[Guest]

In the following puzzle, your goal is to balance the entire device so every beam is horizontal as shown.

Each character space will represent a distance of 1, if you are not using a fixed-width font, you will have some trouble, so adjust accordingly.

For the letters A-L, replace them with the values of 1-12, using each value only once. 'x' will indicate where an attachment is made, to easier identify the difference. In each case, everything is hanging downwards.

          |

    x---x---x---x

    |   A   |   B

    |   x-------x

    |   |       |

    | x---x-x x---x-x

    | C   D E F   | G

x-x-------x x-------x

H I       J K       L

Answers can be the values of each letter, you don't need to draw it out.

[curr3nt]

Here are some of the equations I have...

#1 3K = L

#2 K+L+2G = F

#3 C+D+E = F+G+K+L

#4 D+2E = C

#1 means K|L can only be 1|3, 2|6, 3|9 or 4|12

Including #2 makes F|G (8|2 or 12|4 if K|L = 1|3) or (10|1 if K|L = 2|6) 3|9 and 4|12 are too large to be used

Including #3 Means C+D+E must equal 14, 20 or 19

#4 Means C|D|E can only be 12|6|3, 8|4|2 or 4|2|1 but only one of these equals 14, 20 or 19. 8+4+2=14 but that makes 2 and 8 used twice.

[Guest]

3*K = L

K +L +2*G = F

D +2*E = C

2*H +I = 3*J

C +D +E = F +G +K +L

A +3*(H +I +J) = C +D +E +F +G +K +L +3*B

so we have 4 possibilities for K,L; 1,3; 2,6; 3,9; 4,12.

we can eliminate 4,12, and 3,9 because then the second equation doesn't work.

so let's try 2,6. then G is 1, making F 10.

the third equation combined with the 5th equation means we have

2*D +3*E = 19

we have for D,E 2,5; 5,3; 8,1. 5,3 is the only one with unused numbers.

this makes C 11.

numbers used so far... 1,2,3,5,6,10,11

with equation 4, we don't have any unused numbers that work.

so back to K,L; let's try 1,3

then G is 2 or 4. let's try 2.

then F is 8.

2*D +3*E = 14

D,E = 1,4; 4,2. niether works.

thus G must be 4.

F = 12.

2*D +3*E = 20

D,E = 1,9; 4,3; 7,2. 7,2 is the only unused numbers.

C = 11.

numbers used so far, 1,2,3,4,7,11,12

going to equation 4,

2*H +I = 3*J

H, I, J = 5, 8, 6; or 9, 6, 8;

let's try 5, 8, 6 then we have

A +3*19 = 40 +3*B

A +17 = 3*B

A = 10, B = 9.

[curr3nt]

The possible values I had for #4 are incomplete...