Solve for ?
| 50 | 15 | 10 | 60 |
| 45 | 31 | 5 | 87 |
| 40 | 22 | 42 | ? |
| 9 | 4 | 3 | 14 |
$\begingroup$
$\endgroup$
1
-
6$\begingroup$ Hi William welcome to PSE could you please give us the source of this puzzle? $\endgroup$– PDTCommented Jun 26 at 14:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
The entry is, or rather may be $85$, with the following way to guess it. Take the first and the third columns. Add them. To the result add four times the second column. We obtain: $$ \begin{bmatrix} 50 + 4\cdot 15+10\\ 45 + 4\cdot 31+5 \\ 40 + 4\cdot 22+42\\ 9 + 4\cdot 4+3 \end{bmatrix} = \begin{bmatrix} 120\\ 174 \\ 170\\ 28 \end{bmatrix} = 2\cdot \begin{bmatrix} 60\\ 87 \\ \bbox[yellow]{\ 85\ }\\ 14 \end{bmatrix} \ . $$ There is a match with the last column at three places...
Note: This is the value that makes the given $4\times 4$ matrix have linearly dependent rows and columns. Well, the combination of columns is still "nice", so this is a puzzle plus for the value. (But one should never mention, here or elsewhere, the combination of the rows, it needs the weights / coefficients $1609$, $805$, $655$, $-15875$, or common multiples, except for the case when still not afraid to endure the consequences.)
