2 + 6 = 8

2 + 9 = 5 + 6

6 + 8 = 5 + 9

2 + 5 + 8 = 6 + 9

For 5 weighings and 114 coins the groups are: 9, 12, 20, 22, 25 and 26. For 6 weighings and 454 coins the groups are: 24, 35, 42, 62, 76, 85 and 130.

]]>1. Divide the 30 coins into 5 groups A4, A5, A6, A7, A8 of 4, 5, 6, 7, 8 coins, respectively.

2. Make the following weighings.

a) A4+A5+A6 vs. A7+A8.

b) A4+A7 vs. A5+A6.

c) A6+A7 vs A5+A8.

d) A4+A8 vs. A5+A7.

3. Assume now the scale balances in all the cases (otherwise, we’ve shown that the set isn’t uniform). Mimicking your approach, let a4, a5, a6, a7, a8 be the number of heavier coins in A4, A5, A6, A7, A8, respectively. Our goal is to prove that they are all zero or that each aj is j, for j=4,5,6,7,8.

4. Let a4 be arbitrary.

– Adding a) and b), we get a8=2*a4.

– Adding a) and c), we get a6=(2*a8-a4)/2=3*a4/2.

– Adding a) and d), we get a7=(2*a4+a6)/2=7*a4/4.

– From a), we get a5=a7+a8-a4-a6=5*a4/4.

5. This means that a4 must me divisible by 4, so the only allowed values are 0 and 4. To complete the proof, observe that the formulas from 4. imply the following.

a) If a4=0, then aj=0, j=4,5,6,7,8.

b) If a4=4, then aj=j, j=4,5,6,7,8.

well’ i just want to thank u, and your son, for this post (and all the others). ]]>