Puzzle of Scale Problem

1) condition given: x1+x2+x3+x4=40,

                             x1<x2<x3<x4 

                          and x1,x2,x3,x4 are whole numbers. 

and the 4 sales can be used on both sides, so the 4 scales will use as   -1, o, 1 time for each weigh. 

-1 means the sale was used on the cookie side, 1 means the sale is used on the opposite of the cookie side. 

So because the lowest weigh is 1g, so the x1=1g, 

for 2g, we can use another one scale or use the x1 on the cookie sides.  In order to use fewer scales to measure,  I choose 2=x2-x1, then x2=3g, 

so for 3g, 3=x2, for 4g, we use x2+x1, both 1 and 3g, 4=x1+x2

for 5g, we put x2 and x1 on the cookie side, and start to use x3,  then x3=5+x2+x1 therefore x3=9g, 

6g=x3-x2, 7g= x3-x2+x1, 8g=x3-x1, 9g=x3, 10g=x3+x1, 11g=x3+x2-x1, 12g=x3+x2, 13g=x3+x2+x1, 

now we think about x4, move to the 14g=x4-x3-x2-x1 so the x4=27g. 

x1+x2+x3+x4=1+3+9+27=40g

so the solution is 1,3,9 and 27g scales. 

2) I only find this solution. 

3) I can ask students to think about if they have 5 different scales, what is the widest range of weight they could measure. 

By using the idea of the original question, we can extend to find the 5th scale, 

by solving the equation 41=x5-40, then x5=81, then the weightiest they can measure should be 81+40=121g.