1. Weighing ABCDEFG vs. HIJKLMN

Case a) ABCDEFG = HIJKLMN This means that there a 6 or 8 (h) coins among the A-N, so O and P are from the same type

2. Weighing AB vs. OP

Case a) AB = OP This means all of ABOP are of the same kind. The third weighing is ABOP vs. HIJK. If ABOP is (l) then

A is (l). If ABOP is (h) then A is (h). A Balance ist not possible since then ABHIJKOP would be of the same kind, a

contradiction to the result of the first weighing

Case b) AB (l) OP. This means there is maximal one (h) coin among A and B. The third weighing is A vs B: If the weighing is

unequal, A’s status is settled. If it is balanced both A and B are (l)

Case c) AB (h) OP. This means there is maximal one (l) coin among A and B. The third weighing is A vs B: If the weighing is

unequal, A’s status is settled. If it is balanced both A and B are (h).

Case b) ABCDEFG (l) HIJKLMN In this case there are maximal 3 (h) coins among ABCDEFG

2. Weighing AB vs. CD

Case a) AB = CD This means there is maximal one (h) coin among A and B (otherwise we would have 4 (h) among ABCD).

The third weighing is A vs B: If the weighing is unequal, A’s status is settled. If it is balanced both A and B are (l)

Case b) AB (l) CD This means there is maximal one (h) coin among A and B. The third weighing is A vs B:

If the weighing is unequal, A’s status is settled. If it is balanced both A and B are (l)

Case c) AB (h) CD This means there is maximal one (l) coin among A and B

The third weighing is A vs B: If the weighing is unequal, A’s status is settled. If it is balanced both A and B are (h)

Case c) ABCDEFG (h) HIJKLMN just swap (h) and (l) in case b)

Test 1 Test 2 Test 3 Result Rationale

ABCDEFG < HIJKLMN AB < CD A ** B A heavier
A = B A lighter A=B lighter according to test 2
AB > CD symmetric case
AB = CD A **

A = B A lighter A=B=C=D all lighter according to test 1 and 2

ABCDEFG > HIJKLMN symmetric case

ABCDEFG = HIJKLMN A O A heavier

A = O AOP BCD A heavier A=O=P

AOP = BCD impossible A=O=P=B=C=D not possible according to test 1

Weighing Plan, where A is the Anniversary coin and B…O are the other coins.

Test 1 Test 2 Test 3 Result Rationale

ABCDEFG < HIJKLMN AB < CD A ** B A heavier
A = B A lighter A=B lighter according to test 2
AB > CD symmetric case
AB = CD A **

A = B A lighter A=B=C=D all lighter according to test 1 and 2

ABCDEFG > HIJKLMN symmetric case

ABCDEFG = HIJKLMN A O A heavier

A = O AOP BCD A heavier A=O=P

AOP = BCD impossible A=O=P=B=C=D not possible according to test 1

You’ve mentioned that this can be done for 10 coins in just three weighings; there’s a strategy where the three weighings will be balanced if and only if all 10 coins have the same weight. So for this problem, take 10 of the 16 coins and follow that strategy, making sure the Anniversary coin is weighed each time. The 10 coins can’t all be the same weight, so there will be an unbalanced weighing, and that will settle the issue.

I’m guessing the intended solution was something tricky like that.

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