6
$\begingroup$

This is not a puzzle, this is a question about a certain puzzle, Dog's mead. Several forms of this puzzle have been known, and I'm referring to this particular form. There are some history on that page of researching for authorship of the puzzle, and how many, slightly varies versions came to be. The author of the puzzle is Bill Williams.

Please help re-tagging the question as appropriate.

ACROSS
1. Area of Dog's Mead in square yards
5. Age of Farmer Dunk's daughter, Martha
6. Difference in yards between length & width of Dog's Mead
7. Number of roods in Dog's Mead times nine down
8. Year that Little Pigley came into occupation by the Dunks
10. Farmer Dunk's age
11. Birth year of Farmer Dunk's youngest child, Mary
14. Perimeter of Dog's Mead in yards
15. Cube of Farmer Dunk's walking speed in miles per hour
16. Fifteen across minus nine down

DOWN
1. Value of Dog's Mead in shillings per acre
2. Square of Farmer Dunk's mother-in-law's age
3. Mary's age
4. Value of Dog's Mead in pound sterling
6. Age of Farmer Dunk's son, Ed, who will be twice as old as Mary next year
7. Width of Dog's Mead in yards squared
8. Length in minutes Farmer Dunk needs to walk one and one-third times around Dog's Mead
9. See ten down
10. Ten across times nine down
12. One more than the sum of the digits in puzzle column two
13. Length of tenure in years of Little Pigley by the Dunk family

enter image description here

I have not found the solution to this particular version anywhere on the internet, solutions to other variations, are available, however I think I was able to solve it myself. My question is about the information missing from this puzzle, that is required to solve it. Many versions of that puzzle given in the internet provide many hints, some of which are not necessary for puzzle solution, but some of them seems to be crucial, e.g. it is not possible to solve the puzzle without this information. I would like your help to figuring out which of these hints really need to be added to the problem statement, which are "nice to have to prove you are on the right track", and which are completely unnecessary. I understand that the "nice to have" part is somewhat subjective, but the rest of it I think we can do.

There are some basic information given, that I think is necessary:

20 shillings = 1 pound sterling
1 acre = 4840 square yards
1 rood = ¼ acre
1 mile = 1760 yards

The rest of the post is under spoiler for the sake of those who want to try the puzzle by themselves first.

  • Hint: no number in the grid can start with 0. This seems to be crucial, if you allow 0 as the first digit you cannot really bootstrap the solution.
  • Hint: All calendar years in the puzzle are roughly contemporary. This can be an alternative to the previous hint. If you know that the year of birth (11 across) starts with 1, you can bootstrap the puzzle. I think that with some reasoning you can rule out 2, and since 0 and 3-9 would not be "contemporary" this hint would work.
  • Hint: You probably should at least be told that dog's mead is a rectangular plot of land In my opinion, this is not required. The clues talk about length and width, which makes it obvious.
  • Hint: One number in the puzzle in the area of Dog's Mead in roods, but it related to something in the puzzle quite different from that area. I think that this one is a "nice to have". It is not required to solve the puzzle but it can be used to prove that you are "on the right track". Specifically, this is number 32, the age of Martha (5 across).
  • Hint: Also, one of the number across is the same as one of the number down. This is number 792 which is 14 across as well as 10 down. Now it seems, like this hit is almost the must, otherwise the farmer's age could be both 72 and 62. If he is 62, one could argue that fathering the first child at the age of 17 is too young, but it is possible, so it looks like this hint is a must to make the puzzle unambiguous.
  • Hint: Current year is 1935. In my view this is completely unnecessary, you can deduce this from the rest of the information given.
  • Hint: Everyone's age is given assuming that they've already had their birthdays this year. This may sound obvious, but to me, this is necessary to state to avoid age ambiguity. Otherwise the age may differ by one and the puzzle won't have a unique solution.
  • Hint: All number in the grid are integers. This seems obvious and unnecessary. On the other hand, stating that all the number in the puzzle are integers, may be necessary, since a fraction multiplied by an integer as some clues ask, can give an integer.
  • Hint: No more than one numeral goes in each square. Not sure about this one. It seems usual for cross-number to always have this rule, but if you see such a puzzle for the first time, may be this needs to be told explicitly?
  • Hint: Edward is the eldest child, Mary is the youngest, and Martha is in the middle. This seems necessary, otherwise Martha's age is ambiguous. Either this, or we have to assess the hint above about the area of Dog's Mead in roods (32) as necessary.

