## 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:

 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.