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?

Question

Adhir Mishra has three children: Urmila, Raghu and Sumit. Sumit married Roma, the eldest daughter of Mr. and Mrs. Mohan. The Mohans married their youngest daughter to the eldest son of Mr. and Mrs. Sharma and they had two children Sandeep and Shaifali. The Mohans have two more children, Roshan and Bimla, both elder to Sheila. Sohan and Shivendar are sons of Sumit and Roma. Leela is the daughter of Sandeep.

Answer 1

Instead of using divisibility rules first, we can find the answer by trying to actually build a valid $3 \times 4$ grid and seeing which number refuses to fit.

Row and column targets: the four fixed numbers 1, 9, 14, 15 sum to 39, while the nine candidates sum to 67. The full grid, missing one candidate $x$, totals $106 - x$. For a whole-number average $M$ across 3 rows and 4 columns, this total must equal $12M$.
Testing $x = 15$: this is not even among the nine candidate numbers offered, so it cannot be the missing one.
Testing $x = 4$ and $x = 7$: these give totals of 102 and 99, and neither divides evenly by 12, so no common integer average is possible in either case.
Testing $x = 10$: this gives a total of 96, and $96 / 12 = 8$ exactly, so a common average of 8 is at least numerically possible.

To be sure $x=10$ truly works and is not just numerically convenient, build the grid directly with average 8 (row sum 32, column sum 24): placing 13, 3 in row 1, 11, 7, 5 in row 2, and 12, 2, 4 in row 3 alongside the fixed numbers gives every row summing to 32 and every column summing to 24, using all nine candidates except 10.

Let's summarize:

Only $x = 10$ produces a total divisible by 12.
A complete, valid grid can actually be built while leaving out 10, confirming it is not just a coincidence of arithmetic.

So the number that must be left out of the grid is 10.

Answer 2

Instead of using divisibility rules first, we can find the answer by trying to actually build a valid $3 \times 4$ grid and seeing which number refuses to fit.

Row and column targets: the four fixed numbers 1, 9, 14, 15 sum to 39, while the nine candidates sum to 67. The full grid, missing one candidate $x$, totals $106 - x$. For a whole-number average $M$ across 3 rows and 4 columns, this total must equal $12M$.
Testing $x = 15$: this is not even among the nine candidate numbers offered, so it cannot be the missing one.
Testing $x = 4$ and $x = 7$: these give totals of 102 and 99, and neither divides evenly by 12, so no common integer average is possible in either case.
Testing $x = 10$: this gives a total of 96, and $96 / 12 = 8$ exactly, so a common average of 8 is at least numerically possible.

To be sure $x=10$ truly works and is not just numerically convenient, build the grid directly with average 8 (row sum 32, column sum 24): placing 13, 3 in row 1, 11, 7, 5 in row 2, and 12, 2, 4 in row 3 alongside the fixed numbers gives every row summing to 32 and every column summing to 24, using all nine candidates except 10.

Let's summarize:

Only $x = 10$ produces a total divisible by 12.
A complete, valid grid can actually be built while leaving out 10, confirming it is not just a coincidence of arithmetic.

So the number that must be left out of the grid is 10.

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 Correct Option is B

Solution and Explanation

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