Solution to: New Year's Eve Party

The first observation should be that since the year of the puzzle is 2000 then the birth years of the children must be nineteen hundred and something. So, the starting two digits of 3 across, 21 across and 7 down must be 19. We can also assume the same for 12 down, the year of Leonard's birth, as 10 down, Leonard's age, is a two-digit number meaning that he too must have been born sometime in the 1900s.

15 down is also a year, but this time in the 1800s, so the first two digits must be 18. With the last digit of the year a 9 (from 21 across) we are now only missing the third digit. Thus, we are looking for a four-digit prime number starting 18 and ending with a 9. A quick internet search, use of a calculator (or by hand if you've got time to kill) reveals that two numbers qualify, 1879 and 1889. Unfortunately, for now, that's as far as you can get with 15 down (the reason being is that 17 across and 18 down are linked and two possible reasonable options exist for both numbers 76 and 67 or 87 and 78).

Whenever you are trying to find unknown numbers in a puzzle, a good way to begin is to try and see whether you can put any limits on them. Look at 19 across, the cube of a number with three digits. Now 4³ = 64, is too small, it only has two digits. And 10³ = 1000 is too large, it has four digits. So, the possibilities for 19 across are either 125, 216, 343, 512 or 729 (5, 6, 7, 8 and 9 cubed). This means that See is either 5, 6, 7, 8 or 9 years old. Can we narrow down these options? Well now look at 20 down, the square of Hear's age, a two-digit square number; the possibilities are 16, 25, 36, 49, 64 and 81. Now note that the ending digit of 19 across (the cube number) is the first digit of 20 down (the square number). This means that we can straight away eliminate 125 and 729. Why? Because the ending digits of these numbers are 5 and 9 and no square number begins with a 5 or a 9.

Thus 19 across is either 216, 343 or 512. Let's assume it's 216. This would make See 6 years old (6³ = 216). The ending digit of 216 is 6 which means that 20 down would have to be 64 (8² = 64). Now 20 down represents Hear's age squared, making Hear 8 years old. This is not allowed as we are told in the question that See is the oldest monkey. Therefore, we can eliminate 216. Next, let's assume that 19 across is 512, making See 8 years old (8³ = 512). The ending digit of 512 is 2 making 20 down 25 (5² = 25) so Hear would be 5 years old which would be fine. However, now look at 22 across, Speak's age, the third monkey, multiplied by 15. If 20 down is 25, with the last digit a 5, then the first digit of 22 across would begin with a 5. However, there is no multiple of 15 that begins with a 5 (15, 30, 45, 60 etc). So, 19 across cannot be 512. This leaves one option for 19 across 343, making See 7 years old (7³ = 343). Here, 20 down would have to be 36 (6² = 36) making Hear 6 years old. This means that 22 across would begin with a 6 making 22 across 60 (4 x 15) meaning Speak would be 4 years old. We now have the ages of the three monkeys. 7 across is the sum of See and Hear's age (7 and 6 respectively) so 13. This reveals 3 down, the age of Manhattan the dog, to be 13.

Next, look at 16 down, which is the same number as 16 across, the number of years that the family has lived in the village. In the question we are told that the family has been living there since the nineteen seventies. With the current year being 2000, this number of years must be twenty something. So, the first digit of 16 down and 16 across is 2 meaning that 16 down and 16 across are 23. This means that the family has lived in the village for 23 years.

With the numbers above entered into the cross-number puzzle, 12 down, the year of Leonard's birth is now complete, 1934. This means that Leonard is 66 years old so 10 down is 66. This makes 14 across 61, so Marianne is 61 years old.

Turning now to the ages of the children, again it is a good idea to try and put limits on them. 1 across is the square of Aqua's age, a three-digit number. Now 31² = 961, a three-digit number but 32² = 1024, a four-digit number. This means that Aqua, the oldest child, can be no more than 31 years old. 9 across is two less than Desert's age and we know that it ends in a 9, from 7 down. So, 9 across must be 19, making Desert 21 years old. If 9 across began with a 2 then Desert would be 31 which is too old, Aqa the eldest is at most 31 years old. We can now fill in 9 down to be 19, as 9 down is the same as 9 across. And this means we can now also fill in 7 down, the year of Desert's birth, to be 1979. This reveals 11 across, Frank's age, to be 97.

