mezzacotta Puzzle Competition

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Solution: 3E. Animal Quarters

We have a selection of sixty-three coins, arranged in order of value. There is a space for a missing coin, which by inference will be either 10, 11, or 12 units. Below the coins, we have a picture of scales balancing two sets of four objects each — clearly meant to be coins — and the indication that this is to be done eight times. That requires a total of sixty-four coins, so we infer that we need to add the missing coin to the set.

For the balancing to be possible, the combined value of all coins must be even. The total of the coins present is odd, so the missing coin must have an odd value, and thus it is the 11-unit coin. It seems plausible that we will have to deduce the animal on this coin (otherwise there is little reason for it to be absent), and this will either be the answer or lead to the answer.

Our first task is to find how to divide up the coins into sets that balance against each other. Most of the coins have values divisible by five; the exceptions are the coins with values 1, 11, 12, and 31. These must be arranged so as to produce a net balance of zero (modulo five) in whatever arrangements they end up with. A little experimentation suffices to become convinced that this can only occur if they are all in the same weighing, on the same side, producing a total of 55 units. The other side of this weighing must then contain the coins with values 5, 10, 20, and 20.

Shifting our attention along one place, we get a similar situation by looking at the tens digit: Most of the remaining coins have values divisible by 50, with the exceptions being those of value 110, 110, 120, 210, 200 000 005, and 210 000 005. Although it is possible to get partial weighings using those coins over 200 million, it seems far more sensible to use the lower values and consign those larger coins to later end up balancing against each other.

Once we do this, we again have a situation where we have digits of 1, 1, 1, and 2 that must be arranged in weighings to cancel mod five. (Here, those digits are the tens digits of 110, 110, 120, and 210.) By the same reasoning as before, they must all be in the same weighing on the same side, giving a total of 550. The coins that must balance against them have denominations 50, 100, 100, and 300.

We can proceed in this way, looking next at the hundred's digits, then the thousands, etc. In each case we get 1, 1, 1, and 2, assign the corresponding coins to the same side of a weighing, and the other side is then easily found. (A little care is needed to notice that the dachshund has value 500 010 000 and so belongs in the ten-thousand pile.) We end up with the following sets of balanced coins:

NEWT15NUMBAT
?1110MONGOOSE
LION1220KOOKABURRA
DUCK3120KOOKABURRA
5555
SHREW11050SNAIL
SHREW110100SHEEP
HYENA120100SHEEP
SPIDER210300PANDA
550550
CAMEL200500CAT
MOOSE11001000TOAD
MOOSE11002000LEOPARD
HIPPOPOTAMUS31002000LEOPARD
55005500
TOAD10005000TOUCAN
ZEBRA1100010000PEACOCK
DONKEY2100020000PANTHER
WARTHOG2200020000PANTHER
5500055000
RABBIT11000050000RACOON
WALLABY120000200000KANGAROO
OSTRICH310000300000ANTELOPE
DACHSHUND500010000500000000CHIHUAHUA
500550000500550000
PIG100000500000PUMA
BEAR11000001000000PENGUIN
MONKEY21000001000000PENGUIN
POSSUM22000003000000HAMSTER
55000005500000
BILBY20000005000000BAT
KOALA1100000010000000FOX
KOALA1100000020000000PLATYPUS
CENTIPEDE3100000020000000PLATYPUS
5500000055000000
FOX2000000050000000FROG
GORILLA120000000100000000HORSE
PORCUPINE210000000200000000GIRAFFE
BUTTERFLY210000005200000005EARTHWORM
550000005550000005

Looking at the balancing sets, one thing which we might notice is that the first animals on each side start with the same letter. Looking a little deeper, we can see that the second letters of the second animals are also the same, then the third of the third, and the fourth of the fourth. In other words, if we read down the diagonal then we get the same four letters in each pair. Moreover, these spell words:

NOOK55
SHED550
COOP5500
TENT55000
RATH500550000
PENS5500000
BOAT55000000
FORT550000005

(This also tells us that the second letter of the mystery coin is an O.)

In fact, these words all refer to enclosures of some kind; with a bit of a stretch in thinking, this is another way of interpreting the "quarters" in the puzzle title. A "rath" is the Irish term for a ringfort.

Looking at the new totals that have been formed, we see that we now have eight sets of words with associated values. So we can apply the recursive step to find another balancing:

NOOK55550SHED
COOP550055000TENT
BOAT550000005500000PENS
RATH500550000550000005FORT
555555555555555555

This time we get different words: NOAH SENT. This refers to the biblical story of the flood and Noah's ark (the ultimate animal quarters!), in which Noah sent out birds to test for dry land after the flood abated. The first bird he sent was a raven, but the second bird he sent, which eventually did not return and thus is missing from the set, was a dove.

The answer is DOVE.

Non-author's Notes

This puzzle owes its existance to a conversation that I had with Sean Gardiner, one of the organisers of the SUMS Puzzle Hunt. He observed to me that the 2012 CiSRA Puzzle Competition had a puzzle named "Animals", and the 2013 CiSRA Puzzle Competition had a puzzle named "Animal Halves", and added that the logical progression would be for the next one to have a puzzle named "Animal Quarters". I mentioned this to the CiSRA team, and Andrew Shellshear was inspired to produce this puzzle.