+0  
 
0
54
2
avatar

\(\begin{array}{| *5{@{}c@{}|}} \hline \qquad & \qquad & \qquad & \textcolor{red}{\heartsuit} & \qquad \\ \hline & 74 & & \qquad & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & & & & \\ \hline \end{array}\)

It is possible to place positive integers into the vacant twenty-one squares of this 5*5 grid so that the numbers in each row and column form an arithmetic sequence. Find the number that must occupy the vacant square marked by the \(\textcolor{red}{\heartsuit}\).

 Dec 6, 2019
 #1
avatar+23812 
+1

It is possible to place positive integers into the vacant twenty-one squares of this 5*5 grid

so that the numbers in each row and column form an arithmetic sequence.
Find the number that must occupy the vacant square marked by the \(\color{red}{\heartsuit}\).

\(\begin{array}{| *5{@{}c@{}|}} \hline \qquad & \qquad & \qquad & \textcolor{red}{\heartsuit} & \qquad \\ \hline & 74 & & \qquad & \\ \hline & & & & 186 \\ \hline & & 103 & & \\ \hline 0 & & & & \\ \hline \end{array}\)

 

\(\text{The distance of arithmetic sequence row $1 = d_1$ } \\ \text{The distance of arithmetic sequence row $2 = d_2$ } \\ \text{The distance of arithmetic sequence row $3 = d_3$ } \\ \text{The distance of arithmetic sequence row $4 = d_4$ } \\ \text{The distance of arithmetic sequence row $5 = d_5$ } \\ \text{The distance of arithmetic sequence column $1 = d_6$ } \\ \text{The distance of arithmetic sequence column $2 = d_7$ } \\ \text{The distance of arithmetic sequence column $3 = d_8$ } \\ \text{The distance of arithmetic sequence column $4 = d_9$ } \\ \text{The distance of arithmetic sequence column $5 = d_{10}$ } \)

 

\(\begin{array}{|c| *5{@{}c@{}|}} \hline \color{blue}\text{distance}& \qquad & \qquad & \qquad & \qquad \\ \color{blue}\text{of the rows }\downarrow \\ \hline & & 4d_6+d_5 & 4d_6+2d_5 & 4d_6+3d_5 & 4d_6+4d_5 \\ \color{blue} d_5 & 4d_6 & = & = &=\mathbf{\textcolor{red}{\heartsuit}}= & = \\ & & d_1+4d_7 & 2d_1+4d_8 & 3d_1+4d_9 & 4d_1+4d_{10}\\ \hline & & 3d_6+d_4 & 3d_6+2d_4 & 3d_6+3d_4 & 3d_6+4d_4 \\ \color{blue} d_4 & 3d_6 & \mathbf{=74=} & = & = & = \\ & & d_1+3d_7 & 2d_1+3d_8 & 3d_1+3d_9 & 4d_1+3d_{10} \\ \hline & & 2d_6+d_3 & 2d_6+2d_3 & 2d_6+3d_3 & 2d_6+4d_3 \\ \color{blue} d_3 & 2d_6 & = & = & = & \mathbf{=186=} \\ & & d_1+2d_7 & 2d_1+2d_8 & 3d_1+2d_9 & 4d_1+2d_{10} \\ \hline & & d_6+d_2 & d_6+2d_2 & d_6+3d_2 & d_6+4d_2 \\ \color{blue} d_2 & d_6 & = & \mathbf{=103=} & = & =\\ & & d_1+d_7 & 2d_1+d_8 & 3d_1+d_9 & 4d_1+d_{10} \\ \hline & & & & & \\ \color{blue} d_1 & 0 & d_1 & 2d_1 & 3d_1 & 4d_1 \\ & & & & & \\ \hline \color{green}\text{distance} & & & & & \\ \color{green}\text{of the columns}\rightarrow & \color{green}d_6 & \color{green}d_7 & \color{green}d_8 & \color{green}d_9 & \color{green}d_{10} \\ \hline \end{array}\)

 

The equations:

\(\begin{array}{|lrcll|} \hline (1) & 3d_6+d_4 &=& 74 \quad \text{or} \quad \mathbf{d_4 = 74 - 3d_6} \\ (2) & d_1+3d_7 &=& 74 \quad \text{or} \quad \mathbf{3d_7 = 74 - d_1} \\ (3) & d_6+2d_2 &=& 103 \quad \text{or} \quad \mathbf{2d_2 = 103 - d_6} \\ (4) & 2d_1+d_8 &=& 103 \quad \text{or} \quad \mathbf{d_8 = 103 - 2d_1} \\ (5) & 2d_6+4d_3 &=& 186 \quad \text{or} \quad \mathbf{4d_3 = 186 - 2d_6} \\ (6) & 4d_1+2d_{10} &=& 186 \quad \text{or} \quad \mathbf{2d_{10} = 186 - 4d_1} \\\\ (7) & \dfrac{(4d_1+d_{10})+(3d_6+4d_4)}{2} &=& 186 \\ & (4d_1+d_{10})+(3d_6+4d_4) &=& 186*2 \\ & 4d_1+\dfrac{186 - 4d_1}{2}+3d_6+4(74 - 3d_6) &=& 372 \\ & \ldots \\ & \mathbf{9d_6-2d_1} &=& \mathbf{17} \qquad (I.) \\\\ (8) & \dfrac{(d_6+4d_2)+(4d_1+3d_{10})}{2} &=& 186 \\ & (d_6+4d_2)+(4d_1+3d_{10}) &=& 186*2 \\ & d_6+4(\dfrac{103 - d_6}{2})+4d_1+3(\dfrac{186 - 4d_1}{2}) &=& 372 \\ & \ldots \\ & \mathbf{d_6+2d_1} &=& \mathbf{113} \qquad (II.) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (I.) + (II.): & 9d_6+d_6 &=& 17+113 \\ & 10d_6 &=& 130 \\ & \mathbf{d_6} &=& \mathbf{13} \\ \hline & 2d_1 &=& 9d_6 - 17 \\ & 2d_1 &=& 9*13 - 17 \\ & 2d_1 &=& 100 \\ & \mathbf{d_1} &=& \mathbf{50} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline d_7 &=& \dfrac{74-d_1}{3} \\\\ d_7 &=& \dfrac{74-50}{3} \\\\ \mathbf{d_7} &=& \mathbf{8} \\ \hline 4d_6+d_5 &=& d_1+4d_7 \\ 4*13+d_5 &=& 50+4*8 \\ 52+d_5 &=& 82 \\ \mathbf{d_5} &=& \mathbf{30} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\textcolor{red}{\heartsuit}} &=& 4d_6+3d_5 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& 4*13+3*30 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& 52+90 \\ \mathbf{\textcolor{red}{\heartsuit}} &=& \mathbf{142} \\ \hline \end{array}\)

 

laugh

 Dec 6, 2019
 #2
avatar
+1

Thanks

Guest Dec 6, 2019

8 Online Users