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Brain Teasers -
favorites from "Fun With Puzzles"

I've already devoted two pages to my favorite brain teasers - the wine in the water problem; and rolling one quarter around another. Lately (March 1999) I borrowed a library book called Fun With Puzzles by Joseph Leeming. It was copyright in 1946.

The main point of this page is to share my four (4) favorite puzzles from that book which were new to me. To get to them you have to wade through or skip over a few screens worth of puzzly prose.

Although the book doesn't claim to be the originator of any of the puzzles in it, so many sound familiar that I wonder if it didn't serve as a major source of brain teasers and puzzles over the last several decades. For example, there's the famous "Sisters and brothers have I none, but that man's father is my father's son" in slightly antiquated speech. (Pssst... it's his son.)

There's the old mind-bender of doubling a piddling penny for each square of a chess board - adding up to a whopping total of over 184 quadrillion dollars. (Admit it - that's still amazing in the computer age.)

There's "What happened to the dollar?", where three men pay $10 apiece for separate rooms in a hotel. The clerk gives a $5 refund to the bellboy to return to the three men. The bellboy gives each one a dollar, and keeps $2 for himself. "Now, each man, after receiving the $1 rebate, paid only $9 for his room. That makes $27 for all three. The bellboy kept the other $2, which makes $29. What happened to the other dollar?" My 5th-grade teacher, Swann King, pulled that one on us. Wasn't life a lot better back in the days when something like that could dumbfound everybody? (In case it still does, click here. When you're done, you'll find a link to get you back here.)

The book also has the "Wine and water" problem. I copied Leeming's solution onto my "Wine in the water" page so you can see the old-fashioned way of doing it.

There's the "Band of steel" problem, where you add an extra length to a steel band wrapped tightly around the earth's equator. You'll find the statement and solution of that one on my "Wine in the water" page, too.

I also found quite a few errors. That's disconcerting for a book of this nature. After all, it's only math and logic. It's bad enough stewing over a problem that might turn out to be a stupid, trick question without having to worry about whether the problem was stated correctly to start with. I'll discuss these "problem problems" at the bottom of this page, in case you ever find the book and want to check me out on this.

One of the most prominent goofs is the "Which is more profitable?" problem. Your starting salary is $1500 and the question is, would you choose a $300 raise every year, or a $75 raise every 6 months? Marilyn vos Savant must keep this book on her reference shelf; she dished up the same wrong answer in her Parade column. Here's a fun discussion of Marilyn vos Savant's blunder, by a real writer (not like me).

Four favorites

Mostly, I just wanted to pass on a few goodies I found in this book that caught my fancy. The four below are the best of the 88 problems in the Brain Twisters and Mathematical Puzzles sections, sez me. If you don't like these, then you can safely ignore the rest of the book. Plus, you can go soak your head in mud.

Actually, there are other sections devoted to completely different types of problems, such as number puzzles, word puzzles, and cut-out-and-put-together puzzles. I had a lot of fun with 2 easy-to-make manipulation type puzzles called the "string and the bead" puzzle and the "bachelor's button" puzzle. After making them myself, I had to stare and think long and hard before figuring them out.


Page 43. Weigh this fish.

"What is the weight of a fish if it weighs 10 pounds, plus half its weight?"

Short and simply stated. I was quite surprised by how willing people were to give the obvious - but wrong - answer. You'd figure that people would be very hesitant to say the obvious answer (15 pounds), and in any case could check to see that that answer, and other guesses like 10, 5 and 7.5, do not check out. Click here if you give up. (I'd like to think that this will be the web's most unclicked link.)


Page 28. The three travelers.

Three travelers met at an inn in Persia. Two of them brought food, but the third did not. He proposed that they should eat together and he would pay the value of his proportion.

This being agreed to, the first man produced 5 loaves and the second man 3 loaves, all of which the three men ate together. [The presumption is that they ate equal amounts.] The third man put down 8 pieces of money as the value of his share. The others were satisfied with this but quarreled about the division of the money. Thereupon the matter was referred to a judge, who decided correctly how the money should be divided. What was his decision?