Let's now return to 1 across, the square of Aqua's age. We know that Aqua is at most 31 years old. Because Desert is 21 years old the youngest that Aqua could be is 24. This is because Carmine has to be at least 22 and Byzantine at least 23 (the question states that all the ages in the puzzle are different). So, let's look at the square numbers between 31 and 24. These are, in descending order, 961, 900, 841, 784, 729, 676, 625 and 576. We can quickly eliminate most of these options by noting that 2 down, the same answer as 6 across, is Byzantine's age, at most a number beginning with a 3. Since the second digit of 1 across, the square numbers above, represent the first digit of 2 down then we can remove 961, 900 (no digit begins with a zero) 841, 784, 676 and 576. We are therefore left with only two possible options for 1 across, 729 or 625. Assume 1 across is 625. Then 2 down begins with a 2 meaning so does 6 across. This would make 1 down, Carmine's age multiplied by three, 62. However this is not possible as 62 is not a multiple of 3 so we are now left with 1 across being 729, making Aqua 27 years old (27² = 729). This means the first digit of 2 down and 6 across is a 2 making 1 down 72. This is divisible by three making Carmine 24 years old (24 x 3 = 72).

As we now know that Aqua is 27 and Carmine 24, then Byzantine must be 25 or 26. This means Byzantine was born in either 1974 or 1975 making 3 across, the year of Byzantine's birth, one of these two numbers. The ending digit of these two numbers, which is also the first digit of 5 down, a multiple of Aqua's age, is either 4 or 5. Since Aqua is 27 then this number must be the 5 (2 x 27 = 54) so 5 down is 54. So, 3 across is 1975 making Byzantine 25 years old. This makes 6 across and 2 down 25.

Now 8 across, a multiple of Emerald's age, is the same number as 5 down, 54. As Desert is 21 years old then Emerald must be less than this, so 20 years old or less. The question tells us that the age range of the children is no more than thirteen years. As Aqua is 27 then the youngest that Flame could be is 14, making Emerald at least 15 years old. So Emerald is somewhere between 15 and 20 years old. If you look at multiples of numbers between 15 and 20 then the only one that makes 54 is 18 x 3. So Emerald is 18 years old.

Going to 4 down, Mick's age multiplied by ten, we now know the first two digits are 7 and 5. So Mick is 75 years old and 4 down is 750.

If we now return to 17 across and 18 down, we had two options for these numbers 76 and 67 or 87 and 78. As the second digit of 18 down is either a 7 or an 8 and this represents the year of Flame's birth, 21 across, then we now know this must be an 8; Flame must have been born sometime in the 1980s because we know Emerald is 18 years old so Flame must be 17 years old, or younger. Thus, 18 down is 78 making 17 across, Pamela's age, 87. This means that the year of the great flood in the village, 15 down, was 1889, one of the only two possibilities: 1879 and 1889.

We are left with 13 down, the square of Flame's age, with the digits adding to nine. We also now know that Flame must be somewhere between 14 and 17 years old. If we look at square numbers in this range, they are 196, 225, 256 and 289. Of these options, only 225 has digits that add to nine. So, Flame is 15 years old and 13 down is therefore 225. This means we can now fill in 21 across, the year of Flame's birth, as 1985.

The completed puzzle:

If you're wondering about the names of everyone at the party, Paul White has provided the following explanation:

Leonard is based on the Canadian singer/songwriter Leonard Cohen. Norman was Leonard Cohen's middle name.

Leonard met Marianne (Jensen) on the Greek Island of Hydra in the nineteen sixties. The village in the puzzle, Hardy, is an anagram of Hydra. "So Long, Marianne" is one of his most famous songs and was written about her. It appeared on his debut album "Songs Of Leonard Cohen".

The real Leonard Cohen had two children; in my puzzle he had six. The names of the children are all colours. Leonard has always struck me as a chilled out hippyish kind of guy, so I thought that those names seemed apt!

Frank is based on a particular favourite of mine, Frank Zappa, an unparalled musical genius of the latter half of the 20th Century. Pamela (Miller) was a nanny of Zappa's and a member of the music group the GTO's (Girls Together Outrageously) that Zappa put together. They weren't great! Under the name Pamela Des Barres she wrote a best-selling book in the eighties "I'm With the Band" that detailed her time in the sixties and seventies as a rock groupie.

Mick is based on Mick Jagger, the lead singer of the Rolling Stones. He famously had a relationship with the singer Marianne Faithful in the sixties (not the same Marianne as Leonard Cohen's girlfriend). Whether Cohen's Marianne and Mick Jagger ever met, I don't know!

Manhattan, the name of the dog, is from a Cohen song "First We Take Manhattan" from his 1988 LP "I'm your Man".

See, Hear and Speak are the three moral Japanese wise monkeys (See No Evil, Hear No Evil and Speak No Evil). Again, I think they would have fitted well into the Cohen household!

Copyright © 2026 Paul White

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