$\endgroup$
  • 1
    $\begingroup$ There’s a 404 error on your link. $\endgroup$ – El-Guest Jul 2 at 2:13
  • 2
    $\begingroup$ @El-Guest thank you, I think, fixed. $\endgroup$ – Andrew Savinykh Jul 2 at 2:15
3
$\begingroup$

About Farmer Dunk's age

It is uncertain whether Farmer Dunk is aged 62 or 72 as both fit the clues.
If he is 62 then the children were born when he was aged 17, 30 and 40.
If he is 72 then the children were born when he was aged 27, 40 and 50.
We know that Mrs Dunk's mother is 86.

The puzzle is set in 1935, so the birthdays are
Mary 22 1913
Martha 32 1903
Edward 45 1890
Mother 86 1849

Farmer 72 1863 or
Farmer 62 1873

The typical reproductive age for women is 12 to 51.
But let's say the lower limit for marriage is 16 and childbirth 17.

So if 17 when her first child Edward was born Mrs Dunk was born in 1873.
And if 51 when her last child Mary was born Mrs Dunk was born in 1862.
That puts Mrs Dunk's birthday between 1862 and 1873.

But Mrs Dunk's mother also can't have been less than 17 when she was born.
So that now puts Mrs Dunk's birthday between 1866 and 1873.
And her age in 1935 must be between 62 and 69.

Mrs Dunk 69 1866 thru
Mrs Dunk 62 1873

If Farmer Dunk is 62 he married a woman the same age or up to 7 years older.
If Farmer Dunk is 72 he married a woman between 3 and 10 years younger.

I don't think there were two Mrs Dunks – that would have been a trick question.
But it changes what I previously thought to be likely and go for 72.
and Farmer Dunk did not find a wife until he was in his late twenties.

Going back to the hints.
You put "two equal answers" as a hint but it is the only essential information.
It should be part of the puzzle (the rest can be reasonable assumptions).
Your original post linked to a form of the puzzle that did state the condition:
Dog’s Mead, an old English puzzle

So in summary:
1) The earlier puzzle link was better.
2) The "two same answers" should be a clue, not a hint.

Previous Update:
I have solved the dilemma for 10 ac/dn, but have not updated the answer grid.
There was one thing I overlooked mentioned indirectly in the hints, which is:

The same answer in a crossword never appears twice
So 10d can't be 792 because it is the answer to 14a.
So 10d is 682 and in 10a Farmer Dunk is 62 years old: the more plausible answer.
Happy to have arrived eventually! A very good puzzle.

Edit:
An almost completed puzzle. I only looked at the hints afterwards, and found
that my first answer was incorrect having misread the clue for 6 down.
Revised –

I made the assumption that all are integer values and none begin with 0.
The divisions must result in whole numbers, quite a restriction.

I solved the clues by a combination of logic and possibility perms in
several steps from the types of clue: dimensions - values - ages - dates.
There is some interrelation between those categories, and so I also had to
go back over through the steps, applying limits from answers known later.

But 10a the farmer's age (and so 10d) is the only one with options:

enter image description here

If Farmer Dunk is 62 he was 17 when he fathered Ed now 45 (oldest known child)
If Farmer Dunk is 72 he was 50 when he fathered Mary now 22 (youngest known child) and did not father Ed until he was 27.

The first 62 is more likely.


About the hints

I found that none of them were really necessary in addition to what I assumed except they tipped me off to my original wrong answer by stating Martha's age.

I did implicitly make the assumption that dates were on the 1st January, but did not notice any possible conflict until reading the hints. I just took the ages and years at face value. I also made no allowance for gestation period, shifting perhaps a year doesn't make much difference.

| improve this answer | |
$\endgroup$
  • $\begingroup$ This hint, you mention at the top of the answer is actually opposite. It says that it does appear twice. For example see a variant here $\endgroup$ – Andrew Savinykh Jul 2 at 20:36
  • $\begingroup$ I noticed that your hints seemed to imply the opposite. I didn't read the hints in the linked question, but my interpretation seems to fit the more realistic age of Farmer Dunk. Your commented link here, seems to think that there should be two answers the same. But that might be a matter of opinion: "OMG two answers are the same!" vs. "No we can't have that!" $\endgroup$ – Weather Vane Jul 2 at 20:39
  • $\begingroup$ Out of curiosity, why do you think that 62 is more plausible than 72? I personally think that 72 is more plausible, because, as I explained in OP, 17 is a bit too young to have a kid. $\endgroup$ – Andrew Savinykh Jul 2 at 20:43
  • $\begingroup$ I explained that in the question: and perhaps Mrs Dunk (assuming he only had one wife) was getting beyond child bearing age for that. It is far more usual to bear children earlier in life, than later, especially in rural communities, although today, one might put that off "for a career". Assuming the one and only Mrs Dunk was only 13 when Ed was born, she would have been 36 when Mary was born. But if she had been 23, then 46 is getting as unrealistic as Farmer Dunk waiting until 27 to have the first child. 17 is not at all young: I had a girlfriend who was a grandmother by 40. $\endgroup$ – Weather Vane Jul 2 at 20:49
  • $\begingroup$ Yeah, not convinced, I assumed that he had children from different wives instead. Thank you for your insight though, it's helpful. $\endgroup$ – Andrew Savinykh Jul 2 at 20:57
0
$\begingroup$