This one doesn't lend itself well to tossing out at a party, but the answer is surprising enough to win it 3 stars, in my book. Leeming supplies a neat solution; click here.


Page 26. How old is Jane?

"The ages of Jane and Mary when added together make 44 years. Now Jane is twice as old as Mary was when Jane was half as old as Mary will be when Mary is three times as old as Jane was when Jane was three times as old as Mary. How old, then, is Jane?"

What makes this so great is the statement of the problem itself. The listener has no idea what he's in for. At first, he'll be listening with determination; then he'll realize he's lost it; and as it goes on and on and on, it actually gets funny. He'll probably start thinking the punchline after all of that will be something like, "What was the color of the bus driver's eyes?"

But the difficulty doesn't come from any complicated mathematics or unfamiliar concepts - it's just a chain of relative ages of 2 people. It's sort of like starting to add up a list of four or five 2-digit numbers, and finding it turn into an impossible task.

The downside to this problem is that the ages involve fractions. I don't know about you, but as far as I'm concerned, people ages are only integers. Assuming integral ages would throw quite a monkey wrench into solving this problem - like making it impossible.

Besides that, using fractional ages would seem to force a precision on the terminology that we don't need or want. Talking about Jane being 24 3/4 when Mary was 13 3/4 would only make sense if Jane and Mary had the same - or very close - birthdays.

Anyhow, I've reworked the problem slightly so that all of the ages are integers. It goes like this:

"The ages of Jane and Mary when added together make 60 years. Now Jane is twice as old as Mary was when Jane was half as old as Mary will be when Mary is five times as old as Jane was when Jane was five times as old as Mary. How old, then, is Jane?"

You can see it retains that nice touch of jumping into the future for Mary, and ends up with the one-two punch of the original, where Mary is N times as old as Jane, and Jane is N times as old as Mary. Yeee-ha! Click here to unboggle your mind.


Page 21. What's your name?

Mr. Jones one day got off a train in Chicago and while passing through the station met a friend he had not seen in years. With his friend was a little girl.

"Well, I certainly am glad to see you," said Mr. Jones.

"Same here," said his friend. "Since I last saw you I've been married - to someone you never knew. This is my little girl."

"I'm glad to meet you," said Mr. Jones. "What's your name?"

"It's the same as my mother's," answered the little girl.

"Oh, then your name is Anne," said Mr. Jones.

How did he know?

This belongs to a class of brain teasers you can smell a mile away, and is guaranteed to make you groan. But this one suckered me in. I had to smile. I don't care whether you fell for it or not, I give it 3 stars. Click here to see what a dummy I am.


Problem problems

I mentioned above that I found some errors, or at least problems, with the problems or their solutions. Man, it's hard enough solving problems that are stated clearly and correctly.


Page 19. The pay check puzzle.

Mr. Brown gave his gardener a check in full payment for some extra work. The check was in three figures and actually was more than the gardener was owed, so he was very pleased.

"Just a minute," said Mr. Brown. "If you will not cash the check, I will give you the difference between the product of the three figures showing the number of dollars, and their sum. I can tell you that this is not a small figure."

The gardener agreed to this apparently generous offer, but was very disappointed when he learned the truth of the matter. How many dollars was the check made out for?

This one has its problems. First of all, why wouldn't the gardener just do the arithmetic himself to figure out which was the better deal? Second of all, the answer is a trick answer - a stupid, dirty trick answer, and one with which you might disagree strongly. Thirdly, I claim there is a non-trick answer which is much more interesting. Click here to continue the discussion.


Page 21. Which was the officer?

An American patrol captured three German prisoners, all in ragged clothes, and without identifying marks. The American sergeant wanted to find out which one was an officer which two were privates. He knew from past experience that the officer would be sure to lie. He also knew that the privates would probably not lie.

