Help me with this really difficult puzzle

Okay, I hope the mods are all right with this homework request, for several reasons:

1. The professor didn’t care whether I got the right answer to the problem or not.
2. The professor hasn’t figured out the answer to the puzzle herself.
3. The class is over and I aced it.
4. With my help, the professor came closer to the answer than she’d gotten in sporadic work over the past several years.

This is a math puzzle, one of two that a math-for-children prof gave us to solve as a way of figuring out how we approach math problems for which we don’t have a ready-to-use algorithm (her idea is that we should understand our own thinking so that we can better help kids develop logical thinking about math). The problem is as follows (paraphrased):

There are nine people:
Bill, Mary, Tom, Eliza, Hank, Susan, Ned, Sarah, and Jake.

They’re subdivided into three families, with a mom, a dad, and a grandkid in each family. Last names are Jones, Brown, and Smith. One parent from each family is related to Grampa Smith: one’s a daughter, one’s a son, one’s a stepson.

Grampa Smith, a loathsome slaver, dies in the nineteenth century. He wills his money to these three families in a complicated fashion, with the following rules:

1. A set number n describes the difference in inheritance between each member of each of the three families. The husband in each family receives n dollars more than his wife, and the wife receives n dollars more than the kid.
2. The inheritance is parceled out exclusively in 1-dollar bills.
3. Each heir receives a packet of envelopes with money in them; the money in each envelope equals the number of envelopes in that heir’s packet.
4. The sum of Mary and Sarah’s inheritances equals the sum of Tom and Bill’s inheritances.
5. The sum of Ned, Bill, and Mary’s inheritances is \$299 more than Hank’s inheritance.
6. The sum of the Jones family’s inheritance is “altogether over one third more than” the Browns’s inheritance.

The quesiton is, what are the last names of the family members, and how much does each receive?

I’ll describe where I’ve gotten with the puzzle later.
Daniel

I wish I could devote time to this, but I depart for Christmas travels early tomorrow. Here’s where I’ve got to in 2 minutes of slapdash cogitation (which may be far off base):

Conditions 2 and 3 seem to require that every person’s inheritance be the square of an integer. If so, the number N in condition 1 looks like it must be zero, for the only way that condition 1 is compatible with squares is when all family members receive the same amount

Sorry if it ain’t much.

WARNING - open spoilers continuing from Xema’s post …
not that this is a solution, anyway.

It seems to me, too, that n cannot be non-zero (it requires a[sup]2[/sup] + c[sup]2[/sup] = 2b[sup]2[/sup] to have integer solutions).

Using 4 & 5 together gives us some starting points. We know that of the four men in those equations, at least two got the same amount.

If To = Ha,

2Bi + Ne - Sa = 299.

If To = Ne,

2 (Bi + To) - (Sa + Ha) = 299 = Bi + To - (Ma + Ha)

If To = Bi,
either Sa & Ma are in different families and Bi & To are in another family, or Sa = Ma and 2Bi + Ne - Ha = 299.

If Bi = Ne,
2Bi + Ma - Ha = 299 = 3Bi + To - (Sa + Ha)

If Bi = Ha,
Ne + Ma = 299.

If Ne = Ha,
Bi + Ma = 299.
If I had time, I’d proceed from here to check which cases yield a valid solution for square numbers (at a glance I know those last two are out).

Lemme give y’all some information about where I got with it.

First, you’re right that each inheritance is a perfect square. However, there are solutions for a single family that use a nonzero n: the first one is 1, 25, and 49. 4, 100, and 196 also work. (Also, the original puzzle stipulates not only that the husband will receive n dollars more than his wife, but that he will receive more than his wife–so n=zero solutions are out).

For complex reasons, the roots of all the inheritances in a single family must be all odd or all even. (If they’re not, they can’t all be separated by the same value, since squares of odd numbers are separated from squares of odd numbers by an even number, as are squares of even numbers from even numbers, but squares of odd are separated from squares of evens by an odd number).

Some other things I figured out:
the difference between any two squares will be the sum of a series of consecutive odd numbers.

If two different series of consecutive odd numbers have the same total of n, they MUST have different numbers of elements in the series.

whatever n is, it must be the sum of six different series of consecutive odd numbers

These six different series of consecutive odd numbers have different numbers of elements in the series.

