Impossible Dutchmen's Wives Puzzle

[@adrianmoraru9480]

Bro said it's impossible and then proceeded to solve it like it was nothing.

[@wesleydeng71]

No, he means the hog deal is impossible.

[@mahadevparmekar2565]

I solved it using the first method too. It's definitely not an impossible or trick question. All the information you need is there is the original problem.

[@jonathanodude6660]

@@mahadevparmekar2565 i only used the difference of squares formula to find 31 and 32. i founf 1,8 and 9,12 by constructing the set of squared numbers, knowing that there was an odd number of differences between squares to add, and looking for combinations that equalled 63. same outcome but not very efficient. i didnt think to factor 63 but then again, it is 2:30am...

[@jessicu511]

@@jonathanodude6660This was my approach too, and I was also very tired when doing it.

[@verkuilb]

It's common for Dutchmen to refer to each other by their last names. In which case, "Hendrick, Elas, and Cornelius" are the Dutchmen's last names. And therefore, their wives names are: Mrs. Hendrick, Mrs. Elas, and Mrs. Cornelius.

[@rc-fannl7364]

Except those names don't really sound like last names.

[@verkuilb]

@@rc-fannl7364Yukon Cornelius would beg to differ.

[@Shadowomega008]

this logic puzzle is from 1740. so yeah its an Alternative logical answer

[@rc-fannl7364]

@@verkuilb Lol, that's a fictional character though

[@joostvanrens]

Only if those men are corpsballen

[@gdwww0]

In the older version of 1739, the wives names were spelt Geertruii, Catriin and Anna, which should have been in proper 18th Century Dutch Geertruij, Catrijn and Anna. Elas was originally Claas.

[@F.E.Terman]

Ah, that explains much. I am Dutch, but recognized only two names as Dutch; three if assuming 'Elias'. The above explains it better!

[@koenth2359]

This makes sense. Hendrick, Cornelius and Anna are common names in Dutch, but the other names from the 1740 version sound more like Old Norse to me.

[@ARex545]

I solved it a bit faster by using a shortcut/trick Here's how I did it: Since it is mentioned in the question that the difference between the spendings of husbands and wives is $63 and we know that $63 is also a difference of 2 squares, I did a quick trial and error. 8² - 1²= 63 (Takes no time to calculate this) 12² - 9² = 63 (Just added 63 to 81 since it was next to 8 to check) Now comes the trick, since it is mentioned in the question that Hendrick or H buys 23 more hogs than Katrün or K; and Elas or E buys 11 more hogs than Gurtrün or G, we can easily deduce that E=12 and K=1. We now need a difference of 23 from 9 or 8 which will be equal to 32 or 31. From this point on the question is pretty straightforward. I didn't use equations to solve for (1,8)(9,12)(31,32). Just some good ol' trial and error and using the info given in the question. Took me about 2-3 mins to solve using this method.

[@axiezimmah]

same, got it within a few minutes, you know all the numbers have to be relatively low anyway so it didn't take long to check. Especially not in Excel where you can make some formulas to do the hard work for you

[@angelicasmodel]

I couldn't think how to solve this algebraically, so I did trial and error, too. I immediately spotted 8 and 1, and from there, quick spotted 12 and 9. Then, I worked my way up to 31 and 32 using a calculator. And yes, I realised after doing this, that I could have done it in Excel. The whole thing probably took me as long as I would have taken algebraically.

[@rebeccalynn7795]

me too lol

[@derrickthewhite1]

@@axiezimmah Yeah, I brute forced it in excel. This is the digital age. Once I knew the three pairs, everything else was easy.

[@Aaarrrgh89]

Yeah, I also used brute force in a similar way. Not too complicated.

[@brunogrieco5146]

I solved the first part of the puzzle, finding the integer squares, by brute force using Excel. The factorization idea was genius.

[@larspos8264]

Its Hendrik with Anna, Elas with Katrün and Cornelius with Gurtrün. They all spend squares (as they buy a whole number of hogs). Only option with a difference of 63 are 8&1, 12&9 or 32&31. To see this: take all possible factorizations of 63 and match them with a-b and a+b, as a^2-b^2=(a-b)(a+b). Then it’s just testing all possible cases.

[@LiteBulb88]

Who is buying 32 hogs for $32 each when you can buy 1 for $1? 😛

[@remytherat1357]

what if you need more than 1 hog

[@brokendoll3368]

The merchants keep your names. So next time you want to buy your second hog, it'll cost $3 so you'll have paid $4 in total.

[@mgg4338]

​@@brokendoll3368after your 7th hog it becomes cheaper to ship them from Belgium instead

[@Makememesandmore]

Rinse and repeat

[@avienated]

A person who can sell 32 hogs for $33 each ;) Or Hendrick could pool the hogs with his wife's and sell all 63 pigs for $63 each ^^

[@superneenjaa718]

(Hendrich, Anna), (Cornelius, Gartrun), (Elas, Katrun) Solved it while traveling by train. It'd have been a lot easier with a pen and paper. I kept calculating and forgetting the numbers.

[@rogercarl3969]

I heard that Hendrick and Anna are having BBQ again. All invited.

[@davidsanders5861]

BYOB though...there's no way they can afford beer after blowing that much on hogs.

[@mofleh177]

Possibly they wanted to keep the hogs as Pets all 63 of them!

[@thepotatooverlord2275]

hendrick after spending $1024 for 32 hogs

[@eren_gamer_2009]

i am broke and her wife left her. + you forgot the payment of his wife hogs by him.

[@londonalicante]

Alternative solution: none of them are married. The wives threw the husbands out for filling the house with overpriced hogs.

