Key Concepts and Formulas

• Identity: For any two real numbers $a$ and $b$, $\min\{a,b\} + \max\{a,b\} = a+b$.
• Properties of Double Summation:
  • Linearity: $\sum_{i} \sum_{j} (x_{ij} + y_{ij}) = \sum_{i} \sum_{j} x_{ij} + \sum_{i} \sum_{j} y_{ij}$
  • Constant Factor: $\sum_{i} \sum_{j} c \cdot x_{ij} = c \cdot \sum_{i} \sum_{j} x_{ij}$
  • Summation of a constant: $\sum_{k=1}^{n} c = nc$
• Sum of First $n$ Natural Numbers: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Step-by-Step Solution

We are given two double summations: $A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } }$ $B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } }$ We need to find the value of $A+B$.

Step 1: Combine the summations and apply the key identity. Since both $A$ and $B$ are double summations over the same ranges for $i$ and $j$, we can combine them using the linearity property of summation. $A + B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } } + \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } }$ $A + B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\left( \min \,\{ i,j\} + \max \,\{ i,j\} \right)} }$ Now, we apply the fundamental identity $\min\{a,b\} + \max\{a,b\} = a+b$. For each term in the summation, $a=i$ and $b=j$. $\min\{i,j\} + \max\{i,j\} = i+j$ Substituting this into the summation: $A + B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {(i+j)} }$ Reasoning: Combining the sums and applying the identity $\min\{a,b\} + \max\{a,b\} = a+b$ simplifies the problem significantly by replacing the conditional min/max functions with a simple arithmetic expression.

Step 2: Split the combined summation. We can further decompose the summation of $(i+j)$ into two separate summations using the linearity property. $A + B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {i} } + \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {j} }$ Reasoning: Separating the terms $i$ and $j$ allows us to evaluate each double summation independently, making the calculation more manageable.

Step 3: Evaluate the first double summation: $\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {i} }$. We first evaluate the inner summation with respect to $j$. Here, $i$ is treated as a constant. $\sum\limits_{j = 1}^{10} {i} = i \cdot \sum\limits_{j = 1}^{10} {1} = i \cdot 10$ Now, we substitute this result back into the outer summation with respect to $i$: $\sum\limits_{i = 1}^{10} {(10i)} = 10 \cdot \sum\limits_{i = 1}^{10} {i}$ Using the formula for the sum of the first $n$ natural numbers, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, with $n=10$: $\sum\limits_{i = 1}^{10} {i} = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55$ Therefore, the value of the first double summation is: $10 \cdot 55 = 550$ Reasoning: This step involves standard summation techniques. The inner sum treats $i$ as a constant, and the outer sum uses the known formula for the sum of an arithmetic progression.

Step 4: Evaluate the second double summation: $\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {j} }$. We first evaluate the inner summation with respect to $j$. This is the sum of the first 10 natural numbers: $\sum\limits_{j = 1}^{10} {j} = \frac{10(10+1)}{2} = 55$ Now, we substitute this result back into the outer summation with respect to $i$. Here, $55$ is treated as a constant. $\sum\limits_{i = 1}^{10} {55} = 55 \cdot \sum\limits_{i = 1}^{10} {1} = 55 \cdot 10 = 550$ Reasoning: Similar to Step 3, this involves evaluating a sum of constants and using the formula for the sum of the first $n$ natural numbers. Notice the symmetry: the result is identical to the first double summation, as expected since the ranges for $i$ and $j$ are the same.

Step 5: Calculate the final sum $A+B$. We add the results from Step 3 and Step 4. $A + B = 550 + 550 = 1100$ Reasoning: This is the final step where we combine the evaluated parts to obtain the total sum $A+B$.

Common Mistakes & Tips

• Direct Calculation: Avoid attempting to calculate each $\min\{i,j\}$ and $\max\{i,j\}$ individually for all 100 pairs. This is extremely time-consuming and error-prone. The identity $\min\{a,b\} + \max\{a,b\} = a+b$ is designed to prevent this.
• Summation Index Confusion: Be careful to correctly identify the summation index in nested summations. When summing over $j$, treat $i$ as a constant, and vice-versa.
• Recognizing Symmetry: The problem exhibits symmetry in the summation ranges for $i$ and $j$. This often implies that parts of the calculation will yield identical results, which can be a good check for your work.

Summary

The problem involves calculating the sum of two double summations containing $\min$ and $\max$ functions. The key to solving this problem efficiently is the algebraic identity $\min\{a,b\} + \max\{a,b\} = a+b$. By applying this identity, we transform the expression into a sum of simple arithmetic terms. Combining the summations and then splitting them into sums of $i$ and sums of $j$, we evaluate each part using the formula for the sum of the first $n$ natural numbers. The total sum $A+B$ is found to be $1100$.

The final answer is $\boxed{1100}$.