Let the positive integers be written in the form : If the k^ row contains exactly k number…

Question

Let the positive integers be written in the form :

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

Answer: 103

Accepted Answer

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 ${k}^{}$ 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.

\ldots

${k}^{}$ row: Contains $k$ numbers.

Total Numbers Up to the ${k}^{}$ Row
The total number of integers up to the ${k}^{}$ row is given by the sum of the first $k$ natural numbers:

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

Finding the Row Containing 5310
We need to find the smallest integer $k$ such that $S(k-1)$\frac{(k-1)k}{2} Estimate $k$
Let's approximate $k$ by solving the inequality:

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

Multiply both sides by 2:

$k(k+1)\ge 10620$

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

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

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

For $k=102$:

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

For $k=103$:

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

Determine the Correct Row: Since $S(102)=5253Conclusion
The number $5310$ will be in the ${103}^{}$ row.

Solutions · 1