The first prisoner questioned murmured some guttural words and fainted. The second prisoner pointed to the first and said, "He said he's a private. That's true, both he and I are privates."

The third prisoner pointed to the second and said, "He's a liar."

Despite the prisoners' statements, the American sergeant soon figured out which German was the officer and which two were privates. How did he do it?

Now what the heck does "probably not lie" mean? That the privates are truth-tellers? Or, sometimes they lie but more frequently tell the truth? Figuring that there is no way to distinguish a lying private from a lying officer, I proceeded on the assumption that the privates were truth-tellers.

The next problem is the statement by prisoner 2. If he is a liar, does that mean that every sentence he says is false, or is it enough that the sum total of what he says is false? Supposing the former, will the listener be able to distinguish 2 sentences separated by a period from one sentence with a dash in the middle?

Anyhow, my method for solving this sort of problem is to simply write out all of the possibilities of truth-telling and lying and see which one doesn't give rise to a contradiction. This cuts down drastically on painful thinking. Here are all 8 permutations of True and False as applied to the statements of the three prisoners.


                   Possibility
             1  2  3  4  5  6  7  8
             ----------------------
Prisoner 1   T  T  T  T  F  F  F  F  
Prisoner 2   T  T  F  F  T  T  F  F
Prisoner 3   T  F  T  F  T  F  T  F

We know (we think) that the solution has to involve 1 F and 2 Ts. This eliminates all but possibilities 2, 3, and 5.

Since prisoner 2 and 3 contradict each other, one must be a liar - they can't both be truth-tellers. That eliminates possibility 5.

That's as far as I could go. I don't see any contradiction arising from either possibility 2 or 3. Possibility 2 says that prisoner 3 is the lying officer, possibility 3 says prisoner 2 is. Click here for the official lowdown.


Page 22. Alone, alone, all all alone.

A young man, otherwise quite sane, once said: "I had luncheon today with my father's mother-in-law's husband, my step-brother's nephew's father and my step-mother's father-in-law, yet I ate by myself."

How do you think this was possible?

Click here.


There is a class of puzzles which I'll call "match-up puzzles". Typically, you are given a list of N people, say, and you need to make one-to-one match-ups with N "things", such as occupations, characteristics, spouses, or what-not. You are given a collection of statements which eventually lead to the solution, generally by telling you directly or indirectly what can't be.

These puzzles are sort of fun, but almost all of them in this book have mistakes or ambiguities. In "Who's who on the ball team?" (page 20), the problem starts:

Listed below are some comments about the 9 members of a baseball team. With a little studying and reasoning, you should be able to figure out what position each holds on the team.

Then there is a list of statements such as: Bob, Joe, Frank and the catcher were beaten at golf by the second baseman.

Everything goes smoothly until you get to comment 15: Bill, Bob, and Jack, the center fielder, and the right fielder were bachelors. The others were married.

Yikes. Does that syntax and punctuation mean Jack is the center fielder? I eventually forged ahead assuming that Jack was not the center fielder; that the first "and" in that statement was extraneous and misleading. This led to the desired final answer, so it must have been right.

What I like about these match-up puzzles is that you can usually figure them out without using any brains. Just make a grid with the items you are trying to match up along the 2 axes. Scribble out the grid cells that are found out to be impossibilities. When you do find a match-up, scribble out all the other cells in that row and column.

