Q76. Let pp, qq and rr be 2 digit numbers where p < q < r. If pp + qq + rr =tt0, where tt0 is a 3-digit number ending with zero, consider the following statements:
1. The number of possible values of p is 5.
2. The number of possible values of q is 6
Which of the above statements is/are correct?

The numbers pp, qq, rr can be written as 11p, 11q, 11r. So, 11p + 11q + 11r = 11(p+q+r) = tt0. Since tt0 is a 3-digit number ending with zero and a multiple of 11, it must be 110 or 220 (as max sum of p,q,r is 7+8+9=24, so 11*24=264; min sum is 1+2+3=6, so 11*6=66).

Case 1: p+q+r = 10 (if tt0=110). Given p<q<r, and p,q,r are digits 1-9. Possible (p,q,r) triplets: (1,2,7), (1,3,6), (1,4,5), (2,3,5). Here, p can be 1 or 2 (2 values). q can be 2, 3, 4, 5 (3 distinct values: 2,3,4,5, but 5 is not possible for q as q<r, so q can be 2,3,4).

Case 2: p+q+r = 20 (if tt0=220). Given p<q<r. Possible (p,q,r) triplets: (3,8,9), (4,7,9), (5,6,9), (5,7,8). Here, p can be 3, 4, or 5 (3 values). q can be 6, 7, 8, 9 (3 distinct values: 6,7,8,9, but 9 is not possible for q as q<r, so q can be 6,7,8).

Statement 1: Possible values for p are {1, 2} (from Case 1) and {3, 4, 5} (from Case 2). Total = 2+3 = 5 values. Statement 1 is correct. Statement 2: Possible values for q are {2, 3, 4} (from Case 1) and {6, 7, 8} (from Case 2). Total = 3+3 = 6 values. Statement 2 is correct. Both statements are correct.