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Two people, Matthew and William, needed to pay a mysteriously talking dog $11 in order for it to rent them an electric sled. Unfortunately, the two didn't have any money, so they offered the dog treats instead, and it agreed.

There were three types of the treats the two had: red, yellow, and blue.

Eleven red treats are worth $12.

Eleven yellow treats are worth $13.

And finally, eleven blue treats are worth $14.

The dog only accepted exactly $11 worth of 10 treats.

To solve this problem, you need to:

  • Calculate the number of red, yellow, and blue treats that Matthew and William needed to give the dog to satisfy its requirements.

Hint 1

How much is one of each treat?

Hint 2

The formula in order to calculate the treats is: $\frac{12r+13y+14b}{11}$
($r$ being the price of one red treat, $y$ being the price of one yellow treat, and $b$ substituting the price of one blue treat.)

This puzzle was based on the comic series: '수학도둑', Issue 27 by: Yeo-un-bak

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    $\begingroup$ Welcome to Puzzling, take our tour! Could you please provide proper attribution for this question? $\endgroup$
    – Alex Ravsky
    Commented Nov 24 at 8:21

3 Answers 3

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Let they have to give $r$, $y$, and $b$ red, yellow, and blue treads, respectively.

Then we have to find a solution of the following system in nonnegative integers.

$$\begin{cases} r+y+b=10 \\12r+13y+14b=11\cdot 11.\end{cases} $$

Subtracting the first equality multiplied by $12$ from the second, we obtain $y+2b=1$.

So $y=1$, $b=0$, and thus $r=9$.

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A simpler idea:

If we give the dog ten red treats, we'll be short $\\\$\frac{1}{11}$. We can make up this difference by replacing one red treat with a yellow treat, so one yellow and nine red treats will satisfy the dog.

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  • $\begingroup$ nice answer not requiring symbolic mathametics, just a clever simple practical pragmatic idea and approach and some basic primary school arithmetic calculations $\endgroup$
    – FirstName LastName
    Commented Nov 26 at 14:06
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Just to compliment existing answers.

Nice puzzle IMO because of two aspects

    1. question is entertaining but also a bit misleading and teasing

eleven blue treats are worth $14

or, if you want, any subsequent extra

eleven green treats are worth $15

and so on

do not matter

    1. question is recreational : no mathematical formula is required to find solution

That said: as in fact even

eleven does not matter

underlying mathematics is that

$f(x)=(((x-2)/x)*(x+1))+((1/x)*(x+2))$

is just a fancy way to write

$x$ (as one can simplify)

here

[x=11] : $(9/11) * 12 + (1/11) * 13 = 11$

so

exactly $n worth of n-1 treats always works

and one can also verify

 julia> f(x) = (((x - 2) / x) * (x + 1)) + ((1 / x) * (x + 2))
 f (generic function with 1 method)

 julia> f(11)
 11.0

 julia> f(1)
 1.0

 julia> f(1000)
 1000.0

 julia> f(-0.5)
 -0.5

 etc...

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