SUBIECTUL II

a) Find the number of natural divisors of the number $15^{51}$ that are multiples of the number $225^{20}$.

b) Show that the number $A=\frac{21^{n}+23^{n}-2^{2 n}+2^{n+1} \cdot 3^{2}}{38}$ is a natural number for any non-zero natural number $n$.

Supliment Gazeta Matematică $3 / 2013$
## SUBJECT II

a) Find the number of natural divisors of the number $15^{51}$ that are multiples of the number $225^{20}$.

b) Show that the number $A=\frac{21^{n}+23^{n}-2^{2 n}+2^{n+1} \cdot 3^{2}}{38}$ is a natural number for any non-zero natural number $n$.

| SOLUTION | SCORE |
| :---: | :---: |
| $225^{20}=15^{40}$ | $1 \mathrm{p}$ |
| $15^{51}: 225^{20}=15^{11}=3^{11} \cdot 5^{11}$ | $1 \mathrm{p}$ |
| Multiples of $15^{40}$, divisors of $15^{51}$ are of the form " $15^{40}$ times divisors of $15^{11}$", in number  of | $1 \mathrm{p}$ |
| $(11+1) \cdot(11+1)=144$ |  |
| a) $(21)^{n}=(19+2)^{n}=M_{19}+2^{n}, \quad(23)^{n}=(19+4)^{n}=M_{19}+4^{n}$ | $1 \mathrm{p}$ |
| $2^{2 n}=\left(2^{2}\right)^{n}=4^{n}, 2^{n+1} \cdot 3^{2}=2^{n} \cdot 2 \cdot 9=2^{n} \cdot 18=(19-1) \cdot 2^{n}=M_{19} \cdot 2^{n}-$ | $1 \mathrm{p}$ |


| $2^{n}$ |  |
| :--- | :--- |
| $A=\frac{M_{19}+2^{n}+M_{19}+4^{n}-4^{n}+M_{19} \cdot 2^{n}-2^{n}}{38}=\frac{2 \cdot M_{19}+2^{n} \cdot M_{19}}{38}=\frac{M_{38}}{38} \in N$ | $2 p$ |