Let the positive integers be written in the form:

If the $$k^{th}$$ row contains exactly $$k$$ numbers for every natural number $$k$$, then the row in which the number 5310 will be, is ________

Notice the pattern of the last number in each row:

• Row 1 ends with: $$1$$
• Row 2 ends with: $$3$$ (which is $$1 + 2$$)
• Row 3 ends with: $$6$$ (which is $$1 + 2 + 3$$)
• Row 4 ends with: $$10$$ (which is $$1 + 2 + 3 + 4$$)

The last number in the $$k^{th}$$ row is simply the sum of the number of elements in the first $$k$$ rows. Using the standard summation formula for the first $$k$$ natural numbers, the last term of the $$k^{th}$$ row is:

$$L_k = \frac{k(k+1)}{2}$$

We need to find the row $$k$$ that contains the number $$5310$$. This means $$5310$$ must be strictly greater than the last number of the $$(k-1)^{th}$$ row and less than or equal to the last number of the $$k^{th}$$ row:

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

To find an approximate value for $$k$$, let's simplify the inequality to rough squares:

$$\frac{k^2}{2} \approx 5310 \implies k^2 \approx 10620$$

We know that $$100^2 = 10000$$, so $$k$$ must be slightly larger than $$100$$.

Let's test a couple of nearby values for the exact row endings:

• For $$k = 102$$: The last number of the 102nd row is $$\frac{102 \times 103}{2} = 51 \times 103 = 5253$$.
• For $$k = 103$$: The last number of the 103rd row is $$\frac{103 \times 104}{2} = 103 \times 52 = 5356$$.

Since $$5253 < 5310 \le 5356$$, the number $$5310$$ appears after the 102nd row has ended, placing it securely inside the 103rd row.

The row is $$103$$.