The Anniversary Coin

Konstantin Knop, the world’s top authority on coin-weighing puzzles, suggested the following problem for the 2019 Russian Math Olympiad.

Puzzle. Eight out of sixteen coins are heavier than the rest and weigh 11 grams each. The other eight coins weigh 10 grams each. We do not know which coin is which, but one coin is conspicuously marked as an “Anniversary” coin. Can you figure out whether the Anniversary coin is heavier or lighter using a balance scale at most three times?

Share:Facebooktwitterredditpinterestlinkedinmail

5 Comments

  1. Joseph:

    This puzzle seems to be related to one you’ve blogged about before: given a set of coins, each of weight either x or y, figure out whether the coins are all the same weight.

    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.

  2. Joseph:

    Meh. Maybe I was too hasty with my previous comment. Even if all three weighings are unbalanced, it may still not be clear whether the Anniversary coin is heavier or lighter. But I’m still guessing there’s a connection to the “all the same weight or not” problem.

  3. Robert L:

    The key is to observe that a result of three equal balances must be avoided, because otherwise a switch of all 16 coins would lead to the same result.
    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
    B A heavier
    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

  4. Robert L:

    Sorry tabs did not work

    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
    B A heavier
    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

  5. Gennardo:

    (l) means light (coin) or ligther, (h) means heavy (coin) or heavier. The coins are A-P, and A is the special coin

    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)

Leave a comment