To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.

Understanding the Pattern

First row ($k = 1$): Contains 1 number.

Second row ($k = 2$): Contains 2 numbers.

Third row ($k = 3$): Contains 3 numbers.

…

$k^\text{th}$ row: Contains $k$ numbers.

Total Numbers Up to the $k^\text{th}$ Row

The total number of integers up to the $k^\text{th}$ row is given by the sum of the first $k$ natural numbers:

$ S(k) = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} $

Finding the Row Containing 5310

We need to find the smallest integer $k$ such that $S(k-1) < 5310 \leq S(k)$. This means:

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

Step 1: Estimate $k$

Let's approximate $k$ by solving the inequality:

$ \frac{k(k+1)}{2} \geq 5310 $

Multiply both sides by 2:

$ k(k+1) \geq 10620 $

This is a quadratic inequality. We can approximate $k$ by taking the square root:

$ k \approx \sqrt{10620} \approx 103 $

Step 2: Calculate $S(k)$ for $k = 102$ and $k = 103$

For $k = 102$:

$ S(102) = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253 $

For $k = 103$:

$ S(103) = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356 $

Step 3: Determine the Correct Row

Since $S(102) = 5253 < 5310 \leq 5356 = S(103)$, the number $5310$ lies between $5254$ and $5356$, inclusive. Therefore, it is in the $103^\text{rd}$ row.

Conclusion

The number $5310$ will be in the $103^\text{rd}$ row.