Mr. A and Mr. B are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1, 2, 3, .., 160. Mr. A chose to do it row wise, so that first row is numbered 1, 2, 3, ..., 16 (in order), then second row is numbered 17, 18, ..., 32 and so on. Mr. B chose to do it column wise so that first column is numbered 1, 2, 3, ..., 10 (in order), then second column is numbered 11, 12, 13,..., 20 and so on. Let N be the sum of numbers which occupy same position in both arrangements, find the value of n/7
Answers
Given: Mr. A and Mr. B are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1, 2, 3, .., 160. Mr. A chose to do it row wise, so that first row is numbered 1, 2, 3, ..., 16 (in order), then second row is numbered 17, 18, ..., 32 and so on. Mr. B chose to do it column wise so that first column is numbered 1, 2, 3, ..., 10 (in order), then second column is numbered 11, 12, 13,..., 20 and so on.
Let N be the sum of numbers which occupy same position in both arrangements
To Find : , find the value of n/7
Solution:
Number which occupy the same positions are
1 , 54 , 107 , 160
Sum of numbers = 1 + 54 + 107 + 160 = 322
322/7 = 46
value of n/7 = 46
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