Now, I got really fancy when, playing around with the numbers, I realized that the following formula would be helpful:
tr+t(t-1)=wr+ (t+t-1+w)

t=the difference between the square roots of the kid and the mom
w=the difference between the square roots of the mom and the dad
r=the odd number that separates the child’s inheritance from the perfect square immediately above it.
(Yeah, I know this is really clunky. Cut me some slack, I’ve not had algebra in 15 years, and I was making this up from scratch).

The nice thing about this formula is that you can plug it into Excel and get a series of cells that, ultimately, allow you to plug in a t and a w, and have everything else figured out–i.e., it tells you the possible values of the child’s, mom’s, and dad’s inheritances given those values for t and w.

I did this, plugging in a little over 300 values methodically, and found that where n=3360, there were three family combinations:
4, 3364, and 6724
2116, 5476, and 8836
9409, 12769, and 16129.

These are all perfect squares, and in each family, the family members are separated by 3360.

Looking further, 16129+299=2116+5476+8836. Hank must be 16,129, with the other three numbers in some order equalling Ned, Bill, and Mary.

This is about as far as I got. I had some more speculation, but I can’t explain it well at this point: I think that Mary=5476, Sarah=3364, and both are wives. I also think that either
Bill=2116, Tom=6724.
OR
Bill=8836, Tom=4.

Unfortunately, I was unable to come up with rules for parsing these individuals into families, and assigning inheritances to the numbers. This is where I got stuck. And at some point while working on the puzzle I remember googling it and finding a solution but being too honest to look at it; now when I google it, I get nothing.

Daniel

I’d think that, a more apt description of this is a logic or computer science problem (the thriving field of constraint satisfaction problems – CSP). I just TA’d for an AI class where one of the projects was to solve these types of problems (albeit simpler) as they appear on the LSAT logic section. What that came down to was parsing the language of the problem to extract the “slots”, their possible “values”, and the relations (or constraints) imposed on the values. These are then put in a restricted language format (a custom Backaus-Naur form) which is amenable to standard search algorithms. With that done, it ends up being a plug-n-chug search for a solution. They even incorporated a hierarchy of sorts into the solution process, such that a subsolution for a certain set of constraints could then be used in the overall solution.

Pretty cool project, although it had issues with converting the semantics of natural language of the questions into BNF form. For instance, in the test problem I made up, it interpreted “dish” (as in a dinner plate) as a verb. Nonetheless, looking at it in this way seems to describe what goes on in my head when I see these problems. As with others, I wish I had the time right now to play around with this…

One more not: it’s from the book More Mathematical Puzzles of Sam Loyd, Martin Gardner (editor), published 1960. If someone has this book on their shelf, that might help!

Daniel

Ummmmm . . . what? Condition 1 simply means that you have to have 3 perfect squares in an arithmetic series like 1, 25, 49 (all separated by 24). It’s not impossible and in fact would benefit us by severely limiting the number of possibilities that can be used in the puzzle.

More food for thought: Condition 4 needs a number to be written as the sum of two squares (at least) two different ways. There’s a well known way to find these numbers (check out Mathworld)

Indeed: given the 9 perfect squares that I think equal the nine different inheritances, there are two different ways to satisfy the fourth condition. My notes are at home, but if nobody else can figure them out, I can post them when I get home tonight.

Daniel

You are certainly right - disregard my poorly reasoned response above.

So, I wrote up a quick program to look at what possibilities are out there. Firstly, do we know for sure that we are dealing with three different sets of squares? Is it possible that two families have the same breakdown? We know that at least two of them, the Joneses and the Browns, are different, but does the math confirm the Smiths must be different? Will have to look at that.

With the assumption that the child and wife each get less than 100 million I get 150 sets of squares that would work for all the families getting a different amount, and 761 more if we can count only two familes getting different amounts. Ech. I do confirm that n=3364 is the lowest n that works for three sets of squares.
If I limit it to 1 million, a little more reasonable, we still have 11 and 59 combinations to look at. The answer may not fall in that n=3364 solution. I will start looking at the other conditions a little closer.

I’m not sure what I was thinking; I realized my mistake this morning but I see you’ve posted more complete information.

