Let A = sum_{i=1}^{10} sum_{j=1}^{10} min {i, | vedclass

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(D) We know that for any two numbers $i$ and $j$,$\min \{i, j\} + \max \{i, j\} = i + j$.Therefore,$A + B = \sum_{i=1}^{10} \sum_{j=1}^{10} (\min \{i,

Vedclass · Accepted Answer

$1100$

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(D) We know that for any two numbers $i$ and $j$,$\min \{i, j\} + \max \{i, j\} = i + j$.Therefore,$A + B = \sum_{i=1}^{10} \sum_{j=1}^{10} (\min \{i, j\} + \max \{i, j\}) = \sum_{i=1}^{10} \sum_{j=1}^{10} (i + j)$.Expanding the sum: $A + B = \sum_{i=1}^{10} (\sum_{j=1}^{10} i + \sum_{j=1}^{10} j) = \sum_{i=1}^{10} (10i + \frac{10 	imes 11}{2}) = \sum_{i=1}^{10} (10i + 55)$.$A + B = 10 \sum_{i=1}^{10} i + \sum_{i=1}^{10} 55 = 10 	imes \frac{10 	imes 11}{2} + 10 	imes 55$.$A + B = 10 	imes 55 + 550 = 550 + 550 = 1100$.

Let $A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$ and $B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$. Then $A + B$ is equal to

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