[@benhur4154]

I dunno... even if you bought 32 of them, $32 for a whole hog is pretty damn cheap.😁

[@bobli840]

Well, someone bought a hog for a single dollar, so 32 is quite a distance from it. Besides, 1740 was a long time ago. $32 could be quite a lot back then

[@furbyfubar]

Anna has no real case against her husband then, seeing as she just bought 31 hogs for $31 each herself.

[@sabyasachirimpa]

You could not solve the problem, so here you trying to become smart.

[@Friend_of_the_One-Eyed_Ladies]

Who the hell keeps hogs in the house?

[@TheFman2010]

I used an Excel spreadsheet to work out the squares and found three pairs of numbers whose squares differ by 63: 1 and 8, 9 and 12, 31 and 32. Not as elegant as your method, but works just as well.

[@nijucow]

That's the worst deal I've ever heard of

[@a.hardin620]

Yes Sir. It’s some kinda of bizarro economics! 😂

[@hhadzimu]

But you HAVE heard of it.

[@HenrikMyrhaug]

That's got to be the best deal I've ever seen. So it would seem...

[@ericherde1]

Weirdest deal, but I’m pretty sure you’d have to buy a lot of hogs for it to not be an amazing deal. I’m not involved in the agricultural market, but surely a hog normally costs way more than a dollar, right?

[@acarbonbasedlifeform70]

The worst trade deal in the history of trade deals. ... Maybe ever.

[@alexbenjaminrosenthal3989]

I finally got one of these! Easier than I thought. Once you look through a perfect square table its easy to find the ones 63 apart

[@daniel-mk5lz]

I was able to do it without the equations, just sketched it up and tried a few possibilities .

[@laomedondardanos5658]

I found the solution a bit more experimental: First I tried a few differnt numbers to get a feeling for the problem and found the (8, 1) pair as a solution. Then I thought about the clues and deduced that if 1 is a solution for a women, I can find other solutionions exists with 11 more or 23 more. So I tested 12 and 24 and found the (12, 9) pair. Then I thought that because I already found a possibility for the "Elas has 11 more hogs than Gurtrün" clue, I might try the other clue that Hendrick has 23 hogs more and tested 9+23=32 which led to the third solution (32, 31) and then I just needed to connect the names.

[@voided_sun]

When they say school teaches a lot of useless stuff. These special rules for solving quadratic equations really takes the cake.

[@NimrodClover]

I paused the video right after the question was posed... Isn't the question, "What is the name of each man's wife?" So, isn't the answer, "Gurtrun, Katrun, and Anna" ? That is their names, correct ? It says nothing about the name being the specific woman married to each man, but rather the name of each man's wife. It you just say their names again, we know that the women are married to the men in some way as the text says they are wives of each man, thus that is the answer to the question asked... Gurtrun, Katrun, and Anna. Simple. Now if they had asked to match each man to his wife, that would have been a totally different question.

[@mathmethman]

The solution as presented makes the assumption that each of the six people buys a different number of hogs. This is not stated in the question, but needs to be deduced. Having established possible pairings as (32,31) (12,9) and (8,1) it is clear that Hendrick must buy at least 23 hogs and therefore buys 32. (Katrun therefore buys 9.) What about Elas? He buys at least 11 hogs, so he *could* buy either 32 or 12. However 32 is 11 more than 21 and nobody buys 21 hogs. So Elas buys 12. (Gurtun therefore buys 1.) Now we can deduce that Hendrick's wife buys 31 hogs and Gurtrun's husband buys 8. So in fact the six DO all buy a different number. But you can't assume this to begin with.

[@lafiosca]

I wondered if someone else would post this.

[@skz01_]

There were no assumptions made. We can deduct that Elas bought 12 and Gurtrun bought 1 because there are the only pair of possibles values of husband bought and wives bought that differs by 11. Moreover as you said, Elas cant buy 32 because 32 is 11 more than 21 and 21 is not a possible mumber of hogs bought by the wives.

[@mathmethman]

@@skz01_ At 7:04 the commentary says "As 32 is already taken..." This phrase surely assumes that everyone bought a different number of hogs and the rest of the sentence proceeds on that assumption. (It turns out the assumption is correct, but it's still an assumption at this point.)

[@nkrln9648]

The problem itself stated that A buys more than B, and each pair pf couple has 63$ difference. So in the beginning its already safe to assume that the number of each person is not known exactly, either same of different. So we must work out the method of all possibilities. Even if you make the assumption that one person could buy the same as the other wont make the answer/method easier. No assumption is made, i work it out using almost the same method as presh did, so it is the way it should be done whether they have same number of each people same or not. The pairing of 32 and 21 etc comes from a logical method, you can try that no people with same number could result in solving the problem anyway. No answer is available also other than the pairings that is stated, so some people have or dont have the same hogs doesnt matter

[@mathmethman]

@@nkrln9648 I think that last sentence of yours is sort of what I was trying to say. It doesn't matter whether or not the puzzle includes the fact that everyone buys a different number of hogs. It happens to be true, because there is only one solution to all the other given facts, and that solution requires each person to purchase a different number. My point is that we don't KNOW it is true until we have worked out the answer, and yet, as stated above (at 7:04) the fact that "32 is already taken" is used to eliminate the possibility that Elas bought 32 hogs. This is false logic. Where does "32 is already taken" come from? The speaker is ASSUMING that Elas can't have bought the same number of hogs as Hendrick.

[@FatherEli]

One that I actually got! I didn't use either method shown in the video, but they're both superior in terms of time spent, I reckon. I got as far as determining 'Right, they've each spent a square number amount of money, let's just check the first few square numbers and see which ones are exactly 63 apart" I could definitely have saved -some- time after finding the second pair if I'd thought to add 23 to the lower parts of each pair and check those, but that didn't occur to me until after.

[@prufrock1977]

I got this right on instinct. Logic of the size comparison helps to reveal the answer.

