Let the positive integers be written in the form :
1; 2, 3; 4, 5, 6; 7, 8, 9, 10; ..., ⋱
If the kᵗʰ row contains exactly k numbers for every natural number k, then the row in which the number 5310 will be, is __________.

Solution

Maths sir · 4 steps

Do not count rows one by one: row k ends at the triangular number k(k+1)/2.

1. Where rows end. Rows 1 to k hold 1 + 2 + … + k numbers in total. So row k ends at Tₖ = k(k+1)/2. Check: the rows end at 1, 3, 6, 10, which matches the pattern.

   (k-1)k/2 < 5310 ≤ k(k+1)/2

2. Estimate k. Set k(k+1)/2 ≈ 5310, so k² + k ≈ 10620. That gives k ≈ 102.6, so the answer is either row 102 or row 103.
3. Check both rows. Row 102 ends at (102 × 103)/2 = 5253. Row 103 ends at (103 × 104)/2 = 5356. So row 103 runs from 5254 to 5356.
4. So. Since 5253 < 5310 ≤ 5356, the number 5310 sits in row 103.

The idea to keep. In a triangle of consecutive numbers, each row ends at a triangular number. Trap the given number between two consecutive triangular numbers, and the larger one tells you the row.

Answer from the NTA final answer key. Solution by PrepWiser.