Romanian Mathematics Competitions // Математически състезания от Румъния

 Click on the image to download. The password is   ogeometrie

You can find other RMCs here.

Other Magazines can be found here.

          Example problem : 

                    "The natural numbers $x,\;y,\;z\;$ satisfy the equation

$$4\cdot x+7 \cdot z=7 \cdot y +9.$$

                      a) Show that $x+2\cdot y+3 \cdot z +4\;$ is a multiple of $5$.

                      b) Determine the remainder of dividing $2 \cdot x+y+5 \cdot z$ by $6$."

(The intercounty mathematics and computer science competition "Marian Țarină", 2024 p. 187)

ANSWER CiP

a) $x=7\cdot t +4,\;y-z=4 \cdot t+1,\;\;t\in \mathbb{N};$

                                    b) $3.$

Solution CiP

                    From $4 \cdot x+7 \cdot z=7 \cdot y+9$ we deduce $4 \cdot x-2=7 \cdot y-7 \cdot z+7$ and see that $4 \cdot x-2=7 \cdot a,\;a \in \mathbb{N}^*.$ 

          Expressing the number $x$ we ​​see from the calculation below that the number $a$ must be even, $a=2 \cdot a^{'}$ then that the number $a^{'}$ must be odd, $a^{'}=2 \cdot a^{"}+1$. 

$$x=\frac{7 \cdot a +2}{4}=\frac{7 \cdot 2a^{'}+2}{4}=\frac{7 \cdot a^{'}+1}{2}=\frac{7(2a^{"}+1)+1}{2}=7 \cdot a^{"}+4.$$

We write $a^{"}=t$ and get 

$$x=7 \cdot t+4,\;\;y-z=4 \cdot t +1, \;\;t \in \mathbb{N}.$$

               a) Substituting $z=y-4t-1$ we have

$$x+2y+3z+4=(7t+4)+2y+3(y-4t-1)+4=5y-5t+5,$$

obviously a multiple of $5$.

               b) Substituting $y=z+4t+1$ we have 

$$2x+y+5z=2(7t+4)+(z+4t+1)+5z=6z+18t+9=6 \cdot (z+3t+1)+3,$$

hence the remainder of dividing this number by $6$ is $3$.

$\blacksquare$