[@Angi_Mathochist]

I just solved it with a spreadsheet. Put husband numbers across the top, wife numbers down the side, subtraction in the middle, use conditional formatting to highlight all 63's, easily see that beyond a certain point everything is increasing so there are only 3 solutions. Save everything to values, delete all non-essential rows and columns to see more easily what we have, match husbands and wives to remaining clues, done.

[@Arken_games]

the squares really help shrink the possible combinations. then the differences help to make best guess

[@a.hardin620]

You created that fancy equation and I was expecting you to solve it to get the answers but then you simply did a guess and check knowing that the difference of squares had to be 63! 😅 Great puzzle!

[@andreasboe4509]

Great and simple. Awesome. A real classic.

[@wombat4191]

I used this very informal way to solve this in my head. We know that for subsequent whole numbers, the difference of their squares is always their sum: m^2 - n^2 = n+m, where m = n+1. I guess that would require proving, and while that shouldn't be too hard to do, I won't bother here. Based on that, I determined that 32 and 31 is the largest pair of whole numbers that have their squares differ by exactly 63. So I know that one husband bought 32 hogs, and his wife bought 31. Now, I can also determine that for a non-subsequent pairs of positive whole numbers, the difference of their squares is just sum of all the subsequent pairs between them, which works out to (n+k)^2 - n^2 = (n+k)+2(n+k-1)+2(n+k-2)...+2(n+2)+2(n+1)+n, where n and k are whole positive numbers (this actually also works out to be k^2 + 2nk, which Presh also got in the second method in the video). This is saying that for example with 13 and 16, the difference of their squares is 13+14+14+15+15+16. That isn't really that useful for big k values, but it lets me see that the average between n+k and n should be the difference divided by 2k, and that odd difference means that k is odd, and even difference means that k is even. Now that all seems needlessly complicated, but I didn't think of any of those equations in my head, it's just a formal way of writing it out. I just intuitively thought of the two findings in the end of the previous paragraph, and based on them I could approximate that the only way for Hendrick to buy more than 23 hogs is if he buys 32. That means that Katrün buys 9. Now, what value of k would make it so that 2k*9 is fairly close to 63 but not over it (because of the average of n and n+k in the previous paragraph)? That's k=3, so I check if the difference of the squares of 9 and 12 is 63: 9+10+10+11+11+12 = 63, so 9 and 12 is a valid pair. I also saw that 1 and 8 are an obvious pair, as it's just 64-1=63. So Elas bought 12 hogs and Gurtrün bought 1, since those make the difference if 11. Therefore, Elas and Katrün are a pair, and since Hendrick and Gurtrün are obviously not, it means that Hendrick and Anna, and Cornelius and Gurtrün are pairs. The only question remains, why do Hendrick and Anna need so many hogs?

[@agytjax]

This is definitely not an "impossible" puzzle. I could solve it using excel - finding the squares, their difference which would be 63 and their originals that would be the difference of 23 and 11 !

[@melissascheid841]

This is what I did. I had it calculate the square root of X^2 -63. Solutions were when the result was a natural number

[@hhgygy]

Exactly the same method I used 😎

[@benhur4154]

Chalk up one more for Excel here

[@foogod4237]

To be fair, the readers of The Ladies' Diary in 1743 probably didn't have easy access to Excel...

[@Posby95]

@@foogod4237 True, but it can be done with pen and paper too.

[@cesardelgadillo4139]

Hendrick and Anna are some real hog lovers, no wonder they are together

[@beakt]

I paused at 2:10 and solved it using Notepad and Calculator in five minutes. I love puzzles like these.

[@willemakkermans4067]

Same here, used the back of a shop receipt, and just used a calculator to quickly square 31 and 32. Turned out I used method 1 to the letter.

[@DemoniqueLewis]

I simply guessed the correct answer given the fact it is a riddle and only four names were given. Nice to know there is a way to actually figure it out.

[@jeremiahlyleseditor437]

This was a clever little word problem. It's a format I've never seen before even in school

[@Posby95]

I think I solved it! This all revolves around square numbers; especifically, those from 1 to 1024. "Each husband has spent 63 dollars more than his wife", and this is true when: - She spent 1 dollar (1 hog) and he spent 64 dollars (8 hogs). - She spent 81 dollars (9 hogs) and he spent 144 dollars (12 hogs). - She spent 961 dollars (31 hogs) and he spent 1024 dollars (32 hogs). From the difference in the number of hogs, we deduce that: - Katrün bought 9 hogs and Hendrick bought 32. - Elas bought 12 hogs and Gurtrün bought 1. We combine this new info with the old one, and by process of elimination we infer that Anna buys 31 hogs and Cornelius buys 8. Now we have all couples: Gurtrün/Cornelius, Katrün/Elas, and Anna/Hendrick. This was fun to figure out. Now I'll watch the video to see if MYD solved it the same way.

[@josephprozinski9311]

Excel is your friend in figuring this out quickly doing it in a brute force way.

[@manudude02]

I was tired when working this out, I used the fact that the difference between x^2 and (x+1)^2 was 2x+1, and then combined it with the arithmetic sum formula. Difference of 2 squares is so much easier lol. At least it still leads to the right answer....

[@daniellenze7321]

I googled perfect squares 1 through 1000, and then calculated the next one which is 32^2. I stopped at 32 because the difference between 32^2 and 33^2 is greater than 63 so there wouldn't be anymore possibilities. Then I just subtracted 63 from every number on the list and if the difference was a perfect square then I knew I had a pair of numbers that would work. I got the three pairs that MYD got in the video and used the same logic to get the answer. It isn't a very pretty solution but it works.

[@elf4914]

One quick note: The riddle never stated that no person bought the same number of hogs, so at 5:52, the statement is incorrect. We don't actually know that each pair is unique, just that these are the possible pairs, but it could be that two husband-wife pairs bought the same arrangement of pigs. Of course, when you work out all the way to the answer that is the case, but if it hadn't been, then that assumption could throw off the entire attempt to solve the riddle.

