1. a) (3p) Let the fraction: $\frac{1+2+3+\ldots+n}{650-649+648-647+\ldots+4-3+2-1}, n \in N^{*}$. Determine $n \in N^{*}$ such that the fraction is unitary.

## Prof. Constantin Ciobîcă

b) (4p) Determine the natural numbers $m, n$, and $p$, non-zero and distinct, knowing that:

$$
\frac{4}{1+m}+\frac{9}{m+n}+\frac{30}{m+n+p} \geq 10
$$

Prof. Dorel Ispăşoiu
## Solution:

a) $650-649+648-647+\ldots+4-3+2-1=\underbrace{1+1+\ldots+1}_{\text {325 times }}=325$

$1+2+3+\ldots+n=\frac{n \cdot(n+1)}{2}$

The fraction becomes $\frac{n \cdot(n+1)}{650}$. It is a unit fraction if $n \cdot(n+1)=650$.

$650=2 \cdot 13 \cdot 5^{2}=25 \cdot 26 \Rightarrow n=25$.

b) We have $m, n, p \in N^{*}, m+1 \geq 2$, so $\frac{4}{1+m} \leq \frac{4}{2}=2$. Also $m+n \geq 3$, so $\frac{9}{m+n} \leq \frac{9}{3}=3$;

$m+n+p \geq 6$, so $\frac{30}{m+n+p} \leq \frac{30}{6}=5$

Adding the three inequalities member by member, we get:

$\frac{4}{1+m}+\frac{9}{m+n}+\frac{30}{m+n+p} \leq 10$, so $\frac{4}{1+m}+\frac{9}{m+n}+\frac{30}{m+n+p}=10$

This means that in each case, the equalities must be satisfied. Therefore, $m=1, n=2$ and $p=3$.

## Scoring:

| Calculation of the denominator | $\mathbf{1 p}$ |
| :--- | :--- |
| Calculation of the numerator | $\mathbf{1 p}$ |
| Determination of $n$ | $\mathbf{1 p}$ |
| $m+1 \geq 2$, so $\frac{4}{1+m} \leq \frac{4}{2}=2$ | $\mathbf{2 p}$ |
| $m+n \geq 3$, so $\frac{9}{m+n} \leq \frac{9}{3}=3$ |  |


| $m+n+p \geq 6$, so $\frac{30}{m+n+p} \leq \frac{30}{6}=5$ |  |
| :--- | :--- |
| Adding the three inequalities member by member, we get: |  |
| $\frac{4}{1+m}+\frac{9}{m+n}+\frac{30}{m+n+p} \leq 10$, so $\frac{4}{1+m}+\frac{9}{m+n}+\frac{30}{m+n+p}=10$ | $\mathbf{1} \mathbf{~ p}$ |
| Finalization | $\mathbf{1} \mathbf{p}$ |