Another fun, first grade "challenge" problem. We've already solved this one, can you???

Magic Y

X--------X
--X-----X
---X---X
-----X
-----X
-----X
-----X

Use the numbers 1 - 10 so that each row equals 23.

whats the answer?! i can't figure it out!

Neither can I

Use the numbers 1 - 10 in the places marked with an X? In place of the hyphens (dashes? minus signs?)?

...so that each row equals 23 when the numbers are added? multiplied? This "first grade 'challenge' problem" cannot be solved because there isn't enough instruction given--unless using creativity to decide on the rules for the problem so it can be solved is the point of the "challenge".

Using addition, which is the only way it is solvable, gives:
7 is in the center.
The branches are permutations of (10,5,1), (4,9,2), and (2,6,8).

It's all very well giving the answer, but if those who have tried to solve it don't see HOW it was done, they aren't going to learn.

So... the numbers 1-10 sum to 55 (sum 1->n = n.(n+1)/2)
23 in each of 3 rows sums to 69.
Since every number is counted once, except the central number, which is counted 3 times (2 more than the others), and the difference we have to make up is 14 (69-55), the central number must be 7.

So now we need to make 3 sets of 3 numbers to make the balance of 16 (23-7).
Simple trial and error gets us there quite quickly. (choose the 'hardest' numbers to fit first, if there's a choice).

Whichever branch the 1 is on, it will need 15 more from 2 numbers: 10+5, 9+7 or 8+7. but 7 is already used. So let's pick 1,6,9 for one branch.
2 needs 14 more. Only 14 sum possible from the remaining numbers is 10+4, so 2,4,10 for another branch.
We can now be fairly confident the remaining answers will sum to 16, without looking. 3,5,8 do indeed add up, and we are done!

Note the alternate choice to use at the start also implies a solution: 1,5,10 + 2,6,8 + 3,4,9. (You will find the best puzzles have only 1 solution...)