[@KindredBrujah]

That's incorrect. We get enough information to know the quantities purchased by one out of each husband-wife pairing, so given we know that each husband spent exactly the same amount more than their wife, we know all three sets must be occupied.

[@pbenikovszky1]

9:18 PRESH NOOOO, NO GUESSING :D :D just say that x is a whole number, therefore (63-n^2)/2n is a whole number too, so 63-n^2 must be even, and n < 8, so it can be 1, 3, 5 or 7, and we can rule out 5 easily, as 63 and 5 are relative primes, this means that 2n needs to be a divisor of 63 in order for the whole expression to be a whole number, but10 does not divide 63, so it is ruled out ;)

[@57thorns]

Why would the Dutch pay in dollars? So, diving into the fascinating world of old currencies, yes, it would be possible: The Netherlands introduced its own dollars in the 16th century: the Burgundian Cross Thaler (Bourgondrische Kruisdaalder), the German-inspired Rijksdaalder, and the Dutch lion dollar (leeuwendaalder). The latter coin was used for Dutch trade in the Middle East, in the Dutch East Indies and West Indies, and in the Thirteen Colonies of North America. But the English word "dollar" was first use in America for the Spanish piece-of-eight (peso).

[@sahibpreetsingh838]

Nice bring these type of puzzles more often

[@antonymartin8704]

The 2 centre segments are clearly greater than half the area of the cirle. Thus the side segments then must sum to less than half the are and devided by 2 means the shaded area must be less than a quarter.

[@Makememesandmore]

I thought this was trial and error, then you reminded me algebra exists

[@tooba6290]

Haha same

[@CornerCaseGirl]

Me too! Just calc`d all the perfect squares and their differences in a reasonable range. As rusty as my algebra is, that was probably faster!😕

[@emeraldaly7646]

The hog salesman has a weird pricing scheme, but I guess it worked out. Sold a crapton of hogs 😂

[@graysonjudd6634]

I did it the long way and wrote out squares until I got to 32 squared and saw it was 63 off from 31 squared. Then I looked for other squares that had a difference of 63. 8 and 1 stood out immediately, and 12 and 9 took a second to find. I sorted out what husband matched which wife the same way. But I have the added knowledge of knowing how much each couple spent. C&G spent $65, E&K spent $225, and H&A spent $1987.

[@henrydenner5448]

I solved it by using a mix of the first and 2nd methods. It is a cool problem. Enjoyed it.

[@YaztromoX]

I solved this, but got a bit lazy and instead of factoring I just setup a spreadsheet that mapped the values 8 through 50 with sqrt(A1^2-63), used auto-fill to copy down, and looked for where the solution was an integer. 😊

[@Khantia]

I solved it, but with a bit of brute forcing. I think I could've figured out a more delicate solution, but the brute forcing worked, so I didn't bother trying.

[@tremkl]

Fun note on the “guess and check” portion of the second solution is you can use modular arithmetic to know which need checking, and it helps build an intuition for why they’re the same process. Since you have 2n in the denominator, you know the numerator needs to be divisible by 2n. Since 2n is always even, and 63 is odd, we know n must be odd to generate a whole number and can throw out any odd n. Since 8^2 is 64, any number of 8 or higher will generate a negative number and must also be thrown out. Next, n^2 will always be evenly divisible by n, and so 63 must be evenly divisible by n to generate a whole number. Combining these, the only odd devisors of 63 less than 8 are 1, 3, and 7, and so we have 3 solutions.

[@cesardelgadillo4139]

How do you eliminate 5 from the possible options, through guess and check?

[@tremkl]

@@cesardelgadillo4139 Because 63 isn’t divisible by 5. If 63 isn’t divisible by n, then 63 - n^2 won’t be divisible by n, and therefore it won’t be divisible by 2n.

[@cesardelgadillo4139]

@@tremkl got it👍

[@TheBlobik]

I have cheated a bit with Excel spreadsheet that just quickly computed sqares of a range of numbers, added 63 and computed a square root of that to see which would come out as whole numbers, and once I got the pairs the rest was trivial. Still happy i solved it on my own ;)

[@qc1okay]

He made an error at 6:00 in the video. The question didn't prevent people from buying the same number of hogs as someone else. It turns out that such a scenario turns out not to work, but it still had to be considered. For example, we had to consider that two husband-wife pairs both bought 32 and 31. That's quickly eliminatable, but it still needed to be considered.

[@foogod4237]

That _technically_ was prohibited implicitly, because if two couples bought the same numbers, then it would be impossible to tell which of the wives was married to which of the husbands, and the problem would be unsolvable (even if the math worked). Therefore, the presumption that the problem has a solution means that each of the couples must have bought a different number of hogs. But I agree that for the sake of completeness, this should at least have been mentioned...

[@qc1okay]

@@foogod4237 Why would two couples buying the same numbers of hogs prevent a solution? Only if one does some math do we know that that cannot happen, and math is how to get the solution, so you're saying that solving the problem means that we know that duplicates aren't right? I agree that if we solve the problem, we do know the answer, and the answer happens not to use dupes.

[@michaeldegrave5905]

@@qc1okayIt makes a unique solution impossible, so it would be unsolvable. Therefore it’s not an answer to the puzzle.

[@jdogg5639]

It's explicitly not possible for duplicate pairs to be the case after you've found the 3 possible husband/wife number pairs, due to the differences in number of hogs purchased between the people that are stated. The 23 and 11 differences force Elas and Katrun to be the 12,9 pair with no alternate option, and then Gurtrun needs a husband on the 8,1 pair and Hendrick needs a wife on the 32,31 pair. In other words, it doesn't rely on assuming they are 3 unique pairs - the 3 unique pairs are a direct logical conclusion of the givens

[@johntippin]

Yes, thank you!

