JEE Main 2024 — 8 April, Shift 1 — Question 25

The positive integers are written in rows such that the $k^{\text{th}}$ row contains exactly $k$ numbers.

Step 1: Last number of the $k^{\text{th}}$ row

The total numbers up to the $k^{\text{th}}$ row are

$$T_k = 1+2+3+\cdots+k = \frac{k(k+1)}{2}.$$

Thus, the last number in the $k^{\text{th}}$ row is $\dfrac{k(k+1)}{2}$.

Step 2: Locate the row containing 5310

The number 5310 lies in the $k^{\text{th}}$ row if

$$\frac{k(k-1)}{2} < 5310 \le \frac{k(k+1)}{2}.$$

To estimate $k$, solve

$$\frac{k(k+1)}{2} \approx 5310.$$ $$k^2 + k - 10620 = 0.$$

Step 3: Solve the quadratic equation

$$k = \frac{-1 + \sqrt{1 + 4 \cdot 10620}}{2} = \frac{-1 + \sqrt{42481}}{2} \approx 102.5.$$

Step 4: Check nearby integers

$$\frac{102 \cdot 103}{2} = 5253, \qquad \frac{103 \cdot 104}{2} = 5356.$$

Since

$$5253 < 5310 \le 5356,$$

the number $5310$ lies in the $103^{\text{rd}}$ row.

$$\boxed{103}$$