In the final of a ballroom dance competition, each of the twenty-five judges had to rate five dance couples with grades 1 to 5 as in school, with each grade being used exactly once. After the competition, a table was published showing the average grades for each couple:

| couple $\mathrm{A}$ | couple $\mathrm{B}$ | couple $\mathrm{C}$ | couple $\mathrm{D}$ | couple $\mathrm{E}$ |
| :---: | :---: | :---: | :---: | :---: |
| 4.68 | 3.86 | 3.36 | 1.44 | 1.60 |

It soon became clear that exactly one of the averages in the table was incorrect. Determine which average is incorrect and correct it.

One of the judges was the uncle of a dancer from couple E. However, he rated the couples honestly and without bias in the following manner: 1 for $\mathrm{D}$, 2 for $\mathrm{E}$, 3 for $\mathrm{B}$, 4 for $\mathrm{C}$, and 5 for $\mathrm{A}$. Nevertheless, after the corrected results were announced, he wondered whether he could have moved couple $\mathrm{E}$ to the first place by dishonest grading. Determine whether he could have achieved this. 

(L. Šimünek)
The average of the grades for each pair is equal to the sum of all assigned grades divided by 25. The sums of the grades for each pair should thus be:

$$
4.68 \cdot 25 = 117, \quad 3.86 \cdot 25 = 96.5, \quad 3.36 \cdot 25 = 84, \quad 1.44 \cdot 25 = 36, \quad 1.60 \cdot 25 = 40.
$$

We see that the sum of the grades for pair $\mathrm{B}$ is not a natural number, so the result for this pair was incorrect. Now we will correct it. The sum of all grades awarded in the competition was

$$
25 \cdot (1 + 2 + 3 + 4 + 5) = 25 \cdot 15 = 375.
$$

Subtracting the sums of the grades awarded to pairs $\mathrm{A}, \mathrm{C}, \mathrm{D}, \mathrm{E}$, we get the sum of the grades awarded to pair B:

$$
375 - 117 - 84 - 36 - 40 = 375 - 277 = 98.
$$

The correct average of the grades awarded to pair $\mathrm{B}$ was thus

$$
98 : 25 = 3.92.
$$

If the mentioned judge wanted to benefit pair $\mathrm{E}$ as much as possible, he could have given a 1 instead of a 2. This would reduce the sum of the grades for pair $\mathrm{E}$ from 40 to 39. To harm pair $\mathrm{D}$ as much as possible, he could have given a 5 instead of a 1, increasing the sum of their grades from 36 to 40. In this case, the result for pair $\mathrm{D}$ would be worse than the result for pair $\mathrm{E}$. The mentioned judge could thus move pair $\mathrm{E}$ to the first place by unfair grading.

Scoring. 2 points for identifying the incorrect average; 2 points for correcting it; 2 points for solving the remaining part of the problem.

Note. Multiplying by 25 is the same as multiplying by 100 followed by dividing by 4. The incorrect average can thus be detected by checking whether the two digits after the decimal point are divisible by 4. The correct average for pair $\mathrm{B}$ can be determined (without multiplying by 25 and then dividing by the same number) as follows:

$$
15 - 4.68 - 3.36 - 1.44 - 1.60 = 3.92.
$$