[@JDrapic]

I cheated a bit. Used an excel sheet with =(row()-1)^2 in column A and =(column()-1)^2 in row 1 and expanded it out. Then I put in B2 =$A2-B$1 and pasted that formula until I'd made a square of values to AG33. Next I set conditional formatting to highlight 63 and -63, which highlighted 3 pairs of results (one positive, one negative, which would correspond to the same absolute values). I now had my lowest possible sets of values, which would be 1 less than the row / column they were in. I came up with 8, 1; 12, 9; 32, 31. From there, it was filling in what numbers matched. 32-9=23, so that solved for H and K. 12-1=9, so that solved for E and G. After that, fill in the blanks for C and A and you get the pairs of H, A; E, K; ad C, G.

[@RichCale]

My daughter paused the video to see if she could figure it out, but she couldn't without writing stuff down. I looked it over and gave the answer before the math was done. I just used a bit of logical thinking given the two lines about he had x more than her.

[@mikerapp8163]

Good illustration. I need to brush up on my algebra 😮

[@quentind1924]

If i was one of them, i’d buy one hog, come back 15 minutes later, buy another and repeat until i have enough...

[@foogod4237]

Realistically, "market" sales for livestock like this were (and are) usually done as auctions, so (a) you can't just come back later, and (b) the price depends on what other people are bidding, it's not set by the seller. It seems like some of them were just luckier (or more savvy) than others in which lots they bid on..

[@willemakkermans4067]

Or we can assume for the sake of making it make sense that they had one very sickly hog to sell, some groups of slightly better hogs that they sell as groups (of 8, 9, and the even better looking group of 12), and 2 top groups of 31 & 32 healthy hogs. Only I just had word that the healthy ones were used as barbecue meat.

[@DANIELMABUSE]

This would be a good puzzle for GCSE maths students.

[@jacobgoldman5780]

Before watching the solution here is my attempt: Let h=# of hogs Hendrick buys, e= # of hogs Elas buys, c=# of hogs Cornelius buys, g=# of hogs Gurtrun buys, k=# of hogs Katrun buys, and a=# of hogs Anna buys. Then Hendrick pays h^2 for his hogs, Elas pays e^2 for his hogs, etc. Also from given information h=k+23 since Hendrick buys 23 more hogs than Katrun, and e=g+11 since Elas buys 11 more hogs than Gurtrun. Next since we see that each husband spent $63 more than his wife it seems important to find perfect squares that differ by 63 since that's what the price totals are. That would mean x^2-y^2=63 and we can factor a difference of squares as (x+y)(x-y)=63 and since 63=63x1=21x3=9x7 we could have x+y=63, x-y=1 so x=32, y=31 is a solution and x+y=9, x-y=7 means x=8, y=1 or finally x+y=21, x-y=3 so x=12, y=9 Thus each husband, wife pairing is either 32 hogs and 31 hogs respectively, 12 hogs and 9 hogs or 8 hogs and 1 hog. In order to have 23 hogs more than someone else Hendrick must have either 32 or 31 hogs, but if he has 31 hogs the 8 hogs person would also be a husband which is a contradiction so we know that Hendrick has 32 hogs and Katrun has 32-23=9 hogs. The only difference of 11 hogs is between 12 hogs and 1 hog, so Elas has 12 hogs and Gurtrun has 1 hog. The only husband left is Cornelius, so he must have 8 hogs to match with the 1 hog of Gurtrun, and the last wife is Anna, so she must have the unclaimed 31 hogs with Hendrick. Solution: Hendrick's wife is Anna, Elas' wife is Katrun, and Cornelius' wife is Gurtrun.

[@pietergeerkens6324]

There are only three ways to have a difference of squares equaling 63: 9, seven times, yields: 1² + (3 + 5 + 7 + 9 + 11 + 13 + 15) = 8² 21, three times, yields: 9² + (19 + 21 + 23) = 12² 63, one times, yields: 31² + (63) = 32² Note that 9, 21, and 63 are the only odd factors of 63 greater than its square root; while 7, 3, and 1 are its only odd factors less than its square root. In each case the middle parenthesized number, times the count of parenthesized numbers, equals (the required difference of) 63. That's Geerkens' Gnomon Guideline.

[@lucasoorschot780]

Nice puzzle, though it could have used the actual currency of the time, the "dutch guilder" (abbreviated as "fl", short for the alternate name "florijn") and more realistic names (in Dutch the only - and extremely rare - use of ü would be to indicate the start of a new syllable if there is any ambiguity, which is not the case in Katrün and Gurtrün).

[@yurenchu]

The names were probably meant to be spelled _Katrÿn_ (or _Katrijn_ ) and _Geertruÿ_ (or _Geertruij_ ), but the non-Dutch (possibly Slavic) submitter of this puzzle couldn't type _ÿ_ and also presumed that ÿ is identical to ü (because the Cyrillic letter y is called _upsilon_ , and is transcribed into Western/Latin script as u ).

[@brokendoll3368]

Okay. If you pay n per hog for n hogs (the prices were definitely made up by the merchants), then the total amount payed by one person is n², a perfect square. Each perfect square n² is equal to the sum of all odd numbers from 1 to 2*n-1. Therefor, beyond 32², the difference between to prices is too high, since 2*32-1 is exactly 63. So one pair might have bought 31 hogs for the woman and 32 for the husband. Making a list with all squares till 32² and all possible differences, there are only three options: (31|32), (9|12) and (1|8). Hendrick bought 23 more than Katrün. Therefor Hendrick must have bought 32 and Katrün bought 9. Elas bought 11 more than Gurtrün, that must be 12 and 1 respectively. That leaves 8 for Cornelius and 31 for Anna. Therefor, Anna is married to Hendrick, Katrün is married to Elas and Gurtrün is married to Cornelius. Also, Gurtrün and Cornelius are the only ones who could deduct this like us, the others money spending habits do not seem too smart.

