# The coach, the shoes and the whistle

I recently received the following mathematical puzzle in a Whatsapp group:

After a lot of debating, the group got to the conclusion that the answer was

11

I however, was skeptical. After a while I realized that ? can, in fact, be any number.

You have to show this. For simplicity, you need to show that the solution can be both the number above (i.e. what the Whatsapp group agreed), and 8.162 (rounded to the third decimal).

After doing the above, it becomes evident that any number can solve the problem.

Note: a 50 reputation bonus will go to the one who can mathematically show that any number can solve the problem.

• Could you please clarify: do you assume, that since neither man without whistle, nor single shoe nor single whistle appear in the first three lines, we can chose to ignore the first three lines altogether and define the three icons in the last line in any way we like (which would be the equivalent to defining them as arbitrary functions of the related icons from the first three lines)? If that's what you mean, then obviously (by definition of "arbitrary") the answer can be any number, but as obviously that's not what the author of the original puzzle meant. – Andrew Savinykh Feb 26 '18 at 2:57
• @AndrewSavinykh I guess I am assuming that there is a relationship between one whistle and two whistles, one shoe and two shoes, and the man with the whistle and the man without the whistle and the whistle. Treating the last three symbols as whatever would of course make the puzzle a non-puzzle, rather an annoying thing. – luchonacho Feb 26 '18 at 9:22
• Why dismiss the most natural interpretation of the problem as an "assumption" only to make assumptions anyway, and arbitrary ones at that? For instance, why are you assuming that the answer need be a number? – noedne May 1 '18 at 3:55

I guess it is something like that:

Who said a pair of shoes value is the sum of two shoes value? For an example, I will say values of two objects together multiply.
Then we have 1 shoe value is sqrt (10). A whistle is 2 and a man is 5/2.

Making the result:

sqrt (10) + 5/2 * 2 = 8.16

• Very nice, I think you've nailed it. Interesting that the whistle results in the same irrespective of the operation used due to its value, so this difference plays into the shoes only. – Phylyp Feb 25 '18 at 16:59
• Precisely! Well done! Most of the people just goes for the most intuitive assumption, but this remains an assumption! The 50 rep bounty (coming) is still open for whoever demonstrates that actually, any number could solve the problem. – luchonacho Feb 25 '18 at 19:56

The problem does not specify how to compute the value of multiple objects combined into a single picture, and so multiple choices are possible. As other answers have explained, using addition gives an answer of 11, whereas multiplication gives 8.162.

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However, these aren't the only two possible choices. Arguably, the only hard requirement is that the valuation for a pair of objects be a commutative function, as there is no clear ordering of the pair of objects being combined. This allows for an infinite number of possible functions, so it is reasonable that we can come up with one to get any desired answer.

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One family of such functions would be to take the sum but scale by an arbitrarily chosen constant, i.e. $f(x, y)=\frac{x + y}{k}$. (It makes the math simpler later to divide rather than multiply.) For $k=1$, this is the same as just adding the components, and we would get the answer 11. But there is no reason we can’t set $k$ to other values — any nonzero value we choose for $k$ would still be consistent with the given facts; it would just scale the “base value” of the standalone symbols. Specifically, the base values for one shoe, a man without a whistle, and a whistle would be $5k$, $3k$, and $2k$.

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The final line then becomes $5k + 3k\times2k =$ ?

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This allows us to choose a $k$ so as to give any desired value for the ?: Let’s call our desired answer $d$. We then get $6k^2 + 5k - d = 0$, which we can solve for $k$ using the binomial theorem:

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$k=\frac{-5 \pm\sqrt{(5)^2 -4(6)(-d)}}{2(6)} =$ $\frac{-5 \pm\sqrt{25 + 24d}}{12}$

This has at least one nonzero solution for all real values of $d$, demonstrating that any real number is a valid answer to the puzzle.

Using algebra, we arrive at:

A single shoe = 5 (and a pair = 10)
A coach = 5
A whistle = 2 (and a pair = 4)

However, note that:

The coach has a whistle around his neck in the first 3 occurrences
That implies that coach + whistle = 5
And in turn, coach without whistle = 3

That would result in:

? = shoe + (coach without whistle) x whistle = 5 + (3) * 2
Resulting in ? = 11

This covers the first part of the answer, matching what was originally agreed in your WhatsApp group.

Building on Untitpoi's answer, I'd say that we can say that the question mark can be any number because:

The principle can be extended as one likes with general functions.

For instance, if a single shoe is $x$ then a pair of shoes can be viewed as $f(x)$.
My answer above treated $f(x) = 2 \cdot x$, Untitpoi showed how $f(x) = x^2$ gives a different answer.

