Tuesday, September 2, 2008

Problem of the Day

Problem of the Day

Take a three by three square grid which has the number '1' already inserted in the third row, middle square. You have to complete it by putting eight different prime numbers in the remaining eight empty squares, so that the rows, columns and diagonals add up to the same total; and it must be the smallest possible total under the conditions. Also, the number in the middle square is the average of the two numbers directly above and directly below it and the third largest number is not in the right-hand column, and every square contains one or two digits.

1 comment:

Achilles's Heel said...

A11= 7
A12 =73
A13 = 31
A21= 61
A22= 37
A23=13
A31=43
A32=1
A33=67