The symbol \begin{tabular}{l|l}
3 \& 4 
\hline 5 \& 6 
\hline

 is evaluated as $3 \times 6+4 \times 5=38$. If 

\hline 2 \& 6 
\hline 1 \& 
\hline
\end{tabular} is evaluated as 16 , then the number that should be placed in the empty space is

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Solution 1

When we calculate the value of the symbol, we add the product of the numbers on each of the two diagonals.

The product of the entries on the diagonal with the 1 and the 6 is 6 .

Since the symbol is evaluated as 16 , then the product of the entries on the other diagonal is 10 . Since one of the entries on the other diagonal is 2 , then the missing entry must be 5 .

## Solution 2

Let the missing number be $x$.

Using the definition for the evaluation of the symbol, we know that $2 \times x+1 \times 6=16$ or $2 x+6=16$ or $2 x=10$ or $x=5$.

ANSWER: (E)