[@TheRMeerkerk]

If you buy x hogs, then total price you pay is x^2. The easiest solution to the equation x^2-y^2=63 would x=8 and y=1, meaning husband buys 8 hogs and wife buys 1 hog. The puzzle does not state whether there is a man who bought 7 more hogs than a woman, so let's find some more solutions to that equation. Since we know we found a difference in hog count of 7, we can simply check if lower differences are possible as well. Hog count difference of 6 for example cannot be the case between a husband and wife, because: (x+6)^2 - x^2 = 12x + 36 = 63 => 12x = 27 => x = 9/4. If we generalize for all hog count differences z, we get: (x+z)^2 - x^2 = 63 => 2zx+z^2 = 63 => x = (63-z^2)/(2z). If we plug in all z less or equal to 7, we get these x: 1, 9/4, 19/5, 47/8, 9, 59/4, 31. In other words there were a husband and wife who bought 8 and 1 hogs respectively, 12 and 9 hogs respectively, and 32 and 31 hogs respectively. 32-9=23 and 12-1=11. So Cornelius bought 8 hogs and his wife Gurtrün 1 hog, Elas bought 12 hogs and his wife Katrün 9 hogs, and Hendrick bought 32 hogs and his wife Anna 31 hogs.

[@TheOtherBill]

I would have made the last line of the puzzle "What is Gurtrün's husband's name?" just because it makes it seem much harder (to me, at least).

[@manishadhikari4132]

At @7:10 you said “32 is already taken so” but nowhere the question indicates that two couples cannot have same pair of number of hogs. We know that Elas could not have 32 hogs because then Gurtrun would have 21 hogs but I think your logic was flawed even if the conclusion is correct

[@antoniosamaranchi3223]

Agree

[@pauljomerluz8058]

Its implicitly stated that "two couples cannot have the same pair of number of hogs", here's why; Lets breakdown the clue in two statements: Statement 1: Hendrick buys 23 more hogs than Katrün Based on this, its obvious that Hendrick bought 32 hogs, right? Because of that logic, his wife must've bought 31 hogs in order for the condition "all husbands spent 63$ more than its wife" to be true. Since Katrün's hogs can't be more than 23, either Gurtrün or Anna is the wife of Hendrick. Based on this let's look at what we have established. 1. Hendrick has 32 hogs while his wife has 31 hogs. 2. Katrün's husband must've bought 12 hogs, since we know Katrün bought 9 hogs. (32 - 23 = 9) Statement 2: Elas bought 11 more hogs than Gurtrün. Based on this, Elas could not have bought 32 hogs (u already know why), he also could not have bought 8 hogs, since he bought 11 more hogs than Gurtrün, and she can't have a negative amount of hogs. Continuing on what we have established on S1, we have determined based on S2 that: 3. Elas is the husband of Katrün, since based on S2, he could've only bought 12 hogs. 4. Gurtrün must've only bought 1 hog (12-11 =1), which in turn making Anna the wife of Hendrick since she is the only possible wife who must've bought 31 hogs. 5. By process of elimination, the only possible husband of Gurtrün is Cornelius, who must've bought 8 hogs (again to satisfy the condition of 63$ more dollars spent than its wife), Due to the conditions, it logically can't have two couples who have the same number of hogs, otherwise the statement 1 and statement 2 will not be true.

[@manishadhikari4132]

@@pauljomerluz8058 that’s true. That’s what I concluded as well. I am just saying that the reason is not “32 is already taken”.

[@Friend_of_the_One-Eyed_Ladies]

Pretty obvious and simple puzzle. I didn't bother with all the formulas. Obvious that to have a difference of 63, one number of each pair must be 8 or larger. So I just started subtracting 63 from each square starting from 8², and making a list of those that had results that were also squares (e.g., 32²-63=31²). Once I found 3 pairs, I applied the last 2 pieces of information. Child's play.

[@Jupiterninja95]

I found it but it took some trial and error to find difference of squares

[@arctic_haze]

I misunderstood the pricing oof hogs but came with the same pairings of Durch marriages as at the end of the video. But mine solution took me 15 seconds. Sometimes it is good to misunderstand the instructions.

[@mieskeb8095]

Presh: * Dutchmen's riddle * Also Presh: "Sorry for my pronounciation" My Dutch brain: hehe this will be funny Presh proceeds saying German names Sadsmileyface Still nice puzzle though

[@axiezimmah]

Yeah, these names aren't Dutch, classic Americans don't know the difference between Dutch and German

[@FatherEli]

​@@axiezimmah To be fair to Presh, that's almost certainly an error from the source material. The English-speaking world of the 18th century was largely peopled with folk laboring under the belief that their ignorance of other cultures was no such thing.

[@MarkHaanen]

Given that the 'Dutch girl' in Friends had an obvious Russian / Eastern European accent, this is fine. Plus, the "Pennsylvania Dutch" are German, so it's an easy mistake to make.

[@skoosharama]

As Presh noted, this puzzle is from a journal published in 1740. Back then, English-speakers still referred to both Dutch and Germans with the word "Dutch", and distinguished between them, if they felt it necessary, with the terms "Low Dutch" and "High Dutch". Eventually, "High Dutch" got replaced by the term "German", and then "Low Dutch" just got shortened to "Dutch", because there was no longer any need to contrast it with "High Dutch." But that process had not been completed in 1740. Source: a post titled, "How exactly did the demonym "Dutch" become associated with the Netherlands rather than Germany?" on the AskHistorians subreddit.

[@PADABOUM]

It is quite easy and I did not need to use paper. As the trick is to realise you have X2- Y2 = (X-Y)x (X+Y) i imagine it would have much more difficult for people to,e at the time as very few would have the basic mathematical education we have now commonly.

