A: Draw a 3x3 grid, with the boxes labeled 1 through 9 in the usual order (left to right; top to bottom).

B:For each of the following instructions you must write a number, greater than 10, starting in one box and going across (left to right) or down, with one digit to each box. Your answers should fill the grid, with no two answers overlapping.

C:Starting with a box whose number is a square, write a square number.

D:Starting with a box whose number is a cube, write a cube number.

E:Starting in a box whose number is prime, write a prime number whose digits add up to an even number.

F:Starting with a box whose number is even, write an even number

Unfortunately, one of the copies of my instructions had part F completely missing. Each child turned in a 3x3 grid which was correct for the version they had been given. By coincidence, they turned in identical grids.

How did they fill out their grids?

Let us say the niece's instructions were missing part F

Since the problem suggests a unique solution,  Let us discount the cases which have the same 3-digit cube across the top (boxes 1,2,3) because they could then be any three digit cube and would not be unique, they would simply have to agree between the niece and nephew. 

Then boxes 8 and 9 would have to contain the digits of a two digit cube and would have to be odd - namely 27.  Then box 7 must contain either a 1 or a 7 in order to satisfy the niece's prime number requirement.

This means that boxes 4,7 do not contain an even number and instead boxes 1,4,7 contain a 3-digit square. Since no square numbers end in 7 box 7 must contain a 1.  Now we already know that boxes 1,2,3 contain a 3-digit cube (125, 216, 343, 512, or 729)

So, looking for 3 digit squares for boxes 1,4,7 which begin with 1, 2, 3, 5, or 7 and end in 1 gives us 121 or 361.

If 361 then Boxes 1,2,3 must contain cube 343 meaning box 5 must be even and box 6 must be odd and satisfy boxes 3,6 being prime (31,37) but there are no square numbers which begin with 6 and end with 1 or 7 so this is not possible

If 121, however, boxes 1,2,3 would contain 125 and box 5 would be even with box 6 being odd and satisfying boxes 3,6 being prime (53,57,59). 289 is the only square which satisfies this case giving a viable solution of:

125
289
127

It would take more work to prove that there were no solutions amoung the cases where the niece and nephew pick the same 3-digit cube, but I will leave that to someone else.

Posted by Eric on 2005-03-14 23:03:06