SUPLIMENTUL cu EXERCIȚII al GMB N0 1/2020

 

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GAZETA MATEMATICĂ Seria B N0 1/2019

 

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SUPLIMENTUL cu EXERCIȚII al GMB N0 1/2019

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Dedicated to CONSTANTIN STELESCU on its 60th anniversary //Problem 111 Crux Mathematicorum v2n2 (Feb 1976)

      In the magazine

in Volume 2 Number 2 (February, 1976) page 25

     
     Two solutions of the exercise appear in Number 5 (May, 1976 page 95-96)

        Solution CiP

          The numbers $a,b,c$ can be considered as the roots of a third degree equation

$$x^{3}-px^{2}+qx-r=0.$$

 With the numbers $x_{1}=b-c,\;x_{2}=c-a,\;x_{3}=a-b$ we will form a third degree equation that they check. We will first calculate 

$$x_{1}+x_{2}+x_{3}=0$$

$$x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2} \cdot x_{3}=$$

  $=(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)=bc+ca+ab-a^{2}-b^{2}-c^{2}=$

$$=3\cdot q-p^{2}\overset{\underset{\mathrm{def}}{}}{=}Q$$

the last equality being in accordance with the Vieta's formulas,

$$x_{1} \cdot x_{2} \cdot x_{3}=(b-c)(c-a)(a-b)\overset{\underset{\mathrm{def}}{}}{=}R.$$

 Again with the Viète's formulas (for those who prefer the original French name François Viète instead of the Latinised form of his name, "Franciscus Vieta"), the numbers $b-c,\;c-a,\;a-b$ will be the roots of the equation

$$x^{3}-0\cdot x^{2}+Q\cdot x-R=0.$$

 But then $\frac{1}{x_{1}},\;\frac{1}{x_{2}},\;\frac{1}{x_{3}}$ will be the roots of the "written backwards" equation

$$R\cdot x^{3}-Q\cdot x^{2}-1=0$$

 and the Vieta's formulas give

$\sum_{cycl}\frac{1}{x_{1}} =\frac{Q}{R}$,            $\sum_{cycl} \frac{1}{x_{1}}\cdot \frac{1}{x_{2}}=0$.

     Now $\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}+\frac{1}{(a-b)^{2}}=\sum \frac {1}{x_{1}^{2}}=(\sum \frac {1}{x_{1}})^{2}-2 \cdot \sum \frac {1}{x_{1}\cdot x_{2}}=(\frac{Q}{R})^{2}-2\cdot 0=$

$$=(\sum_{cycl} \frac {1}{b-a})^{2}$$

that is, a square number.

$\blacksquare$

 
 

 

 

T I T U L A R I Z A R E - programa noua incepand cu 2021

Programa de titularizare

 multumita https://profesorjitaruionel.com/wp-content/uploads/2020/12/Programa-Matematica-Titularizare2021.pdf

NUMEROTAREA unei Formule/Ecuatii

$\frac{1}{2}$

\begin{align}

ax+by=c

\end{align}

PROBLEM MA95 - Crux Mathematicorum V46N9

      The statement of the problem

En(p. 436)

Fr(p. 437)

 ANSWER CiP

Smallest prime factor 3; largest prime factor 673

    Solution CiP

    According to the formula

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

the number is 

 $M=2(1+2+\cdots +2018)+2019=2018 \cdot 2019+2019=2019^{2}$.

Because 2019 has the decomposition into prime factors $3 \cdot 673$ we get the answer.

$\blacksquare$ 

=======================================================================

Added May 19, 2021

    
Good answer see V47n04, pages  176-177

          And they get the result $M=2019^2$ and after applying the divisibility with 3 $(2+0+1+9=12=4 \cdot 3 )$ they obtain $2019^2=3^2 \cdot 673^2$. They further show that 637 is a prime number: check the divisibility criterions with 2, 3, 5 and 11; for 7, 13, 17, 19 and 23 (because $25^2=625 < 673< 676=26^2$) apply the division with the remainder ...

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