For example, here's how the grid for this particular problem might look after you've eliminated some possibilities and finally nailed down one of the match-ups - that Frank is the center fielder.


          p  c  1  2  3 ss lf cf rf 
        ----------------------------
Joe     |  |XX|  |XX|XX|  |XX|XX|XX|
        ----------------------------
Bob     |XX|XX|  |XX|  |XX|  |XX|XX|
        ----------------------------
Frank   |XX|XX|XX|XX|XX|XX|XX|()|XX|
        ----------------------------
Ed      |  |XX|  |  |  |XX|XX|XX|XX|
        ----------------------------
Bill    |  |  |  |XX|  |  |XX|XX|XX|
        ----------------------------
Harry   |XX|  |  |  |  |  |XX|XX|XX|
        ----------------------------
Tom     |XX|XX|XX|XX|XX|XX|  |XX|  |
        ----------------------------
Jim     |  |  |  |  |  |  |XX|XX|XX|
        ----------------------------
Jack    |  |  |XX|  |  |  |  |XX|XX|
        ----------------------------

In "The six college men" (page 25) we are supposed to match each man with his college. I claim there is not enough information given. There is an odd statement, "Smith knows Miss Snow, but she won't go out with him." I don't see that this tells us anything. We've already been told who Smith and Miss Snow go with.

In "Dinner at the club" (page 26) we are supposed to match 6 men with 6 identifying characteristics - fat, brilliant, millionaire, etc. Morris and Conway are not mentioned by name anywhere in the list of statements, so how could we hope to distinguish those two?

And even in "The six shoppers" (page 46) they throw us a curve. We're supposed to match six shopper's with what they bought - a ring, a book, a dress, etc. Down in the clue statements, they mention "rugs are sold on the ground floor". It's not immediately obvious, but it turns out that "ring" was, in fact, a typo for "rug".


Page 37. How many dinners?

There were 7 men who decided that they would dine together every evening, provided they could seat themselves in a different arrangement at the table each time they met. How many dinners would they eat before exhausting all possible arrangements?

Click here for the answer(s).

 


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Answer to: What happened to the dollar?

"It's all in the way you figure it. The clerk gave the bellboy $5 and kept $25. The bellboy gave each man $1 and kept $2. Each man paid $9, less $2 to the boy, leaving $25."

I see two ways of looking at it. The most boring is that there's $30 floating around in the picture, of which the clerk has $25, the bellboy $2, and the men $3.

The more fun way is to balance the ledger sheet. Each man really did pay out $9, for a total of $27. The clerk received $25 and the bellboy received $2, for a total of $27. Check.

If you carefully add what the bellboy has to what the men paid, you haven't calculated anything useful. "It's all in the way you figure it", indeed.

Back to the questions.

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Answer to: Which is more profitable?

"The $75 raise every 6 months is a better choice. Over a 3-year period, for example, the total wages paid would be $5625 as compared with $5400 for the other arrangement."

Now here is Charlie Bryant's tour de force regarding Marilyn vos Savant's answer to this brain teaser. Charlie's article was titled, "Intelligence is in the eye of the beholder, or reader." It appeared in the The Journal newspaper (Montgomery County, Maryland, July 2 1992.)

Serious newpaper readers among you may have seen a recent issue of Parade magazine, which contained, as it often does, a column by Marilyn vos Savant. Parade says the Guiness Book of World Records says Marilyn Vos Savant has the highest IQ in the world. Well, old Marilyn may be smart, but she's wrong.

Some weeks back, Marilyn vos Savant (who along with the world's highest IQ also has the world's most suspicious name) posed this question: Say you're making $10,000 a year. Your boss offers you a choice between a $1,000 raise once a year and a $300 raise every six months. Which do you choose?

Marilyn says you're better off with the $300 raise every six months. She says it will accumulate faster than the $1,000 raise. She is, as I may have pointed out already, wrong.

On Sunday her column had a letter from an accounting professor at Texas Christian University. He said, in essence, how can this be? It seems wrong, but I'm not sure why. Please explain it to us again, smartest person in the world.

He didn't seem to be able to figure it out, though, which makes me glad I studied English at [the University of] Maryland and not accounting at TCU. But this is why Marilyn and I are discussing large sums of money for a general readership and he's writing letters asking us to explain. So explain we will. First I'll give Marilyn's high-IQ explanation, which is wrong. Next will come my explanation of how and why she is wrong, which is right.