[@yurenchu]

It would have been more fun if the question had instead stated (for example): "Each husband spent $3900 more than his wife. Hendrick bought 14 more hogs than Katrün ; Elas bought 18 more hogs than Gurtrün."

[@PeteOHair]

I get the solution, but it is very counterintuitive to see that cost per hog is vastly different between the couples (husbands spends $63more) Hendrick-Anna 1 hog, Elas-Katrun 3 hogs, and finally Cornelius Gurtrun 7 hogs so I get what author of this exercise intended but this defies any logic of shopping (to me)

[@vladimirrainish841]

Hendrik is married to Anna, Elas is married to Katrün, and Cornelius us married to Gurtrün. There are 3 possible solutions in integer numbers to x² - y² = 63 equation: 32, 31 12, 9 8, 1 So that Hendrik bought 32, 23 more than Katrün who bought 9 and Elas bought 12, 11 more than Gurtrün who bought 1 x² - y² = (x+y)(x-y) = 63 = 32² - 31² = 12² - 9² = 8² - 1² = 63*1= 21*3= 9*7 since 63 = 7*3*3*1

[@azrobbins01]

Is there a graphical representation of this problem with intersecting lines?

[@vegveg582]

no

[@MoonGlow22]

K 23 H 29 G 11 E 23 A 29 C Best visual ı can do on comments

[@vladspellbinder]

I'm going to ignore the improper use of "impossible" here and instead focus on why Hendrick and Anna need a total of sixty-three hogs and why they have enough disposable income to spend a total of one thousand nine hundred and eighty-five dollars on them.

[@ChristofAbsolution]

This is not a math problem. You can solve this fastest by just using a pinch of logic and no math: Hendrick has 23 hogs more than Katrun (which is not the difference we're looking for) so we know Hendrick must be married to either Gurtrun or Anna. Elas has 11 more than Gurtrun (which is not the difference we're looking for) so we know he must be married to Katrun or Anna. Since both men can't be married to the same person (assumed) they both can't be married to Anna, but one of them HAS to be. This leaves the only 2 possibilities: Hendrick married to Anna, Elas married to Katrun, and Cornelius to Gurtrun or Hendrick married to Gurtrun, Elas married to Anna, and Cornelius to Katrun At this point we KNOW that Hendrick is hog rich (based purely on the difference in hogs), so he can't be married to Gurtrun - leaving Anna (with her unknown number of hogs) as the only possibility.

[@christopherwatson2772]

Did anyone else just list out all the squares up to 12 and check for differences that make 63? Got 2 couples that way, and then got Hendrick from Katrun. Not elegant, but it worked 😅

[@hhgygy]

Without watching the video (and reading the comments). I solved it like this. First of all, I used Excel to see which square numbers have a difference of 63. There were 1+64, 961+1024,81+144 meaning 1 hog + 8 hogs, 31 hogs + 32 hogs, 9 hogs + 12 hogs, respectively for each pair. Now, we know that H had 23 more than K and E had 11 more than G From this we can make a matrix whereby C=8 and his wife is G with 1 H=32 and his wife is A with 31 E=12 and his wife is K with 9 A rare case where I found MYD very easy. Now let's watch the video. LoL, I solved it much faster by just looking at two columns in Excel: x² and x²-63. 😎 Update: now I see others used Excel too. 😎

[@masscreationbroadcasts]

0:41 Ah yes, the illusive reverse reverse discount.

[@RoderickEtheria]

Perfect squares with differences of $63, 31 and 32, 9 and 12, 1 and 8. Hendrick bought 32 hogs. Katrun bought 9 hogs. Elas bought 12 hogs. Gertrun bought 1 hog. Gertrun is married to Cornelius. Elas is married to Katrun. Hendrick is married to Anna.

[@markmajkowski9545]

Wow - HARD WAY. Note 63 difference only allowed limited ways!!! Since 63 is the difference. Only certain combinations are allowed. 8/1 with the highest 32/31 and the differences most be odd of 1,3,5,7 only possible to consider 63 = x^2 - (x-3 then 5)^2 - 63 = 6x-9 or 63= 10x-25 - so 12 implying 9 for wife and 8.8 - not allowed at all- so H1 32 W1 31 H2 12 W2 9 H3 8 W1 1 - So H1 Hendrick bought 32 W2 Katrun bought 9 - Elias bought 12 (must be H2 and 11>8) Gurtrun therefore 1 married 8 buying unnamed man Cornelius H3 - Gurtrun W3. So Anna leftover must be W1.

[@coreyyanofsky]

without watching: - each person's total amount spent must be a square (N hogs at $N each totals up to $N²) - let the number of hogs bought by a wife be n and the *increment* over that bought by a husband be k; then (n + k)² - n² = 2nk + k² = 63 - clearly k must be odd and less than 8, hence in the set {1, 3, 5, 7}; solving for n for each of these values gives (n, n + k) in {(1, 8}, (9, 12), (31, 32)} (k = 5 does not give an integer value for n) - 32 - 9 = 23, hence Hendrik bought 32 and Katrün bought 9; likewise 12 - 1 = 11, hence Elas bought 12 and Gurtrün bought 1 - therefore Anna bought 31 and is married to Hendrik, Cornelius bought 8 and is married to Gurtrün , and the remaining couple is Elas and Katrün

[@michaelz6555]

Beavis is looking pretty dapper these days.

[@SheepLoveParrot]

The difference between 32th and 31th perfect square is just 63.. so each of them bought 32 or less hogs (and at least 1 ofc) H bought at least 24 and K bought 9 or less.. Similar... E >= 12 and G

[@TeiuAl]

Hendrick is living the dream

[@hamishbindrinkin]

Hendrick needs to buy at least 23 hogs... closest square where 63 less than it is also a square is 32. Repeat process for Elas getting 12. You now know that Elas and Katrun are married and that Hendrich can't be married to Gurtrun as the price is too far apart, therefore must be Anna and Cornelius is lucky enough to be married to Gurtrun who is seemingly the only one with enough sense to not spend ridiculously on this terrible deal.

