Wait! What Weights

10, 5, 8, 6, 4, 7, 9, 3, 1, 2

Here's the reasoning:

55 = a+b+c+...+h+i+j

3a + 2b = c+d+...+h+i+j

55 = a+b+3a+2b

55 = 4a+3b

There are only 2 sets of integral values for a and b, namely (7,9) and (10,5)

From the bottom end

g = h + 2(i + j)

j = 2i

g = h + 2(i + 2i)

g = h + 6i

There's only 1 possible value for i , namely 1. Thus j = 2.

Only possible values for g and h are 10 and 4 or 9 and 3. Carry on....

You need to know how torque works before working on this question. It is easier to work from bottom to the top. By simple inspection, the bottom weights i and j can only be 1 and 2 respectively, because the weight g cannot support any values larger than that. The next step is to determine g and h. The only possibilities are (g=9, h=3) or (g=10, h=4) as h cannot be higher since g cannot bear a value higher than 4 for h, and other values like 1 and 2 are used up in i and j weights. You have to assume one of the values for g and h and work upwards in the same reasoning and you'll eventually get the values for all the weights.

SPOILER: a=10, b=5, c=8, d=6, e=4, f=7, g=9, h=3, i=1, j=2

Bottom fulcrum 2i = j

4th fulcrum g = h+2(i+j) = h+3j since g <= 10, then j <= 3 and from 2i = j, j has to be even number, so we have j = 2 and i = 1 and we have g = h+6, h <= 4 since 1 and 2 are used, h = 3 or 4 and g = 9 or 10

3rd fulcrum 2e+f = g+h+i+j = g+h+3 = 2h+9 2(e-h) = 9-f so f <9 and has to be odd number (5 or 7)

Top fulcrum, we have 2 equations

3a+2b = c+d+...+j = 55 - (a+b) --> 4a+3b = 55 2 possibilities; a = 10, b = 5 or a = 7, b =9 and 3a+2b = 2c+4d = 2(c+2d) so a has to be even number (10 only) Finally, we have b = 5, g = 9 (10 is used), h = 3, f = 7 (5 is used) and for 2e+f = 2h+9 --> we have e = 4

e+f+g = 4+7+9 = 20

Brute force in Python! Mainly because this was giving me a headache.

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from itertools import permutations
from string import lowercase

def w ( letter ) :
    global w_perm
    return w_perm [ lowercase . index ( letter ) ]

def sum_left ( n ) :
    global w_perm
    return sum ( w_perm [ n : ] )

for w_perm in permutations ( range ( 1, 11 ) ) :
    if 2 * w ( 'i' ) == w ( 'j' ) \
        and w ( 'g' ) == w ( 'h' ) + 2 * ( w ( 'i' ) + w ( 'j' ) ) \
        and 2 * w ( 'e' ) + w ( 'f' ) == sum_left ( 6 )  \
        and 3 * w ( 'd' ) + w ( 'c' ) == sum_left ( 4 ) \
        and 3 * w ( 'a' ) + 2 * w ( 'b' ) == sum_left ( 2 ) :
        print w ( 'e' ) + w ( 'f' ) + w ( 'g' ) 

solve the problem from the bottom...

at the 5th level (bottom) we have 2i = j

at the 4th level we have g = h+2(i+j) <-(i,j assumed as one element)

at the 3rd level we have 2e+f = (g+h+i+j)

at the 2nd level we have c+3d = (e+f+g+h+i+j)

at the 1st level we have 3a+2b = (c+d+e+f+g+h+i+j)

the first equation you must to solve is the 5th level equation as the bottom of system because when we solve (the equation is balanced) the bottom we can ignore it (is it balanced or not) in the next equation above it, just sum up all of the weight in previous level and assume it as one element in the next equation

so, because the bottom level equation is

2i = j

we only have 5 possibilities

1) i=1,j=2

2) i=2,j=4

3) i=3,i=6

4) i=4,i=8

5) i=5, i=10

try the smallest possible number first

you have to put the smallest possible number first in the farthest place from pivot (torque effect)

so, try the first pair..and ...

5th)

    i=1,j=2

4th)

    g = h+2(i+j)

    g = 3+2(1+2) <- the smallest possible number is 3

    9 = 3+2.3

    g=9,h=3

3rd)

     2e+f = g+h+i+j

     2.4+f = 9+3+1+2 <- i put 4(the smallest possibility) in the farthest place from pivot)

    8+f = 15

    e=4,f=7

2nd)

    c+3d = e+f+g+h+i+j

     c+3d = 4+7+9+3+1+2

     c+3d = 26

     the only possible number: c=8,d=6

1st)

    3a+2b = c+d+e+f+g+h+i+j

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

    3a+2b = 40

    the only possible number is 10 and 5... 

    a=10,b=5

SOLVE!

so, now the answer...

   e+f+g = 4+7+9=20

1) first separate the given numbers into two groups which are equal..

what i did was through trial and error. 9 , 7 , 6 , 3 , 2 = 10 , 8 , 5 , 4

2) the next step was to arrange all the available spots for the masses on straight balancing bar.. where some points on the bar would take two sets of masses.

3) then, again by trial and error i moved the masses around and calculated its moment about the hanging point. (pivot)

• moment on right side + (10) x 1 + ( 7 + 2 ) x2 + (1 + 3 ) x 4

• moment on left side - ( 4) x 3 - ( 8 + 5) x 2 - (6) x 1

• the weight at the balancing point itself has no moment, perpendicular distance from pivot = 0 . also i took a clockwise rotation as positive and anticlockwise as negative.

  the resultant moment is 0 , therefore the system is balanced

Lets look at all the equations- 1)j=2i 2)g= 2(i+j)+h 3)2e+f=g+h+i+j 4)3d+c=e+f+...+j 5)3a+2b=c+d+...+j

(i)i and j cannot be greater than 1 and 2 because this is the maximum (and only)value for which g is <=10..so i=1 and j=2

(ii)only values satisfying equn 2 (after putting the values of i and j) are g=7 and h=1 and g=9 and h=3..but since 1 is already used g=9 and h=3

(iii) only values satisfying equn 3 ----2e+f= 15 (after putting g,h,i,j) are e=4 and f=7 This part is enough to get the answer (20) but i am gonna find out all of the values to verify the results

(iv) the only values that satisfy the equn4 ---3d+c=26 are 6 and 8 so d=6 and c=8

(v) that leaves us with two no.s 10 and 5 and putting them in the equn 5 we get 30 + 10=6+8+26...s o we can see that the equn is satisfied and so the values are correct..

So the answer id e+f+g=4+7+9=20