OK. Marilyn vos Savant and Charlie Bryant are columnists. Each makes $10000 a year. Starting Jan. 1, 1991, Marilyn takes the $300 raise every six months and Charlie takes the $1000 raise once a year. In the first six months we both make $5000. In the next six months, says Marilyn vos Savant, Marilyn makes $5300 and Charlie makes $5000 for a total on the year of $10300 and $10000, respectively.

On Jan. 1, 1992, says Brainiac (I got tired of typing Marilyn vos Savant), we both get raises - $300 for her, $1000 for me. Her pay for the next six months is $5600, mine is $5500. On July 1, she gets a raise and I don't; for the next six months, her pay is $5900 and mine is $5500. After two years she's made $21800 and I've made $21000. In the first six months of 1993, she says, Brainiac makes $6200 and I make $6000; the last six months of 1993 brings her $6500 and me $6000, for a grand total of $34500 for her and only $33000 for me.

Says her.

[Here is a summary of vos Savant's calculations in table form. DS]

                         - - - Brainiac - - -         - - - Charlie - - - 

                       Pay for Pay for Running      Pay for Pay for Running
          Period       half yr   year    total      half yr   year    total
      --------------   ------- ------- -------      ------- ------- -------
      1991, 1st half     $5000                        $5000                
            2nd half     $5300  $10300  $10300        $5000  $10000  $10000
      1992, 1st half     $5600                        $5500                
            2nd half     $5900  $11500  $21800        $5500  $11000  $21000
      1993, 1st half     $6200                        $6000                
            2nd half     $6500  $12700  $34500        $6000  $12000  $33000

All right. Has anyone figured out where Brainiac went wrong? Does anyone here have the raw cerebral power, the active intelligence, the turbo-powered surplus IQ points, to show just how Marilyn vos Einstein is duping the gullible American newspaper-reading public?

Allow me.

I don't know how Brainiac gets paid; maybe Parade magazine mails gold ingots to her post office box. But in the real world, in which people with salaries get paid every two weeks, you're going to get a depressing surprise in your July 15 paycheck on the $300 plan.

You seem to think you're getting 300 bucks as soon as you take your first increase. But if you take that $300 raise starting July 1, you won't make an extra $5300 in the next six months. You'll make $5150.

Why? Didn't you just get a $300 raise? Yes, but it's a raise, not a bonus. A raise is an increase in salary, and your new annual salary, including your $300 raise is, is $10300. So for the next six months, your pay will be half of $10300, or $5150.

Try it this way: $10300 divided by 26 (the number of biweekly paychecks a year) is $396.15. That figure times 13 (the number of biweekly paychecks Brainiac gets in the next six months) is $5150. For 1991, then, Brainiac's pay is $10150. Charlie's is $10000. On Jan. 1 we both get a raise. Brainiac's is, as usual, $300; Charlie's is, as we agreed, $1000. Now her annual salary is $10600; mine is $11000. For this six months she gets $5300; I get $5500. Add these figures to our pay for 1991: As of June 30, 1992, Brainiac has made $15450 and I've made $15500.

On July 1, Brainiac gets her $300 raise to an annual salary of $10900. In the second half of the year she makes $5450 and I make $5500 again. As of Dec. 31, 1992, Brainiac has made $20900 and I've made $21000.

On Jan. 1, 1993, we get our raises again, $300 for Brainiac and $1000 for me. Her salary is now $11,200; July 1 it gets raised - there's that word again - to $11500. She'll make $11350 in 1993 and I'll make $12000. I think you can figure it out from here.

