fanceerootbeer

ok so
O O O O O O six circles
the 2 ends have to equal the same as the middle 2
the last number on the right has to be 1 less that the one next to it
and the last number has to equal the 2nd number from the right minus the 3rd number from the right

these are the numbers you can use 1, 2, 4, 5, 7, 9

Robby

111121

fenchurchthesane

Okay, let's call them a,b,c,d,e,f

a+f=c+d
Possible pairs: 1 and 5, 2 and 4, or 2 and 7, 4 and 5, or 2 and 9, 4 and 7
f=e-1
Possible values for f:
For the 1 and 5, 2 and 4 situation:
Not 1, because 2 is either an end or a middle, as we just established.
Not 4, because 5 is the same.
Not 2 or 5, because 3 and 6 are not in the puzzle.
But it could be be 2 and 7, 4 and 5! And f is...
Not 2, 5, or 7, because 3, 6, and 8 are not in the puzzle.
Not 4, because if 4 is f than 5 is a.
So let's try 2 and 9, 4 and 7. F =
Not 2, 9, or 7, because 3, 10, and 8 are not in this puzzle.
So it must be 4!
So e is 5!
And a is 7!
7 b c d 5 4
You know from the 3rd condition that d is 1, because the second condition says f = e - 1.
That is a shame, because it was previously determined that c and d are 2 and 9.
Something is wrong here, or you said it wrong or something.

Oh, wait, I just read the answer above me. He is right, it is a trick question, as they never said you had to use all the answers. What a cop-out.