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QJEEM22Ma7f8c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma7f8c Solution: Let and . Then A + B is equal to _____________.

Let A=∑i=110∑j=110min⁡ {i,j}A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\min \,\{ i,j\} } } and B=∑i=110∑j=110max⁡ {i,j}B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \,\{ i,j\} } } . Then A + B is equal to _____________.

Step-by-Step Verified Solution

∑i=110∑j=110{i, j}\sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\{ i,\,j\} } }

= {1, 1} {1, 2} {1, 3} ..... {1, 10}

{2, 1} {2, 2} {2, 3} ..... {2, 10}

{3, 1} {3, 2} {3, 3} ..... {3, 10}

⋮\vdots

{10, 1} {10, 2} {10, 3} ..... {10, 10}

Now, A=∑i=110∑j=110min{i, j}A = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {min\{ i,\,j\} } }

= minimum between i and j in all sets and summation of all those values.

and B=∑i=110∑j=110max⁡{i, j}B = \sum\limits_{i = 1}^{10} {\sum\limits_{j = 1}^{10} {\max \{ i,\,j\} } }

= maximum between i and j in all sets and summation of all those values.

For 1 :

1 is minimum in sets =

{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {1, 9}, {1, 10}, {2, 1}, {3, 1}, {4, 1}, {5, 1}, {6, 1}, {7, 1}, {8, 1}, {9, 1}, {10, 1}

∴\therefore 1 is minimum in 19 sets

1 is maximum in {1, 1} sets.

∴\therefore 1 is maximum and minimum in total 20 sets.

∴\therefore Sum of 1 in all those sets = 1 ×\times 20 = 20

For 2 :

2 is minimum in sets =

{2, 2}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {2, 9}, {2, 10}, {3, 2}, {4, 2}, {5, 2}, {6, 2}, {7, 2}, {8, 2}, {9, 2}, {10, 2}

∴\therefore 2 is minimum in 17 sets

2 is maximum in sets = {1, 2}, {2, 1}, {2, 2}

∴\therefore 2 is maximum and minimum in 20 sets.

∴\therefore Sum of 2 in all those sets = 2 ×\times 20 = 40

Similarly 3 is maximum and minimum in 20 sets.

∴\therefore Sum of 3 in all those sets = 20 ×\times 3 = 60

⋮\vdots

Similarly, 10 is maximum and minimum in 20 sets.

∴\therefore Sum of 10 in all those sets = 20 ×\times 10 = 200

∴\therefore A + B = 20 + 20 ×\times 2 + 20 ×\times 3 + ....... + 20 ×\times 10

= 20(1 + 2 + 3 + ...... + 10)

= 20 ×\times 10×112{{10 \times 11} \over 2}

= 1100

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