[Here is a summary of Charlie's calculations in table form. DS]

                    - - - - -  Brainiac  - - - - -       - - - - -  Charlie  - - - - -

                           Pay for  Pay for Running             Pay for Pay for Running
      Period       Salary  half yr    year    total     Salary  half yr   year    total
  --------------   ------  -------  ------- -------     ------  ------- ------- -------
  1991, 1st half   $10000    $5000                      $10000    $5000                
        2nd half   $10300    $5150   $10150  $10150        "      $5000  $10000  $10000
  1992, 1st half   $10600    $5300           $15450     $11000    $5500          $15500        
        2nd half   $10900    $5450   $10750  $20900        "      $5500  $11000  $21000
  1993, 1st half   $11200    $5600                      $12000    $6000                
        2nd half   $11500    $5750   $11350  $32250        "      $6000  $12000  $33000

So what happened? As manager of the payroll department, it seems, Brainiac was giving Charlie a $1000 raise and handing it out as $500 every six months. Meantime, she was giving herself a $300 raise and taking it as $300 every six months. What she was actually doing was getting a $600 raise in her annual salary every six months and calling it a $300 raise.

By her method of accounting, her salary for 1991 [i.e., her total pay for 1991] was $10300 and for 1992 it was $11500. In one year her two $300 "raises" increased her annual salary by $1200. So in the most widely read Sunday supplement in America, the most intelligent person in the world proved that raising your salary by $1200 every year is better than raising it $1000 every year.

Nice work if you can get it.

So Charlie's made it clear how, if you take a twice-a-year "raise" the wishful-thinking way, you are really taking the equivalent of a once-a-year raise that is four times as big! Neither Leeming or vos Savant got it.

Even so, the attentive reader might be wondering, how is it that the $75 twice-a-year "increase" in Joseph Leeming's problem (figured the wrong way) can still beat out the $300 once-a-year raise? Shouldn't that be a wash, since 4x75=300? Good question.

It has to do with the first increase coming along sooner in the twice-a-year scheme. Suppose JL went for the $75 twice-a-year deal, and Charlie got the $300/year raise. JL gets his first $75 increase in the second half of the first year. Thus, his total pay in the first year is $1575, while Charlie's is $1500. From then on, both JL and Charlie will be making a $300 yearly increase ("raise"), but JL's $75 headstart will keep him ahead of Charlie by that much in every following year. So (figured the wrong way), after 3 years JL will be ahead of Charlie by 3 x $75 = $225, which is the difference shown in Leeming's answer.

It's as if two employees were given the exact same yearly raise, but one started with a higher salary. In spite of the increases, that initial difference will always be maintained, and will add up year by year.

I took this one step farther and discovered something that surprised me. I figured if both plans were scheduled to kick in at the beginning of an upcoming year, then the twice-a-year $75 increase, without the benefit of a headstart, would work out identically to the $300 raise. In other words, if Leeming had stated his problem, "Your current salary is $1500...", as opposed to, "Your starting salary...", then the two plans would be a wash. But, under these circumstances, the $300 yearly raise plan is the winner! In the yearly raise plan, you will make $1500 + $300 = $1800 in the first year. In the twice-a-year increase plan, you will make ($750 + $75) + ($750 + $75 + $75) = $1725 in the first year. So you see the shoe is on the other foot; the yearly raise plan jumps out to a $75 lead in the first year, and that difference remains the same year after year as the total pay increases $300 a year in each plan.

It might seem kind of funny that there are only three $75's piled on in the first year, while in every year thereafter there are four more $75's than in the previous year. Thinking in terms of $75 "units", this is what's going on:

                         Year 1   Year 2   Year 3   Year 4   Year 5 . . . 

                        1st 2nd  1st 2nd  1st 2nd  1st 2nd  1st 2nd  
                        --- ---  --- ---  --- ---  --- ---  --- ---  

$75 units above base
  for each half year:    1 + 2    3 + 4    5 + 6    7 + 8    9 + 10 . . .

     Total $75 units:      3        7        11       15       19

Increase in $75 units 
    from year before:      3        4        4        4        4

A mind with a feel for math should look at that and say, "That's weird. Simple, repetitive addition like this shouldn't give rise to one, lone 3 and an infinity of 4s. Math is too well-behaved and regular. " The explanation, or at least, my explanation, is that we have a discontinuity in the pattern right at the start of Year 1. The twice-a-year $75 raise was not in effect in the previous year ("Year 0"). If we extended the pattern backwards, Year 0 would appear in the chart with -1 + 0 = -1 total $75 units, and you see tthat Year 1 would then present four $75 units more than the year before.

