number of set A could t, a product of exactly one prime number from the original set and another prime number which is not from the original set. If none of the prime numbers is used on more than one occasion then the set of numbers will be co-primes and hence satisfy the coordination for set A. Each member must represent one prime number from the original set. So There are exactly 11 elements.

∠6 = 720 and this is 3 digited; because it is given that ∠A + ∠B + ∠C = ABC, value of one of A or B or C shall have to be 7; and then the same will have more 3 – digits. Hence ∠6 is out of consideration.

    ∠5 = 120, and is 3 digited. This could be one of the numbers.

    ∠4 = 24; and when this is added to ∠5, the value is 144 which is 3 digited.

    As the sum is 144, ∠1 shall be taken into account. 

    Then ∠1 + ∠4 + ∠5 = 1 + 24 + 120 = 145 and the given condition is satisfied. A + B + C = 1 + 4 + 5 = 10.

The minimum value will occur when a = b = c = d = 1.

The new product must be a multiple of 53. Only one choice fulfills this requirement.