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Problem 610

AIME late

Combinatorics Difficulty 5.0 Find the answer

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

Your answer

Official solution

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)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.