Enough of this?

Back to the questions.

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Answer to: Weigh this fish.

"The fish weighs 20 pounds."

But how did we arrive at this amazing answer? I suspect in 1946 people used trial and error. Since then, somebody invented algebra.

Let W = weight of the fish.

"A fish weighs 10 pounds plus half its weight" translates into


                 W = 10 + W/2

Solving for W,


                 W/2 = 10

                 W = 20

Hooo-ray for math.

Back to the questions.

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Answer to: The three travelers.

"At first sight it would seem that the money should be divided according to the bread furnished. However, since the three men ate 8 loaves, each one ate 2 2/3 loaves of the bread furnished. Taking this from 5 would leave 2 1/3 loaves furnished the third traveler by the first one. Taking it from 3 loaves leaves 1/3 of a loaf furnished by the second traveler. Hence 2 1/3 to 1/3, or 7 to 1 is the ratio in which the money is to be divided."

If that's not instantly clear, what he's saying is that the first and second man each ate 2 2/3 loaves out of what he himself brought, the two remainders going to the third man.

Back to the questions.

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Answer to: How old is Jane?

"Jane is 27 1/2 years old and Mary is 16 1/2. This will be clear if you trace the question backward, for when Mary was 5 1/2 years old, Jane was 16 1/2. Now when Mary is three times that age she will be 49 1/2 years old. Half of this is 24 3/4, and when Jane was that age Mary was 13 3/4. Accordingly, Jane's age was twice this, or 27 1/2.

"Just the same, it's a heck of a puzzle and a lot of people would like to lay hands on the person who invented it."

Yeah, if it weren't for those fractional years, I'd like to shake his hand. About that "solution", though - it's really just a check on the answer. But how is one supposed to have found the answer? (And is it obvious from the above that the given answer is unique?)

In my integer-only version of the puzzle, you can start with "when Jane was five times as old as Mary" and just work from different starting points until you find one that checks out. The next step is to examine their ages when they add up to 60 and then step through the statement of the puzzle to see if everything fits properly.

Try,

    Mary    1  2  3  4  5  6  7  8  9 10 11 12 13... 
    Jane    5  6  7  8  9 10 11 12 13 14 15 16 17...

and,

    Mary    2  3  4  5  6  7  8  9 10 11 12 13 14... 
    Jane   10 11 12 13 14 15 16 17 18 19 20 21 22...

and,

    Mary    3  4  5  6  7  8  9 10 11 12 13 14 15...
    Jane   15 16 17 18 19 20 21 22 23 24 25 26 27...

and so on. Yer on yer own.

Back to the questions.

The mathematically-inclined may stick around. Up above I asked, "But how is one supposed to have found the answer? And is it obvious from the above that the given answer is unique?" Thanks to Albert Hines (Muskegon, Michigan), I now (writing in August 2002) have the mathematical solution. I'm very impressed with how simply and clearly Albert laid it out. This is what he wrote in January 2002:

    The ages of Jane and Mary when added together 
    make 44 years                                        1.  J+M=44
    Now Jane is twice as old as Mary was when            2.  J=2*M1
    Jane was half as old as Mary will be when            3.  J1=M2/2
    Mary is three times as old as Jane was when          4.  M2=3*J2
    Jane was three times as old as Mary.                 5.  J2=3*M3
    How old, then, is Jane?                              Solve for J.
    
