Previous Article Previous Page Part a An examination of the arrangement reveals that all of the numbers 1-6 will be used as addends one time and that the number used in the intersection will count twice, once for each arm. After placing 1 in the intersection, the horizontal arm can be completed to have a sum of 11 by using 4 and 6 in the other two spaces. Three in the intersection of the horizontal arm is used with 5 and 4 to make a sum of 12.
Part b A similar process to the above is used here.
The sum of the six numbers is 27. Since 27 is odd, only odd number replacements for n will result in an even number sum. Once this is known it is easy to complete the solutions. Part c Part c may prove to be an interesting adventure for students. Try as they might, they will find no solution since the horizontal arm contains three odd numbers, always with an odd number sum, and the vertical arm four odd numbers, always with an even number sum.