Eight of the nine numbers \(2, 3, 4, 5, 7, 10, 11, 12, 13\) are to be placed in the empty cells of the \(3 \times 4\) array (with \(1, 9, 14, 15\) already placed) so that the arithmetic average of the numbers in each row and each column is the same integer. Which one number must be left out?
Show Hint
The total of all 12 numbers in the grid must be a multiple of 12, since it equals 3 times the row sum. Test each option against this rule before trying to build the grid.
Step 1: Set up what the common average tells us.
Call the shared average \(M\). The grid has 3 rows of 4 numbers each, so every row must sum to \(4M\).
The grid has 4 columns of 3 numbers each, so every column must sum to \(3M\).
Adding the three row sums gives the grand total of all 12 numbers in the grid:
\[ \text{Total} = 3 \times 4M = 12M \]
So the total of all 12 entries must be a multiple of 12.
Step 2: Work out the total for each possible number left out.
The four fixed numbers already in the grid sum to \(1 + 9 + 14 + 15 = 39\).
The nine candidate numbers sum to \(2+3+4+5+7+10+11+12+13 = 67\).
If \(x\) is the one number left out, the grid's total is
\[ 39 + (67 - x) = 106 - x \]
Step 3: Find which \(x\) makes the total divisible by 12.
For \(x=4\): \(106-4=102\), not divisible by 12.
For \(x=10\): \(106-10=96 = 12 \times 8\), divisible by 12.
For \(x=15\): 15 is not even one of the nine candidate numbers, so it cannot be the number left out.
For \(x=7\): \(106-7=99\), not divisible by 12.
Only \(x=10\) works, giving total 96 and common average \(M = 96/12 = 8\).
Step 4: Confirm a valid grid actually exists.
With average 8, each row must sum to 32 and each column must sum to 24. One valid filling is:
Row 1: 1, 13, 3, 15 (sum 32). Row 2: 11, 9, 7, 5 (sum 32). Row 3: 12, 2, 14, 4 (sum 32).
Checking columns: \(1+11+12=24\), \(13+9+2=24\), \(3+7+14=24\), \(15+5+4=24\), all correct.
This uses every candidate number except 10, confirming the construction works.
Final Answer:
The number that must be left out is 10.
\[ \boxed{10} \]
