Question

Let the positive integers be written in the form :

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

Answer 1

Correct Answer: 103

1. Each row in the pattern contains numbers equal to the row number. The sequence follows the triangular numbers series.
2. The sum of the first n natural numbers is given by the formula: \( T_n = \frac{n(n+1)}{2} \).
3. To find the row where the number 5310 is located, determine the smallest n such that: \( T_n \geq 5310 \).
4. Starting from the inequality:
   \[ \frac{n(n+1)}{2} \geq 5310 \]
5. Multiply by 2 to clear the fraction: \( n(n+1) \geq 10620 \).
6. Rearrange into a quadratic inequality: \( n^2 + n - 10620 \geq 0 \).
7. Using the quadratic formula:
   \[ n = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ where } a=1, b=1, c=-10620 \]
8. Calculate discriminant: \( b^2 - 4ac = 1 + 42480 = 42481 \).
9. \( \sqrt{42481} \approx 206.12 \).
10. Substitute back:
    \[ n = \frac{-1 \pm 206.12}{2} \]
11. Take the positive root:
    \[ n \approx \frac{205.12}{2} = 102.56 \]
12. Round up to the next integer since \( n \) must be a whole number: \( n = 103 \).
13. Verify: Calculate \( T_{103} = \frac{103 \times 104}{2} = 5356 \), and \( T_{102} = \frac{102 \times 103}{2} = 5253 \).
14. Since \( 5253 < 5310 \leq 5356 \), the number 5310 is in the 103rd row.
15. Confirm the row number falls within the expected range: 103.

Answer 2

The total number of elements in the first n rows is:
\[ S = 1 + 2 + 3 + \dots + T_n = \frac{n(n+1)}{2}. \]
To find the row containing 5310, solve:
\[ \frac{n(n+1)}{2} = 5310. \]
Start testing values:
\[ n = 100, \quad T_n = \frac{100 \cdot 101}{2} = 5050. \]
\[ n = 101, \quad T_n = \frac{101 \cdot 102}{2} = 5151. \]
\[ n = 102, \quad T_n = \frac{102 \cdot 103}{2} = 5253. \]
\[ n = 103, \quad T_n = \frac{103 \cdot 104}{2} = 5356. \]
Since 5310 lies between 5253 and 5356, it is in the 103rd row.
Final Answer: 103.