Math Puzzle???

[54warrior]

My aunt sent me this, any thoughts?

1 2 3 4
5 7 8 9

arrange the eight numbers into two groups of four numbers each.

when you add up each group, the sums must be the same.





My response, the total of all the #'s is 39, which is an odd number, so if you divide it in half, it equals 14.5, which none of the groups could add up to because they are "whole" numbers.

[Blackbeauty]

Now if it were

1 2 3 4
6 7 8 9


it would be easy.... highest + lowest

9 +1 = 8 + 2 = 7 + 3 = 6 + 4


But the 5 throws a monkey wrench into things.....

[dss90c]

If I understand this right, the 2 groups would be:

2, 3, 5, 8 = 18
1, 4, 6, 7 = 18

[Blackbeauty]

Doh!!! Read it wrong.... I'd blame 54Warrior, but IT'S ALL MY FAULT!!!!!

[P's Titan]

Originally Posted by Blackbeauty

but IT'S ALL MY FAULT!!!!!

yes it is!

[SJYokel]

Originally Posted by dss90c

If I understand this right, the 2 groups would be:

2, 3, 5, 8 = 18
1, 4, 6, 7 = 18

But that wasn't the original puzzle...it was:
1 2 3 4
5 7 8 9

I see two answers:
1) It can't be done if you allocate each number to only one group. There are 5 odd numbers, so one of the groups will always be odd and the other will have to be even (the groups could have 1-4, 2-3, 3-2, or 4-1 odd numbers in them).

2) It doesn't say you have to use all of the numbers or use each number only once, so my solution would be:

2 3 5 9 = 19
1 3 7 8 = 19

[dss90c]

Originally Posted by SJYokel

But that wasn't the original puzzle...it was:
1 2 3 4
5 7 8 9

I see two answers:
1) It can't be done if you allocate each number to only one group. There are 5 odd numbers, so one of the groups will always be odd and the other will have to be even (the groups could have 1-4, 2-3, 3-2, or 4-1 odd numbers in them).

2) It doesn't say you have to use all of the numbers or use each number only once, so my solution would be:

2 3 5 9 = 19
1 3 7 8 = 19

No wonder why I thought it was easy! Good catch.