This is a Test of Mathematics Solution Subjective 188 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Consider the squares of an $ 8 X 8 $ chessboard filled with the numbers 1 to 64 as in the figure below. If we choose 8 squares with the property that there is exactly one from each row and exactly one from each column, and add up the numbers in the chosen squares, show that the sum always adds up to 260.
The problem asks us to choose numbers selectively such that they are from unique rows and columns. We write the numbers in the table in such a manner that helps us in our calculations. This is how it will be done:
Let the number in the $ i^{th} $ row and $ j^{th}$ column be $ x$. If we carefully observe the table we find an intuitive way of representing $ x$ as follows:
$ x = 8*(i-1) + j $ if $ x$ is the element in the $ i^{th} $ row and $ j^{th}$ column. Now all that is left to do is sum up all such numbers such that no two $ j^{t} $ or $ i^{th}$ value is same. There for the total sum is: $ (8*(i_1-1)+ j_1) + (8*(i_2-1) + j_2) + . . . + (8*(i_8-1) + j_8) $
that means Sum (S) = $ (8*(i_1 - 1 + i_2 - 1 + . . . + i_8 - 1)) + (j_1 + j_2 + . . . + j_8) $
$ => S = 8*(i_1 + i_2 + i_3 + ... + i_8 - 8) + (j_1 + j_2 + . . . + j_8)$
We know that as all the $ i$ and $ j$ values are different and range from 1 to 8 we can ascertain that
the sum of the $ i$ values $ = $ sum of $ j$ values = $ 1+2+3+...+8 = \frac{8*(8+1)}{2}$
Therefore S = $ 8*(\frac{8*(8+1)}{2} - 8) + \frac{8*(8+1)}{2} $ = $ 8*(36-8) +36 = 260 $
Hence the sum of all the values taken from unique rows and columns is 260.
Hence Proved.
This is a Test of Mathematics Solution Subjective 188 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Consider the squares of an $ 8 X 8 $ chessboard filled with the numbers 1 to 64 as in the figure below. If we choose 8 squares with the property that there is exactly one from each row and exactly one from each column, and add up the numbers in the chosen squares, show that the sum always adds up to 260.
The problem asks us to choose numbers selectively such that they are from unique rows and columns. We write the numbers in the table in such a manner that helps us in our calculations. This is how it will be done:
Let the number in the $ i^{th} $ row and $ j^{th}$ column be $ x$. If we carefully observe the table we find an intuitive way of representing $ x$ as follows:
$ x = 8*(i-1) + j $ if $ x$ is the element in the $ i^{th} $ row and $ j^{th}$ column. Now all that is left to do is sum up all such numbers such that no two $ j^{t} $ or $ i^{th}$ value is same. There for the total sum is: $ (8*(i_1-1)+ j_1) + (8*(i_2-1) + j_2) + . . . + (8*(i_8-1) + j_8) $
that means Sum (S) = $ (8*(i_1 - 1 + i_2 - 1 + . . . + i_8 - 1)) + (j_1 + j_2 + . . . + j_8) $
$ => S = 8*(i_1 + i_2 + i_3 + ... + i_8 - 8) + (j_1 + j_2 + . . . + j_8)$
We know that as all the $ i$ and $ j$ values are different and range from 1 to 8 we can ascertain that
the sum of the $ i$ values $ = $ sum of $ j$ values = $ 1+2+3+...+8 = \frac{8*(8+1)}{2}$
Therefore S = $ 8*(\frac{8*(8+1)}{2} - 8) + \frac{8*(8+1)}{2} $ = $ 8*(36-8) +36 = 260 $
Hence the sum of all the values taken from unique rows and columns is 260.
Hence Proved.
