Question

Six different digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are placed in a figure made of a vertical column of three squares and a horizontal row of four squares that overlap in one shared square, so that the sum of the three entries in the vertical column is $23$ and the sum of the four entries in the horizontal row is $12$ . The sum of the six digits used is

$27$

$29$

$31$

$33$

$35$

Solution:

Three distinct digits from $1$ through $9$ summing to $23$ must be $6, 8, 9$ . The row's other three digits are at least $1 + 2 + 3 = 6$ , so the shared square (belonging to both the column and the row) is at most $12 - 6 = 6$ . Hence the shared digit is $6$ .

The six digits are then $6, 8, 9$ and $1, 2, 3$ , whose sum is $29$ . Equivalently, $23 + 12 - 6 = 29$ .

Thus, the correct answer is B .

1996 AMC 8 Problem 14

Solution:

Problem 14 in Other Years