I've spent some more time on this puzzle, and by and large I agree with Weather Vane's answer, that the only real ambiguity is about Farmer Dunk's age (10 across and 10 down).

My own conclusion, is that it won't hurt to insert these into general directions:

No answer in the grid starts with $0$. One single digit integer per grid cell.

In addition, we need another clue to resolve the ambiguity. Weather Vane suggested one in his answer, I'm suggesting this one, taken from the op:

One of the number across is the same as one of the number down

This will resaulve the ambiguity to make 10 down $792$ and 10 across $72$.

Below are my solution steps for reference.

Solution steps

15 across is a cube of Farmer Dunk's walking speed in miles per hour, and has two digits. It can only be $27$ or $64$. Let's assume it's $27$. This means that Farmer Dunk's speed is $3$ miles per hour or $88$ yards per minute. Let 8 down, length in minutes Farmer Dunk needs to walk one and one-third times around Dog's Mead, be $X$. Then, the perimeter length and and one-third will be $88*X$. yards ($\frac {3*1760} {60} = 88$), and the perimeter itself will have length of $\frac{88*X} {1\frac 1 3}=66*X$. Since $X$, which is 8 down has two digits and $66*X$ which is the perimeter and so it's 14 across has three digits, the only possible values of 8 down and 14 across are: $10$ and $660$, $11$ and $726$, $12$ and $792$, $13$ and $858$, $14$ and $924$ and, finally, $15$ and $990$. We can do similar calculation, assuming that 15 across is $64$. In this case the perimeter will be $88*X$, and possible values of 8 down and 14 across are: $10$ and $880$ and $11$ and $968$. This means that in either case for the two possible values of 15 across, the first digit of 8 down is $1$. We can fill it in the grid.

8 across is the year that Little Pigley came into occupation by the Dunks, and 13 down which is three digits is the length of tenure in years of Little Pigley by the Dunk family. Since 8 across is a four digit number starting with 1, the current year must start with either $1$ or $2$, and if so, then Birth year of Mary, 11 across, must also start with either $1$ or $2$. This means that the last digit of 9 down is $1$ or$2$. The last digit of 16 across, which is fifteen across minus nine down then is $5$ or $6$ if 15 across is $27$, and $2$ or $3$ if 15 across is $64$. Coincidently, this is also the last digit of 7 down, width of Dog's Mead in yards squared. No square ends in $2$ or $3$ though, so we can eliminate $64$ as a possible candidate for 15 across. We can now fill in 15 across as $27$.

Let's go back to possible 8 down and 14 across pairs above. 10 down, which is ten across times nine down ends with $2$. The last digit of 10 across is the same as the last digit of 8 down. 9 down as we learned before ends with $1$ or $2$. Looking at multiplication table the if we multiply $1$ or $2$ by a number, and the result ends with $2$ then this numbers should be either $1*2$, $2*6$. But we learned that 8 down is $10$, $11$, $12$, $13$, $14$ or $15$ so it cannot end with $6$. This means that it's either $11$ or $12$.

We can eliminate $11$ with the following reasoning. If 8 down id $11$, then 9 down ends with $2$ as was explained above. So, according to 16 across, which is fifteen across minus nine down, the last digit of 16 across and 7 down is $7-2=5$. We also know the second from last digit of 7 down, it's the first digit of 14 across, and if 8 down is $11$ as we assumed, 14 across according to the pairs we shown above is $726$. This makes last two digits of 7 down $75$. But 7 down is width of Dog's Mead in yards squared. No square can end with 75. $(10z + 5)^2=100z^2+100z+25=100(z^2+z)+25$, this means that any number ending with $5$ squared will end with $25$, and it is easy to see that if a number does not end with $5$ then squared it will not end with $5$ either. This rules out $11$ as 8 down. This leaves us with a single possible candidate $12$. We can fill it in the grid.

14 across then, as was shown above is $792$. We can fill it in the grid. 9 down now must be $11$. As shown above it must end with $1$ (because 8 down ends with $2$ and 10 down ends with $2$). It cannot start with anything but $1$ due to that 15 across is $27$ and 16 across is fifteen across minus nine down. If it was not $1$ then 16 across would start with $0$ or be negative. We an fill 9 down in the grid.

