Let the positive integers be written in the form:

1

   2 3

  4 5 6

7 8 9 10

If the $k^{t h}$ row contains exactly $k$ numbers for every natural number $k,$ then the row in which the number 5310 will be, is ______.     [2024]

Question

Let the positive integers be written in the form:

1

2 3

4 5 6

7 8 9 10

If the $k^{t h}$ row contains exactly $k$ numbers for every natural number $k,$ then the row in which the number 5310 will be, is ______. [2024]

Smart Achievers · Accepted Answer

(103)

First element of $k^{t h}$ row $= \frac{k (k - 1)}{2} + 1$

Last element of $k^{t h}$ row = $= \frac{k (k + 1)}{2}$

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

$\Rightarrow k (k - 1) + 2 \leq 10620 \leq k (k + 1)$

$\Rightarrow k^{2} - k + 2 \leq 10620 \leq k^{2} + k$

$\Rightarrow - k + 2 \leq 10620 - k^{2} \leq k \Rightarrow 2 \leq 10620 - k^{2} + k \leq 2 k$

$\Rightarrow 1 \leq 5310 - \frac{k^{2}}{2} + \frac{k}{2} \leq k \Rightarrow 5310 - \frac{k^{2}}{2} - \frac{k}{2} \leq 0$

$\Rightarrow k^{2} + k - 5310 \times 2 \geq 0 \Rightarrow k^{2} + k - 10620 \geq 0$

For $k = 100, k^{2} + k - 10620 = 10100 - 10620 < 0$

For $k = 103, k^{2} + k - 10620 = 10609 + 103 - 10620 > 0$

So, the value of $k = 103$

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