At a university in Budapest, the most successful students can apply for two types of scholarships. One of the conditions for the outstanding scholarship is that the average of the grades earned in the last semester must be above 4.5. For the republican scholarship, however, an average of at least 4.51 is required.

How many grades does someone need to earn to have an average above 4.5, but not reach 4.51?
Let the number of grades be $n$, and their sum $s$. The fact that the average is above 4.5 but does not reach 4.51 means that

$$
4.5<\frac{s}{n}<4.51
$$

Multiply both sides of the inequality by $2n$, and subtract $9n$:

$$
9n<2s<9.02n \quad 0<2s-9n<0.02n
$$

The $2s-9n$ in the middle is an integer. The fact that it is greater than 0 means it is at least 1: (1)

$$
1 \leq 2s-9n<0.02n
$$

from which

$$
1<0.02n
$$

that is,

$$
50<n
$$

follows. Since $n$ is a positive integer (the number of grades), this inequality means that $n$ is at least 51.

If $n=51$, then it is possible to choose the grades so that the condition is satisfied. For example, if someone gets 26 fives and 25 fours, then their average is:

$$
\frac{26 \cdot 5 + 25 \cdot 4}{51} = \frac{230}{51} \approx 4.5098
$$

Thus, at least 51 grades are needed for an average between 4.5 and 4.51. (Otherwise, it is quite rare for someone to have so many grades in one semester!)

Gergely Herényi (Budapest, József A. Gymnasium, 3rd grade)

Note. The $n$ cannot be any number greater than 50. If $n$ is even, then $2s-9n$ is a positive even number, so instead of (1) we have

$$
2 \leq 2s-9n<0.02n
$$

from which we get

$$
100<n
$$

The inequality. However, all odd numbers greater than 50 and even numbers greater than 100 are possible, because in these cases the value of $s$ can be chosen so that $2s-9n=1$, or $2s-9n=2$. For example, in the case of an odd $n$, $\frac{n+1}{2}$ fives and $\frac{n-1}{2}$ fours, and in the case of an even $n$, $\frac{n}{2}+1$ fives and $\frac{n}{2}-1$ fours can achieve the desired average.