Thanks to Brian Smith for the original problem and to Leming for the improved solution.
Separate the stacks into two piles of 6 and one of 5 coins.
1. Weigh the two stacks of 6 coins against each other.
Label the heavier stack "A", the lighter stack "B" and the stack of 5 coins "C".
The 35g coins will be one of these combinations:
A B C
5 0 0
4 1 0
4 0 1
3 2 0
3 1 1
3 0 2
2 2 1
2 1 2
1 1 3
1 0 4
0 0 5
Take one coin from B and add it to C.
2. Weigh A vs C. Keep the heavier of A and C. The heavier stack will have 3, 4, or 5 35g coins unless
If A balanced with B and heavier than C or
If A heavier than B and balanced with C.
then there are only two coins in the heavy stack.
3. With 3, 4, or 5 heavy coins out of 6:
i. Weigh 3 vs 3. The heavy (or balanced side) will have at least two 35g coins. Keep the heavy stack.
ii. From the heavy stack of three coins weigh 1 vs 1.
If they are balanced both are heavy. If one is heavier than the other, then that coin and the one not weighed are the two 35g coins.
Total: 4 Weighings.
4. With 2 heavy coins out of 6:
Weigh 3 vs 3.
ii. Unbalanced: The heavy side will have two 35g coins. Weigh 1 vs 1 as above to find the two coins.
Total: 4 Weighings
ii. Balanced: Each side has one heavy coin.
Take one stack and weigh 1 vs 1. If these balance the heavy coin is the 3rd of the three.
If unbalanced the heavy side is the 35g coin.
Repeat for the other stack of 3.
Total: 5 Weighings
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