Suppose that we have boys and girls who competes in a tournament where each of the person will play against each of the person exactly one time and if he wins then he gets $1$ point if he loses then he gets $0$ point and if the result of game is draw then both of players get $0.5$ point. We know that quantity of boys is three times bigger than quantity of girls. And we know that after the end of tournament overall points of boys are $20\%$ bigger than girls overall points.

We need to find how many people could participate in this tournament.

Let $x$ is quantity of girls and then $3x$ is quantity of boys, $4x$ is quantity of people.

I know that overall points of all people is $\frac{4x(4x-1)}{2} = S$

Boys got $20\%$ more points so if $y$ are points which were got by girls then we have $y+1.2y=S \Rightarrow y = \frac{10x(4x-1)}{11}$. At most girls can get $x\cdot3x + \frac{x(x-1)}{2}$ points, because they can win all boys (each girl wins against each boy) and they need to distribute other points beetwen them. So we get inequaltiy: $\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$ and we get from it that $x = 1$ and overall quantity of people is $4$.

I think that there is an error (or errors) in my solution. Can somebody help me find them?


From inequality $\frac{10x(4x-1)}{11} \leq x\cdot3x + \frac{x(x-1)}{2}$ we get that $0 \leq x \leq 3$.

So quantity of people can be $4$, $8$ or $12$.

Am i right now?


1 Answer 1


Addendum added to respond to the comment/question of Arty.

Thanks to Ross Millikan for pointing out:

  • The obvious mistake of my confusing the total number of points scored with the total number of points scored by all of the girls.

  • The more subtle mistake that the average score of all of the girls must not be greater than what their score would be, if every girl beat every boy.

If $x$ girls scored $P$ points, then $3x$ boys scored $1.2P$ points, so $2.2P$ points were scored in total.

Let $~Q = 2.2P = ~$ be the total number of points scored.

Then, the total score of all the boys is $~\dfrac{6Q}{11}~$ and the total score of all the girls is $~\dfrac{5Q}{11},~$ so $~Q~$ must be a multiple of $~11.~$

Also, each game generated $1$ point, distributed as either [1,0] or [1/2,1/2]. Therefore $~Q~ = ~$ the number of games played.

Since there were $~x~$ girls and $~3x~$ boys, you have that

$$Q = \binom{4x}{2} = (2x) \times (4x - 1).$$

So, $~Q~$ must be chosen so that there exists a positive integer $x$ such that
$(2x) \times (4x-1) = Q.$

So, either $2x$ is a multiple of $11$ or $(4x - 1)$ is a multiple of $11$.

The smallest positive integer $x$ that can serve as a $\color{red}{\text{candidate value}}$ is $x = 3.$

Exploring this $\color{red}{\text{candidate value}}$:

  • The total number of games played is
    $\displaystyle \binom{12}{2}~ = 66 = Q.$

  • The total number of girls is $~3~$ and the total number of boys is $~9.$

  • The total number of points collectively scored by all of the girls is $~\dfrac{5Q}{11} = 30.$
    So, the average score of each girl was $~10.$

  • The total number of points collectively scored by all of the boys is $~\dfrac{6Q}{11} = 36.$
    So, the average score of each boy was $~4.$

$\color{red}{\text{Does this work?}}$

Suppose that every girl beat all 9 boys.

Further suppose that every girl:girl game and every boy:boy game ended in a draw.

Then each girl would score
$\displaystyle 9 + \left[2 \times \frac{1}{2}\right] = 10,$

and each boy would score
$\displaystyle 0 + \left[8 \times \frac{1}{2}\right] = 4.$

So, the $\color{red}{\text{candidate value}}$ of $x = 3$ works okay.


Responding to the comment/question of Arty.

First of all, your overall approach was $\color{red}{\text{better}}$ than my $\color{red}{\text{eventual}}$ approach, in one way, and flawed in another way.

That is:

  • You elegantly confronted the issue that the total score of the girls, assuming that each girl beat each boy, had to be at least $~\dfrac{5}{11}~$ of the total games played.

    Initially, I totally overlooked that issue. Then, after Ross Millikan commented, I edited my answer after the fact. However, I did not try for the elegant (and better) approach of establishing an inequality involving $~x.~$

    Instead, I (inelegantly) used the notion of a candidate value to verify that my solution worked.

  • You totally overlooked the constraint that the number of games played, $~\dfrac{4x(4x-1)}{2}~$ has to be a multiple of $~11.~$ This is because the total number of points scored by the girls is
    $\displaystyle y = \frac{5}{11} \times \frac{4x(4x-1)}{2}.$

$\displaystyle \frac{10x(4x - 1)}{11} \leq [x \times 3x] + \frac{x(x-1)}{2}.$

I agree with your updated interpretation of the above inequality. That is you have that:

$\displaystyle \frac{40x^2 - 10x}{11} \leq 3x^2 + \frac{x^2 - x}{2} = \frac{7x^2 - x}{2}.$

Cross multiplying, this implies that

$\displaystyle 80x^2 - 20x \leq 77x^2 - 11x.$

Since $x$ must be positive, (so $x$ is non-zero), you can divide the above inequality through by $x$ to give

$80x - 20 \leq 77x - 11 \implies 3x \leq 9 \implies x \leq 3.$

This begs the question, which value of $x$ is satisfactory.

You need $~\displaystyle \frac{4x(4x-1)}{2} = (2x) \times (4x-1)~$ to be a multiple of $11$.

Of the candidate values given by $~x \in \{1,2,3\}~$, only $~x=3~$ satisfies this constraint.

  • $\begingroup$ Thank you for the answer. I mostly try to understand is my proof correct or not. Can you check it? And only after I understand am i right or not i can look at your proof. Excuse me, maybe i was needed to add a tag "proof verification" to clarify my intentions. $\endgroup$
    – perepelart
    Commented Dec 30, 2022 at 7:29
  • $\begingroup$ @Arty See the Addendum that I have just added to my answer. $\endgroup$ Commented Dec 30, 2022 at 15:19
  • $\begingroup$ Thank you for your answer. Number of games played is needed to be divisible by $11$ because it is natural number and $5 \cdot 4\cdot x(4x-1)$ is not divisible by eleven if $x = 2$ or $x=1$. Did i get it right? $\endgroup$
    – perepelart
    Commented Dec 30, 2022 at 17:58
  • 1
    $\begingroup$ @Arty $(2x) \times (4x-1)$ needs to be divisible by $11.$ When $~x=1~$ or $~x=2,~$ the product is not divisible by $11$. $\endgroup$ Commented Dec 30, 2022 at 21:51

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