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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:

NEWT | 1 | 5 | NUMBAT |

? | 11 | 10 | MONGOOSE |

LION | 12 | 20 | KOOKABURRA |

DUCK | 31 | 20 | KOOKABURRA |

55 | 55 | ||

SHREW | 110 | 50 | SNAIL |

SHREW | 110 | 100 | SHEEP |

HYENA | 120 | 100 | SHEEP |

SPIDER | 210 | 300 | PANDA |

550 | 550 | ||

CAMEL | 200 | 500 | CAT |

MOOSE | 1100 | 1000 | TOAD |

MOOSE | 1100 | 2000 | LEOPARD |

HIPPOPOTAMUS | 3100 | 2000 | LEOPARD |

5500 | 5500 | ||

TOAD | 1000 | 5000 | TOUCAN |

ZEBRA | 11000 | 10000 | PEACOCK |

DONKEY | 21000 | 20000 | PANTHER |

WARTHOG | 22000 | 20000 | PANTHER |

55000 | 55000 | ||

RABBIT | 110000 | 50000 | RACOON |

WALLABY | 120000 | 200000 | KANGAROO |

OSTRICH | 310000 | 300000 | ANTELOPE |

DACHSHUND | 500010000 | 500000000 | CHIHUAHUA |

500550000 | 500550000 | ||

PIG | 100000 | 500000 | PUMA |

BEAR | 1100000 | 1000000 | PENGUIN |

MONKEY | 2100000 | 1000000 | PENGUIN |

POSSUM | 2200000 | 3000000 | HAMSTER |

5500000 | 5500000 | ||

BILBY | 2000000 | 5000000 | BAT |

KOALA | 11000000 | 10000000 | FOX |

KOALA | 11000000 | 20000000 | PLATYPUS |

CENTIPEDE | 31000000 | 20000000 | PLATYPUS |

55000000 | 55000000 | ||

FOX | 20000000 | 50000000 | FROG |

GORILLA | 120000000 | 100000000 | HORSE |

PORCUPINE | 210000000 | 200000000 | GIRAFFE |

BUTTERFLY | 210000005 | 200000005 | EARTHWORM |

550000005 | 550000005 |

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:

NOOK | 55 |

SHED | 550 |

COOP | 5500 |

TENT | 55000 |

RATH | 500550000 |

PENS | 5500000 |

BOAT | 55000000 |

FORT | 550000005 |

(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:

NOOK | 55 | 550 | SHED |

COOP | 5500 | 55000 | TENT |

BOAT | 55000000 | 5500000 | PENS |

RATH | 500550000 | 550000005 | FORT |

555555555 | 555555555 |

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**.

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.