This text is one of many originally written in 2012 but only published years later.

A major revision is the near-removal of the “late chauffeur” (cf. below). Apart from that, some minor fixes, and the odd addendum, the text corresponds to the 2012 version.

Chances are that the originally intended scope was larger than the current text, but, a dozen years later, I can only speculate on my intentions and what other problems might eventually have been included, had I continued work in 2012.

Whether the text will be extended in the future seems less likely than in 2012 (cf. original introduction), because of the sheer number of other texts/topics/projects/whatnot.

Problem solving is a fascinating area. In this article (likely to be expanded as time goes by), I give some of my thoughts on problem solving, with a focus on “entertainment problems”.

Note that these problems are usually stated from memory long after the first encounter or brought to my attention by some individual. For this reason, I cannot provide a source or give credit (barring an after-the-fact search which the reader can do just as well). To boot, the origins of some problems might be clouded by time. For the same reason, there might be some dispute of what the “canonical” formulations and solutions are; however, this is unimportant to my current purposes.

Other articles relating to problem solving include Solving the right problem.

Recently, two colleagues presented an interesting problem. Roughly:

Two mathematicians travel on an airplane.

The first asks the second, “You have three sons, right? How old are they now?”. The second answers, “The product of their ages is 36 and the sum of their ages equals the current date [from context: day of month]”.

The first laments being unable to solve the riddle, and the second adds: My fault, I forgot to say that my oldest son has a dog.

How old are the sons?

Side-note:

While I will give the solution below, I recommend that the reader gives it a try on his own. The same applies to any other problems presented.

A common issue with problems are unstated assumptions. (Even when it comes to more academic problems, cf. a discussion of my experiences at the FUH.) Here we have at least three (not counting the interpretation of “date”):

The first mathematician is sufficiently capable of performing his calculations and considerations without fault. (A very common assumption in this type of trivia problem.)

Age is measured exclusively in whole years, with the implications, e.g., that 16, 1.5, and 1.5 years old is not a valid solution, and that a child older than another is a positive integer number of years older. (In terms of the answer. We might still have two children aged 14 years and 2 resp. 11 months, but these are then considered to both be 14—or, with a non-standard approach, 14 resp. 15.)

This is particularly important with an eye at the dog-clue—if the oldest son is so by dint of, e.g., being born five minutes before a twin, the problem fails.

The mathematician is aware of the date. (Cf. below.)

A better formulated problem would state such assumptions explicitly—and
failure to do so could be a sign of lacking brains or care in the poser of
the problem. However, with problems intended to entertain, not test students,
some compromises might be made in order to not accidentally give
additional clues—and this often results in leading, yet inconclusive,
statements. (A good example is the unstated assumption about the first
*mathematician*: A problem featuring two kindergarten teachers would be
less likely to have the same assumption.)

An interesting special case is the many “not” assumptions typically present. Typical examples include that there is no mirror in a fortunate position, that a character within a problem does not possess information unstated in the problem, that this-or-that is not given away by nervous reactions, ... Obviously, attempting to list all of these would overload the text of any problem; however, it still pays to ensure that it is clear what the rules of the game are (possibly through a mathematical abstraction or a general “spirit of the law” statement). Cf. also the “three light-bulbs” problem below.

Firstly, it is clear (taken the above assumptions into consideration) that the ages are some combination of the factors 2, 2, 3, 3, and an arbitrary number of 1s. (36 can be written as 2 * 2 * 3 * 3 * [any number of 1s], and cannot be reduced into further integer factors).

Addendum:

An interesting question is whether those “any number of 1s” could trip up someone—maybe, especially, someone interested in math, who might jump to “the prime factorization of 36 is 2 * 2 * 3 * 3” and ignore possibilities like a one-year old toddler. (A mistake extremely unlikely in someone who thinks in terms of children first and math second. Cf. my own issue of thinking in terms of math first and light-bulbs second, with the three light-bulbs problem.)

Secondly, one of the sons is strictly older than the other two. (The other
two *might* still be equally old, however.)

Thirdly, it is not possible to solve the problem without knowing the previous item. Indeed, and this is of paramount importance, it is implied that even a mathematician who knows the current date (equal to the sum of the ages) cannot solve the problem without this seemingly trivial information! (But dog ownership is entirely beside the point.)

Now, if the children had been e.g. 18, 2, and 1 years old, the date must have been the 21st—and 18, 2, and 1 is the only combination of ages that makes 21 while fulfilling our other assumptions (barring permutations, e.g. 2, 18, and 1). Indeed, for most dates, the solution is either trivial (for someone good at math) or non-existent (e.g. for the 22nd). Specifically, ignoring permutations, we have the following possible combinations:

36 + 1 + 1 = 38 (> 31 and not a possible day)

18 + 2 + 1 = 21 (see above)

12 + 3 + 1 = 16

9 + 4 + 1 = 14

9 + 2 + 2 = 13 (!)