[@jamessanchez3032]

You could say Hendrick and Anna went hog wild.

[@stevandunn2437]

So much for a bulk buy discount

[@nexen1041]

Absolutely genius

[@AFSMG]

Excelente trabajo

[@kurzackd]

Cool little puzzle !! :) Solved it in like 4-5 minutes. To be fair, the hardest part was actually ... *....making sense of the instruction text !! :D* ... hahaha... no offence, but it was worded *TERRIBLY !! :)* .

[@HarishKini]

This becomes much more easier if the husbands names were A, B and C, and the wifes names were X, Y and Z

[@willemakkermans4067]

I had the husbands down as A1, A2 and A3, the wives as B1, B2 and B3 😄

[@deyesed]

Curiously, the problem also works if each hog is just $1 as I initially read it, and gives the same solution!

[@willemakkermans4067]

Just not with each husband spending $63 more than their wife.

[@yurenchu]

If each hog costs just $1 , then there are multiple solutions, such as: Hendrick buys 71 hogs, his wife Anna buys 8 hogs, Elas buys 111 hogs, his wife Katrün buys 48 hogs, Cornelius buys 163 hogs, his wife Gurtrün buys 100 hogs. (which gives the same pairing of husbands and wives as the video's solution) but also: Hendrick buys 123 hogs , his wife Gurtrün buys 60 hogs Elas buys 71 hogs, his wife Anna buys 8 hogs, Cornelius buys 163 hogs, his wife Katrün buys 100 hogs. (which gives a completely different pairing of husbands and wives.)

[@willemakkermans4067]

@@yurenchu fair play 😄😄

[@graafslis]

Elas, Gurtrün, Katrün... are you sure these people are Dutch? 😄

[@axiezimmah]

More like Deutsch

[@fraggedlocust]

0:10 Don't let the haters win, names are arbitrary. How are you supposed to know how to pronounce Reighfyl?

[@moby4177]

Or you can see that it’s a question that has a discrete answer and realise that, no matter wht those numbers were, as soon as you compared two husbands with two wives, there were only two combinations that were possible solutions from the jnformation given. Then you look at the numbers and realise that the Anna/Elias combination doesn’t work, so it must be Heinrich and the non-mentioned Anna, and the other two combos flow from there.. once you reject the list possibility of Elias and Gutrun, because it wouldn’t give enough information if the twi were married. Of course, Simon from CtC would reject this approach.

[@ewsise]

damn I used sum of odds to find the values of the hogs bought, way more complicated 💀

[@jaredwonnacott9732]

Watch your assumptions! Though it was true that no couple made the same purchases as another couple, it could have been that they all bought the same amount. You said a couple times that one option was already taken, so it must be another number, but theres no reason to assume that. Its easy to prove, but worth the extra 30 seconds to clarify.

[@patsk8872]

Knowing there must be square numbers separated by 63, the only 3 pairings possible are 1 with 8, 9 with 12 and 31 with 32. Then matching based on the other info, Gurtrun is Cornelius' wife (and they bought 1 and 8 respectively), Katrun is Elas' wife (and they bought 9 and 12 respectively) and Anna is Hendrick's wife (and they bought 31 and 32 respectively.) Sorry, had to edit for typo with a name.

[@markjohansen6048]

You assume that each person bought a different number of hogs. For example, the E and C could not both have bought 32 hogs. Nowhere is that condition stated in the problem. It turns out that the only solution is for everyone to have bought a different number , but you can't assume that.

[@yurenchu]

He never assumes that. First he just makes an inventory of what the possibilities are for the husband to spend exactly $63 more than his wife; it turns out there are exactly three possibilities. Then next he determined from the remaining conditions how many hogs Hendrick must have bought, and how many Katrün; and likewise for Elas and Gurtrün. It turns out that they respectively bought 32 , 9 , 12 , and 1 hog(s), which just so happens to match three of the three pairwise possibilities. Since Elas bought 12 and Katrün bought 9, they must be a married pair. Hendrick (buying 32) cannot be married to Gurtrün (buying 1) (because (32,1) is not a pair), so he must be married to Anna; and that means Gurtrün must be matched to Cornelius.

[@penz-yy6oy]

Is this easier with inequalities and using all the possible outcomes.? for ex. if H is Hendrick and K is katrun etc. H^2-K^2=23^2 that is given and we know the amount the husband paid was 63 less than that of the wives. so if K is H's wife H^2-K^2=63, but 23^2 is >63 so K is not H's wife. by applying it to the other given info. E^2-G^2=11^2 , 11^2 > 63 so G is not E's wife. with these the possible outcomes are. 1. H, G in which case , ( E,K C,G ) 2. H, G ( E,G C,K ) 3. H, A ( E,K C,G ) By adding H^2 - K^2 = 23^2 and E^2 - G^2 = 11^2 We get H^2 - K^2 + E^2 - G^2 = 23^2 + 11^2 by further change the above sum H^2 - G^2 + E^2 - K^2 = 23^2 + 11^2 If G was H, wife then H^2 - G^2 = 63 and if K was E's wife then E^2 - K^2 = 63. Because 23^2+ 11^2 >>>126 then the above statement is false. which leaves us with the only possibility which is 3. idk i may be wrong in this reasoning though

[@discerningfreedom4124]

The real lesson learned here is buy the hogs one at a time so it will be much less expensive!

[@Salcor65]

Easy with excel. Got combination 1 and 8, 9 and 12, 31 and 32. Rest is logic.

[@denvercheddie]

Factors of 63 are the key.