Sticking with the numbers given for n=3360 and #5 (which appears to be the only solution for Hank with these numbers), it’s clear that Ned, Bill, and Mary are in the ‘middle’ family (and Mary must be the wife, and get 5476). Combining this with #4 implies Tom & Sarah are in the same family (or at least the difference between them = n).

This leads to what’s already been said, that Sa = 3364, and either Bi = 2116, To = 6724 OR Bi = 8836, To = 4.

Neither of these gives a complete solution, though the second one is better, as it only leaves confusion over whether Eliza is Hank’s wife or daughter (one hopes Hank is not confused).

The last condition would seem to fit two ways, so we have to either speculate that Smith gets the bigger inheritance (as the direct descendant) or that the Jones’s have 1/3 more than the other two families combined (which is not what’s stated in the problem).

The other possibility is that there is another n or another triple that fits this n.

Some help to finding more values may be here

Sorry, I didn’t notice flight’s post (I went several hours between preview & post).

I think we have to assume an unstated “these conditions lead to a unique solution” and therefore assume that we can decide the last names. Thus we should only look at three different values to distinguish the families, and that only one of them is > 1/3 more than another. Any way to check your sets for this condition, flight?

I also noticed that we don’t need to figure out familial relationships, just last names. In that case, n=3360 almost works, except we don’t know how to assign the last names.

:smack: It sure does. That’s one of the big glaring things that I was sure I was missing.

I’m not at all understanding the mathworld link; I’m trying this problem with relatively little higher math, and not much fluency at all in mathspeak (i.e., I can puzzle it out very slowly, but I can’t read it comfortably).

Can you elaborate on what you mean about the last condition? I originally read it to mean that one family received exactly 4/3 what another family received, but my paraphrase is how I currently interpret it: it means one family receives more than 4/3 what another family receives.

One thing that bothers me about this problem is that it was written in 1960, which makes me think there must be a way to approach it that doesn’t involve spreadsheets. My professor didn’t know how to use Excel, which is why she wasn’t able to make as much progress on it as I was; but surely the puzzle designer didn’t use a spreadsheet, either.

Daniel

OK, I am writing this while figuring it out. It seems that the lowest n that works for two sets of squares, 840, has potential.

The sets are:
1, 841, 1681
529, 1369, 2209

We know right off the bat that the higher set must be Jones and the the lowest must be Smith. Jones does come out being more than a third more than Smith, but that phrasing seems very odd to me and like it should mean more. Oh Well.

Statement 4 has 12 unique solutions with these numbers. In all cases one of the guys is a dad and one of them is a child (any combination of dad and child is a possibility). Mary and Sarah are both moms.

Looking at statement 5 with the knowledge Mary is a mom, Hank must be a dad or that side of the equation can’t be large enough. Likewise, at least one of Ned and Bill must be a child.

Crap. It seems it will not work. I can find no solution. I guess I will take a look at your possible solution. If that does not work out, it will be necessary to create some kind of criterion for out numbers based on the 299.

Okay, Google booksearch got me the answer, although the explanation for it is pretty much nonexistent, other than the perfect square business:

Bill Jones 8836
Mary Jones 5476
Ned Jones 2116
Hank Smith 16129
Elizabeth Smith 12769
Susan Smith 9409
Jake Brown 6724
Sarah Brown 3364
Tom Brown 4

Can anyone figure out why these figures are correct?
Daniel

By great than 1/3 more, I assume they mean the sum of one of the sets of squares is greater than 4/3 times another set, right?

The following numbers seem to work:

Ned Jones 196
Mary Jones 100
Bill Jones 4
Jake Brown 49
Eliza Brown 25
Hank Brown 1
Tom Smith 196
Sarah Smith 100
Susan Smith 4

Perhaps by “altogether over one third more than” means that they get more than 1/3 of the total inheritance more than the Browns. This would make the ones who get the most the Browns and the ones who get the leas the Jones.

I believe they mean that n is constant between families, not just within them.

You’re correct: there’s only one value for n, whereas younggun’s figures give two different values (96 and 24).

According to my spoilered answer, which is from the answer section of the book, the Smiths get the most, flight, with the Browns getting the least and the Joneses getting the middle amount.

Daniel