6 + 6 + 1 = 13 (!)

6 + 3 + 2 = 11

4 + 3 + 3 = 10

We can conclude that (for the days at all possible) the problem is only *not*
easy to solve if we assume the 13th. On the 13th, we have the problem of two
combinations (9, 2, 2; 6, 6,
1) being available. We now apply the second clue and find that the answer
must be 9, 2, and 2.

The above solution leads to an important observation about trivia problems: One must pay attention not merely to what one self knows, but also what the characters within the problem know (and the possibility that they face a slightly different problem). Here, the mathematician must know the date at hand or the problem cannot be solved using only the provided clues.

(The paramount example is the three-hats problem discussed below.)

Looking at an alternative approach, we find another principle: Asymmetry. If there is asymmetry of some sort within the space of answers still under consideration, this is usually a very strong indicator of where to look for the solution. Above, e.g., we see that the 13th is “special”.

One of the most famous problems, and one most readers will already be familiar with, runs as follows (with many variations):

Three prisoners sentenced to life in jail are given a last chance. They are shown four hats, two black and two white, and then blindfolded. While unseeing, they are put in a row, the first facing a wall, the second facing his back, the third facing the back of the second. Each has one of the hats placed on his head, is told not to move (on pain of instant death), and the blindfolds are removed. They are now told that the one who first gives the color of his own hat (which he cannot see) is made a free man, while the other two return to jail—but if someone speaks up with the wrong color, he dies (leading to the unstated assumption, cf. above, that no-one will just guess).

Addendum:

Guessing might still be a valid option, as it still is a 50–50 shot.

In my original text, I went as far as noting that the first prisoner might have no other choice (as will be clear from the below, there is no constellation where he can solve the problem by thinking), but I failed to address guessing more generally.

Consider first an alternate problem where a correct answer from the
one leads to the death of the others, instead of their being returned to jail.
With the time factor at hand, just immediately shouting out one of the colors
gives a 50–50 shot at being right. Stop to actually think and death might follow
with approximately a 2/3rds probability: Firstly, thinking will only work if
a solution can be found.
Secondly, the thinker stands the risk of *someone else* guessing,
before a solution has been found. In the overlap, should there be a constellation
where more than one thinker could find the solution, it would be imperative to be
the first to do so—and the immediate guess might still be the best option.

(Here, note that even stopping to think about aspects of the problem could take too long. With thought, we know that mere chance gives the second and third prisoners a 50–50 chance of having a findable solution, while the first never has a findable solution—but putting in that thought takes time that someone else might use to answer, be it through thought or as a guess, and the prisoners might be well advised to just assume equal chances and jump straight to guessing.)

If the conditions in jail are bad enough, a “freedom or death” approach might, even as the problem actually stands, be preferred by some. Likewise, if a prisoner has very important matters to attend to, a “freedom or death” choice might make sense—a dissident against a tyrannical regime, e.g., might be willing to risk death to increase the chance at freedom and more effective work, while a young lover might see that love as worth the risk.

A long silence ensues. Finally, one of the prisoners speaks up, giving the correct answer. Which prisoner?

The long silence (implying another unstated assumption) is the crucial key:

If the two first prisoners (both in the line of sight of the third) were
carrying hats of the same color, the third would trivially be able to deduce
that his own hat was of the other color. His failure to speak up with at
most a *short* pause implies that they wear hats of *different* colors.
This is of no help to the first prisoner, but the second (once realizing
the cause for the silence of the third) easily concludes that if
the first prisoner (whom he can see) wears a white [black] hat, then his own hat is black [white].

Thus: The second prisoner speaks up.

In contrast, if the problem had included one of the prisoners speaking up
*immediately*, it must have been the third prisoner.

By 2024, I have independently addressed this problem in at least one other text, on epiphanies relating to thinking.

(The difference in treatment is in part one of priorities and angle, in part one of memory, in part one of the natural fluctuations that occur when addressing the same topic at different times.)

One of the rare problems that saw me fail utterly and embarrassingly is the following:

A house has three light-bulbs in the attic that (for reasons unmentioned) are controlled by three switches in the basement (with the unstated assumption that it is impossible to observe the lamps while using a switch, etc.), each switch being in an on–off relationship with one of the light-bulbs. Unfortunately, there is no indication which switch controls which light-bulb. (However, it is known that the switches are all initially off.)

What is the minimum number of journeys from basement to attic or attic to basement that it takes to correctly identify which switch does what, where should one ideally start, and how many toggles of switches is the minimum?

Being well versed in math, binary logic, and the like, I more or less immediately concluded that three journeys (basement–attic, attic–basement, basement–attic; obviously, picking the basement as the starting point) would be needed, with two toggles: Toggle one arbitrary switch on, observe which lamp is on, toggle one of the remaining switches on, and observe which additional lamp is now on.

