Key Concepts and Formulas

• Arithmetic Series Sum: The sum of the first $k$ natural numbers is given by the formula $S_k = \sum_{i=1}^{k} i = \frac{k(k+1)}{2}$. This represents the total count of numbers up to and including the $k^{\text{th}}$ row.
• Range of Numbers in a Row: If $S_{k-1}$ is the total count of numbers up to row $k-1$, and $S_k$ is the total count of numbers up to row $k$, then row $k$ contains the integers from $S_{k-1} + 1$ to $S_k$.
• Inequality for Row Determination: A number $N$ will be in row $k$ if $S_{k-1} < N \leq S_k$.

Step-by-Step Solution

Step 1: Understand the Pattern and the Problem Statement The problem describes a triangular arrangement of positive integers where the $k^{\text{th}}$ row contains exactly $k$ numbers. Row 1: 1 (1 number) Row 2: 2, 3 (2 numbers) Row 3: 4, 5, 6 (3 numbers) And so on. We need to find the row number where the integer 5310 is located.

Step 2: Relate the Number of Elements to the Row Number The total number of integers up to the end of the $k^{\text{th}}$ row is the sum of the number of elements in each row from 1 to $k$. This sum is given by the formula for the sum of the first $k$ natural numbers: $S_k = \frac{k(k+1)}{2}$ If a number $N$ is in the $k^{\text{th}}$ row, it means that the total count of numbers up to the end of the $(k-1)^{\text{th}}$ row is less than $N$, and the total count of numbers up to the end of the $k^{\text{th}}$ row is greater than or equal to $N$. Mathematically, this can be expressed as: $S_{k-1} < N \leq S_k$

Step 3: Set up the Inequality for the Given Number We are given the number $N = 5310$. We need to find the row number $k$ such that: $S_{k-1} < 5310 \leq S_k$ Substituting the formula for $S_k$: $\frac{(k-1)k}{2} < 5310 \leq \frac{k(k+1)}{2}$

Step 4: Estimate the Value of k To find an approximate value for $k$, we can focus on the right-hand side of the inequality: $5310 \leq \frac{k(k+1)}{2}$ Multiply both sides by 2: $10620 \leq k(k+1)$ Since $k$ and $k+1$ are consecutive integers, $k(k+1)$ is approximately $k^2$. So, we can estimate $k$ by taking the square root of 10620: $k^2 \approx 10620$ $k \approx \sqrt{10620}$ Let's calculate the square root: $\sqrt{10620} \approx 103.053$ This estimation suggests that the row number $k$ is likely to be around 103.

Step 5: Test Values of k Around the Estimate Since $k$ must be an integer, we will test the integer values around 103, specifically $k=102$ and $k=103$, to see which one satisfies our inequality.

First, let's calculate $S_{102}$: $S_{102} = \frac{102(102+1)}{2} = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253$ This means that the $102^{\text{nd}}$ row ends with the number 5253.

Next, let's calculate $S_{103}$: $S_{103} = \frac{103(103+1)}{2} = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356$ This means that the $103^{\text{rd}}$ row ends with the number 5356.

Step 6: Verify the Inequality and Determine the Row Now we check if our number 5310 falls within the range defined by $S_{102}$ and $S_{103}$: We need to check if $S_{102} < 5310 \leq S_{103}$. Substituting the calculated values: $5253 < 5310 \leq 5356$ This inequality is true. The number 5310 is greater than the last number of the $102^{\text{nd}}$ row (5253) and less than or equal to the last number of the $103^{\text{rd}}$ row (5356). Therefore, 5310 must be in the $103^{\text{rd}}$ row.

Common Mistakes & Tips

• Approximation Accuracy: While $\sqrt{10620} \approx 103.05$ is a good estimate, always verify by calculating the exact sums for the integers around the estimate. Don't just round the square root to the nearest integer without checking.
• Understanding $S_k$: Remember that $S_k$ is the cumulative count of numbers up to the end of row $k$. The numbers in row $k$ are from $S_{k-1} + 1$ to $S_k$.
• Inequality Direction: Ensure the inequality $S_{k-1} < N \leq S_k$ is set up correctly. The number $N$ is included in row $k$ if it is equal to $S_k$.

Summary The problem involves arranging positive integers in rows such that the $k^{\text{th}}$ row has $k$ numbers. The total count of numbers up to the end of the $k^{\text{th}}$ row is given by the sum of the first $k$ natural numbers, $S_k = \frac{k(k+1)}{2}$. To find the row containing the number 5310, we set up the inequality $S_{k-1} < 5310 \leq S_k$. By estimating $k$ using $k^2 \approx 2 \times 5310$, we found $k \approx 103$. Testing $k=102$ and $k=103$, we calculated $S_{102} = 5253$ and $S_{103} = 5356$. Since $5253 < 5310 \leq 5356$, the number 5310 belongs to the $103^{\text{rd}}$ row.

The final answer is $\boxed{103}$.