#### Problem

The numbers 1 to 16 have been arranged randomly in the 2 by 2 grid.

 6 13 4 12
 10 1 7 8
 5 16 11 15
 3 2 9 14

The product of the numbers in each quadrant are 3744, 560, 13200, and 756, respectively; all but the last quadrant are divisible by 16.

How would you arrange the numbers 1 to 16 in the grid, such that the product of the numbers in each quadrant is divisible by 16?

#### Solution

Of course there is no restriction on how many numbers we place in each quadrant, but we must ensure that there is at least a factor of 24 in each quadrant.

• 2, 6, 10, and 14 contain one factor of 2: 2222=24.
• 4 and 12 contain two factors of 2: 2222=24.
• 8 contains three factors of 2: 23.
• 16 contains four factors of 2: 24.

As we are one factor of 2 short, and no more even numbers are available, we prove that it is impossible to fill the grid with at least one number in each quadrant such that the product of numbers in each quadrant contains a factor of 24.

Surprisingly we can solve the problem by bending the "rules" slighty: place 16 in the first quadrant, 8 and 2 in second quadrant, and all the other numbers in the third quadrant. With no numbers in the fourth quadrant, and zero being divisible by 16, we have a solution. However, it could be argued that "the product of numbers" requires at least two numbers in each quadrant; in which case, no solution exists.

Can you arrange the numbers 1 to 20 in the same 2 by 2 grid so that each quadrant contains at least two numbers and the product of numbers in each quadrant is divisible by 20?

Problem ID: 222 (24 May 2005)     Difficulty: 2 Star

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