16 across now is $27-11=16$. We can fill it in the grid.

Let $l$ be Dog's Mead length and $w$ be Dog's Mead width. According to 14 across $l+w=792/2=396$. According to 6 across, difference in yards between length & width of Dog's Mead $10\leq l-w \leq 99$. Combining the two we'll get $10 \leq 396- 2w \leq 99$ or $193 \geq w \geq 149$. Which gives us only eight possible number with squares ending in $6$ (7 down): $154$, $156$, $165$, $166$, $174$, $176$, $184$, $186$. Out of these eight, only $174^2$ and $176^2$ end with $76$. 7 across, number of roods in Dog's Mead times nine down, only work out to be integer, if the width is $176$: the length will be $220$ then, and 7 across is $\frac {220*176*11} {4840} * 4 = 352$. We can fill it in the grid.

We can also fill in 6 across which is $220-176=44$ and 7 down, which is $176^2=30976$. 1 across, area of Dog's Mead in square yards is then $176*220=38720$. We can fill it in the grid.

12 down, one more than the sum of the digits in puzzle column two, will then be $8+1+2+7+1=19$. We can fill it in the grid.

We now know that 6 down, age of Farmer Dunk's son, Ed, who will be twice as old as Mary next year, is $45$. Next year he will be $46$, and Mary will be $23$, which gives us 3 down, Mary's age, $22$. We can fill it in the grid.

Let's now look at 1 down, value of Dog's Mead in shillings per acre, let it be $a$ and 4 down value of Dog's Mead in pound sterling, let it be $b$. We know how many acres Dog's Mead is now from 1 across: $\frac {38720} {4840}=8$. That gives as $a=\frac{b*20}{8}$. $b$ is a three digit number ending in $42$. If the first digit of it is more than one, then $a$ is over $600$, but when know that $a$ starts with $3$. Thus, the only possible value for 4 down is $142$. We can fill it in the grid. 1 down is then $\frac {142*20}{8}=355$. We can fill it in the grid too.

10 across is Farmer Dunk's age, and 10 down is ten across times nine down ($11$). Now because 10 across and 10 down share the first digit, Farmer Dunk's age cannot be $92$ or $82$. He cannot be $52$ or younger, because he has a $45$ years old son. So his age is either $62$ or $72$. This is an unresolved ambiguity. 10 down then is either $792$, or $682$. This does not change the remainder of the solution though.

2 down is square of Farmer Dunk's mother-in-law's age. It's a four digit number starting with $7$. The only possible values for Farmer Dunk's mother-in-law's age and their squares are: $84$ and $7056$, $85$ and $7225$, $86$ and $7396$, $87$ and $7569$, $88$ and $7744$, $89$ and $7921$. The second digit in the square is the first digit in 5 across, age of Farmer Dunk's daughter, Martha. The second digit of Martha's age is $2$ as it is already filled in. Since as we discovered earlier, Farmer Dunk's age cannot be more than $72$, Martha cannot be $92$, or $72$. If Martha is $52$, then the 2 down square is $7569$, which makes 8 across, year that Little Pigley came into occupation by the Dunks $1910$, which cannot be true because 13 down, length of tenure in years of Little Pigley by the Dunk family is a three digit number. Given Mary's age $22$ and partial year of birth $191x$, we can confidently say that it is not possible that three digit number of years has already passed since $1910$. This rules out Martha's years of $52$. The other two possibilities that match the list of squares above are $22$ and $32$. Mary is youngest, and is $22$, so Martha must be $32$ then. We can fill 5 across in now. We can also fill in 2 down with $7396$.

Let the last digit of 13 down, length of tenure in years of Little Pigley by the Dunk family, be $n$, and let it first digit, which is also the last digit in 11 across, birth year of Farmer Dunk's youngest child, Mary, be $m$. Along with 8 across, year that Little Pigley came into occupation by the Dunks, that we now know to be $1610$ and 3 down, Mary's age, $22$ we can write: $1610 + (100m + 20 + n) = (1910 + m) + 22$. Both sides of the equation are the current year. Simplifying we will arrive at $100m+n=302+m$. Since $m$ and $n$ are one digit integers, $m$ has to be $3$, solving for $n$ we will get $5$. 13 down then is $325$, which we can fill into the grid. This completes the puzzle.

Final solution

enter image description here

| improve this answer | |
$\endgroup$
  • $\begingroup$ I think I now agree and have added some more reasoning and comments to my answer. $\endgroup$ – Weather Vane Jul 3 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.