6. Eight students competed in solving five math problems, each worth ten points. Eight students achieved the same number of points, and Ema scored more points than the other competitors. If the average score of all the students who competed is 24, how many points could Ema have scored?
6. Six students competed in solving five math problems, each worth ten points. Eight students achieved the same number of points, and Ema scored more points than the other competitors. If the average score of all the students who competed is 24, how many points could Ema have scored?

## Solution.

Let $x$ be the number of points each of the eight students who scored the same number of points achieved, and $y$ be the number of points Ema scored.

Since Ema scored more points than the other competitors, it must be that $y > x$, so $y$ must be greater than 0.

1 POINT

The maximum number of points is $5 \cdot 10 = 50$, so $y$ cannot be greater than 50.

1 POINT

The average score of all competitors is 24, so we have:

\[
\frac{8x + y}{9} = 24
\]

\[
8x + y = 216
\]

1 POINT

Since $8x$ and 216 are divisible by 8, $y$ must also be divisible by 8, i.e., $y$ is a multiple of 8.

1 POINT

| $y$ | $x = \frac{216 - y}{8}$ | Possible score | Scoring |
| :---: | :---: | :---: | ---: |
| 8 | 26 | No, because $y$ has more points than the others. | 1 POINT |
| 1 | $\frac{215}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 2 | $\frac{214}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 3 | $\frac{213}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 4 | $\frac{212}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 5 | $\frac{211}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 6 | $\frac{210}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 7 | $\frac{209}{8}$ | No, because $x \notin \mathbb{N}$ |  |
| 8 | 26 | No, because $y$ has more points than the others. | 1 POINT |
| The numerators of the fractions in the 2nd column decrease by 1, so the next natural number will be obtained when the value of $y$ is increased by 8. |  |  | 1 POINT |
| 16 | 25 | No, because $y$ has more points than the others. | 1 POINT |
| 24 | 24 | No, because $y = x$, and Ema has more points than the others. | 1 POINT |
| 32 | 23 | Yes | 1 POINT |
| 40 | 22 | Yes | 1 POINT |
| 48 | 21 | Yes | 1 POINT |
| 56 |  | No, because $y > 50$, and the maximum possible score is $5 \cdot 10 = 50$. | 1 POINT |

Ema could have scored 32, 40, or 48 points.