    Since we have more unknowns than equations, we have to note that 
    Jane is always the same number of years older or younger than Mary;    
    
                                                         6.  J-M=J1-M1
                                                         7.  J-M=J2-M3
    Now we 7 unknowns and 7 equations.  
    Here's my solution that does not require any guessing:
    
    Use 5. and 7. to eliminate M3:      J2=3*(J2-J+M)  ->  2*J2=3*J-3*M
    Use 4. to eliminate J2:             2/3*M2=3*J-3*M
    Use 3. to eliminate M2:             4/3*J1=3*J-3*M ->  J1=9/4*J-9/4*M
    Use 2. and 6. to eliminate M1:      J-M=J1-1/2*J   ->  J1=3/2*J-    M
                                                           --------------
    Use previous two equations to eliminate J1:             0=3/4*J-5/4*M
    
    Use 1. to eliminate M:              3/4*J=5/4*(44-J)  ->  J=27 1/2
    
    Hence, 
      J  = 27 1/2
      M  = 16 1/2
      M1 = 13 3/4
      J1 = 24 3/4
      M2 = 49 1/2
      J2 = 16 1/2
      M3 = 5 1/2

Regarding the uniqueness of the answer, after seeing the puzzle solved in this manner using simultaneous equations, I'm now convinced.

Albert also supplied a solution for the problem interpreted in this plausible way:

The ages of Jane and Mary when added together make 44 years. Now Jane is twice as old as Mary was when Jane was half as old as Mary will be when Mary is three times as old as Jane was when Jane was three times as old as Mary [IS RIGHT NOW.] How old, then, is Jane?

For this interpretation, the answer works out to: Jane is 34 4/7 years old, and Mary is 9 3/7 years old. (Equation 5 above becomes J2=3*M, and equation 7 is dropped.) Note that if one interprets the puzzle this way, some of the verb tenses in the statement of the puzzle would not be consistent with the answer. The first 2 instances of "was" would have to be changed to "will be".

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Answer to: What's your name.

First of all, here are some familiar groaners taken from the same book.

"A beggar's brother died. But the man who died had no brother. How is this possible?"

And:

"A big Indian and a little Indian stood on a hill. The little Indian was the big Indian's son, but the big Indian wasn't the little Indian's father. How was this possible?"

Had enough yet? "Mr. Jones's friend was a lady - and her name was Anne! [Emphasis mine.]

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Answer to: The Pay check puzzle.

The answer is $123. The product of 1 times 2 times 3 is 6. The sum of 1 plus 2 plus 3 is also 6. The difference between the product and the sum of the three numbers is therefore zero, which, as the artful Mr. Brown said, is not a small number.

Isn't it? Anyway, what's to distinguish $123 from $321, or $213, etc.? They would all give the same result.

Here's what I think is a much better and more interesting and probably more surprising answer: the gardener would take a loss no matter what the original check was made out for. Try out a few test cases, like $999, or $119 or whatever comes to mind. Certainly any figure with a zero in it is an instant loser.

To prove to yourself that the operation will always yield a smaller number, just think of a number in the 800s, say. For the product to just barely reach the 800s, the 8 would need to be multiplied by two 10s. But each of the other digits is 9 or less. The product alone will be significantly less than the original figure - never mind the subtraction. The same logic applies no matter what digit is in the hundreds column.

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Answer to: Which was the officer?

The third prisoner was lying and therefore was the officer. The second prisoner's statement [note - not statements] must have been true. If it were false, both he and the first prisoner would be officers - and there was only one officer.

Is that an example of pre-computer age logic? In 1946, was NOT(A AND B) equivalent to (NOT A) AND (NOT B)? It seems to me that an officer saying, "both he and I are privates", is lying appropriately.

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Answer to: Alone, alone, all alone.

The young man married the mother of his father's second wife and had a son. His stepmother also had a son.

It seems to me that the young man is his stepmother's stepfather, not stepmother's father-in-law. That would be the father of his stepmother's husband.

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Answer to: How many dinners.

There are 5040 arrangements, so the men would eat together every evening for more than 13 years.

I guess we're supposed to be surprised by how big even the smallest factorials are. Anyway, I claim the statement of the problem is ambiguous. If "arrangement" refers to who sits to the left of whom around the table, there would only be 6! = 720 distinct arrangements.

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