Instead, if $f(x)$ is some other function (say $f(x) = ln(x)$ ), it can result in some other number. The choice of different functions for deriving a pair of shoes from a single shoe, a pair of whistles from a single whistle, and a coach with/without a whistle can all generate different numbers.

• This is the first part. The second on is the key. I will give a hint later on if nothing happens. (there is a typo on your first hidden message, bottom). If I may however, do not go the route of size/orientation. This would make the problem just annoying. This is ultimately a mathematical problem. If it looks like one shoe, it is meant to look like one shoe, of the same "nature" than any other shoe. – luchonacho Feb 25 '18 at 15:45
• Thanking you for pointing out that the size/orientation is not the correct approach, that saves me from going down a rabbit hole :-). Typo fixed as well. – Phylyp Feb 25 '18 at 16:12
• @luchonacho - I've edited my answer to potentially answer the second part of the question, building on Untitpoi's answer. – Phylyp Feb 26 '18 at 3:15

I think I understand what you mean. Let's letter a man with the whistle as $M'$, pair of shoes as $S'$ and pair of whistles as $W'$. If we also letter a man without the whistle as $M$ a single shoe as $S$ and a single whistle as $W$, then most people will assume that: $$S'(S)=2S$$ $$W'(W)=2W$$ $$M'(M,W)=M+W$$

The frame of your question is to challenge the above and claim that any three functions can be used.

Note how the first two functions here depend on one variable, and the last one on two. A function without variables is just a special case of a function with $N$ variables. So we can define $S'$, $W'$ and $M'$ to be constants. Say, $S'$ and $W'$ to be equal zero, and $M'$ to be equal to the desired answer, which can be anything. That gives you the proof, you asked for.

This solution is anticlimactic though. If we go this far, why not also redefine what + and x and = mean or even what 30 and 20 and 13 mean.

I could guess, that you probably had in mind some way of defining these functions according to some set of rules, that you did not publish ("reasonable" assumptions), so that one can get any result, but since you did not publish these it's impossible to confidently guess what you had in mind as what is "reasonable" often highly subjective. Of course if you are allowed to interpret the puzzle in any way you like you can get any result you like.

In any case, this is hardly what the original author of the puzzle meant.

I disagree, the answer will always be 11 as it is a 3 equations 3 unknowns system. The last equation uses the results of the system.

1-> 1 shoe = 5
2-> coach with a whistle = 5
3-> whistle = 2
In the last one the coach doesn't wear a whistle, so coach alone = 5-2=3
5+3*2=11, multiplication is to be done first

• Thanks, but this is not correct. You will see by yourself once the solution is presented. – luchonacho Feb 25 '18 at 15:46
• @luchonacho you can say if this correct or not, if it's clear what the rules are. Under axiomatics assumed in this answer , this is entirely correct, and on the face value, nothing in the question is given contradicts this axiomatics (excluding your conjectures, of course). – Andrew Savinykh Feb 26 '18 at 2:39
• @AndrewSavinykh assumed but not made explicit. That is the error. To give as settled that just one set of assumption was obvious. – luchonacho Feb 26 '18 at 9:23

I am probably very wrong, but I want to point something out that may lead to an answer.

What if you were to look at the picture like this(reading left to right continuously): 3 pairs of shoes = 30(1 pair of shows + 2 coaches) = 20(1 coach + 2 pairs of whistles) = 13(1 pair of shoes +(whistle less coach + 1 whistle) = ?. It could also look like this: 6a = 30(a+2(b+c) = 20(4c + (b+c)) = 13(c+(b*c) = ?. This might allow for the "?" to be anything. I have seen similar riddles and math equations that require you to think this way.

To get a lot of answers, simply observe that:

we have not been shown the coach's feet, and so we cannot make any assumptions about whether each coach has zero, one (left or right) or two shoes on. This is finitely enumerable, with $$4^3=64$$ possible sets of equations to be solved.

• Haha, very true! – luchonacho Jan 8 '19 at 17:30

If Coach with whistle is 5,
and Coach/whistle plus 2 whistles could only mean that the whistles are worth 4 each because 5+4+4=13. So the coach without a whistle has to be 5-4=1. Then the last answer has to be 5+1*4=9

• Have you seen the other answers? – luchonacho Nov 10 '18 at 16:11

First line: a shoe is $$5$$.
Second line: we minus the $$2$$ shoes, giving us $$10$$... so $$2$$ Man and Whistle = $$10$$. This means $$1$$ Man and Whistle is $$5$$... The combination is either $$4 + 1$$ or $$2 + 3$$ to make this.
Third Line: $$13$$ - Man and Whistle ($$5$$) $$= 8$$ then we take $$2$$ whistle to make $$0$$... meaning a Whistle is $$4$$ and a man alone is $$1$$
Fourth Line: we have $$5 + 1 \times 4$$... BODMAS means $$4 \times 1 = 4 + 5 = 9$$.