isi-entrance 2015 QB2

isi-entrance · India · UGB Proof Direct Proof of a Stated Identity or Equality
Consider the squares of an $8 \times 8$ 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 obtained is always 260.
12345678
910111213141516
1718192021222324
2526272829303132
3334353637383940
4142434445464748
4950515253545556
5758596061626364
Let the chosen entries be in the positions $\left( i , a _ { i } \right) , 1 \leq i \leq 8$. Thus $a _ { 1 } , \ldots , a _ { 8 }$ is a permutation of $\{ 1 , \ldots , 8 \}$. The entry in the square corresponding to $( i , j )$th place is $i + 8 ( j - 1 )$. Hence the required sum is $\sum _ { i = 1 } ^ { 8 } \left( i + 8 \left( a _ { j } - 1 \right) \right)$. 
Consider the squares of an $8 \times 8$ 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 obtained is always 260.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\
\hline
17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\
\hline
25 & 26 & 27 & 28 & 29 & 30 & 31 & 32 \\
\hline
33 & 34 & 35 & 36 & 37 & 38 & 39 & 40 \\
\hline
41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 \\
\hline
49 & 50 & 51 & 52 & 53 & 54 & 55 & 56 \\
\hline
57 & 58 & 59 & 60 & 61 & 62 & 63 & 64 \\
\hline
\end{tabular}
\end{center}
Paper Questions
QB1 QB10 QB11 QB2 QB3 QB4 QB5 QB6 QB7 QB8 QB9 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28