I was confounded and unbelieving when I heard that the official solution
was *one* trip. (But still with two toggles and still starting in the
basement.)

This was preposterous! Indeed, it amounts two determining the exact form of a permutation of xyz knowing only, e.g. and hypothetically, that 100 maps to 010—a complete impossibility. Notably, the results for the inputs 000 and 111 are trivially 000 resp. 111 and bring no information whatsoever about the underlying permutation. Further, for reasons of symmetry, it does not matter whether the initial input contains one or two 1s or where the 1 is situated—each of these combinations give the exact same amount of information. Without loss of generality, we can, then, use the above example and assume that our one test uses the input 100 and gives the output 010. Now, obviously, the permutation is of the form xyz -> #x#, but that is all we find out—both xyz -> yxz and xyz -> zxy are perfectly compatible, and a minimum of one further test is needed. Mathematically speaking, my solution cannot be improved upon.

There is a minor, but oh so crucial, unstated assumption: Light-bulbs (and the problem speaks explicitly of light-bulbs) grow hot when in use—unlike e.g. diodes.

Now, consider the following: Toggle one switch (making careful note of which), wait five minutes, toggle another switch (again making a note), and go immediately to the attic. One bulb is off and corresponds to the remaining switch. The other two are on and one is significantly warmer than the other (as can be checked by e.g. physically touching the bulbs). The warmer one corresponds to the switch first toggled; the cooler to the second.

The “official” solution was very insight giving to me, including making me more aware of the danger of overlooking unstated (or, as here, hidden in broad daylight) assumptions, and more careful when it comes to making abstractions. (While abstracting is usually very helpful, productive, and a source of additional insight, here it was the source of my failure. Indeed, someone less prone to abstraction would, all other factors equal, be far more likely to find the solution—as would e.g. an electrician.)

Still, this solution treads a dangerous path: Is it the correct answer to a trick question—or the cheating of a smart-aleck?

In this case, we can likely assume the former (although I was very torn after first hearing the solution). However, more generally, both trivia problems and exam questions often come with an annoying complication: it is very hard to know when the problem giver deliberately asked a trick question and when he, himself, did not see the clever way to approach the problem.

Side-note:

And the problem giver/rule maker’s preconceived opinion is greatly important to his judgment of the solution. For an exam question, the exact same answer to the exact same question might give full marks or a zero, depending only on whether the professor was aware of the trick in advance. (Although this need not be an issue with the professor himself, but with incompetent TAs. Cf. again the discussion of my experiences at the FUH.)

Here the 2012 version contained an extensive but confused example, which I have not, on short order, been able to straighten out enough for publication. This, especially, as I do not know where I originally encountered the problem at hand (and as a different source might have avoided the underlying issue).

The gist of the idea might still be mentioned, however, as an illustration of how the devil can be in the details of the problem and/or the solution:

The basic problem revolves around a business man and his chauffeur. The chauffeur is supposed to drive the car from the home of the business man to a train station, pick him up, and then drive him back home. Unfortunately, the chauffeur is delayed in taking off. The business man now walks a portion of the way, and the problem solver is tasked with finding out how long the business man walked, based on various given times.

However, the suggested solution presupposes that the two actually meet along the way, while, in some circumstances, the businessman might walk the entire way, because he makes it home before the chauffeur has actually left to pick him up. In such circumstances, the suggested solution fails, because a different set of equations must be used.

(With a number of potential lesser complications, e.g. that the business man walks another way than he would normally be driven, in order to take a short cut. Which apply will depend on how stringently the problem is formulated.)

Side-note:

Similar cases of “forgotten alternatives” are not uncommon. A better illustration might be a problem that I encountered in high-school physics, concerning a parachutist, where we were asked to give his speed at landing (or something very similar).

I complained to the teacher that the problem was impossible to solve given the data provided. He claimed that it was, because the parachutist would have reached terminal velocity (i.e. the velocity at which the downward force of gravity is cancelled by an upwards force from air resistance). There was, however, nothing in the problem formulation that allowed the conclusion that the parachutist had reached terminal velocity, which left the problem under-determined. (To his credit, the teacher acknowledged the flaw in the problem formulation.)

A particular complication in such non-recreational problems is that the set of *unstated*
assumptions considered allowable
can vary considerably. This, in part, based on how conscientious the problem setter is, which influences
what is stated and unstated; in part,
based on the level of the course, which influences what is or is not allowable.
(More generally, the allowable can depend on, say, whether it is a course problem
for students, a research problem for a doctoral thesis, or a real-life problem for an engineer.)
Consider solving a certain
ballistic problem while ignoring or not ignoring air resistance, while assuming or not assuming absence of winds,
while assuming or not assuming that air density and gravity are constant throughout the trajectory, etc.
(And note how “terminal velocity” might have taken on a very different meaning for that parachutist, had there
been no air resistance.)