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?
Answer & explanation
Answer: (c) Both 1 and 2
pp + qq + rr = 11(p + q + r) and tt0 = 110t, so p + q + r must be 10 or 20. The triples with p < q < r are (1,2,7), (1,3,6), (1,4,5), (2,3,5), (3,8,9), (4,7,9), (5,6,9) and (5,7,8), giving 5 possible values of p and 6 of q.
- pp = 11p, so pp + qq + rr = 11(p + q + r). Also tt0 = 100t + 10t = 110t.
- 11(p + q + r) = 110t gives p + q + r = 10t. With digits 1 to 9 and p < q < r, the sum is at most 7 + 8 + 9 = 24, so it is 10 or 20.
- Sum 10: (1,2,7), (1,3,6), (1,4,5), (2,3,5). Sum 20: (3,8,9), (4,7,9), (5,6,9), (5,7,8).
- Values of p: 1, 2, 3, 4, 5 → 5. Values of q: 2, 3, 4, 6, 7, 8 → 6.
- Both statements are correct.
- ✓ 1. p can be 1, 2, 3, 4 or 5 — five values.
- ✓ 2. q can be 2, 3, 4, 6, 7 or 8 — six values.
Remember · Repeated-digit numbers factor neatly: aa = 11a and aa0 = 110a. Turn the puzzle into a simple sum condition, then list cases.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·