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 ...

=end added=

PROBLEM MA93 - Crux Mathematicorum V46N9

The statement of the problem

En(p. 436)

Fr(p. 437)

ANSWER CiP

$k=\frac{66}{7}$

Actually $x=2^{-\frac{48}{7}}, y=2^{\frac{36}{7}}, z=2^{\frac{78}{7}}$

 

          Solution CiP

     We use the formula ($a\in \mathbb{R},\;a>0,\;a\neq1$)

$log_{a^{k}}t^{k}=log_{a}t$, where $k\in \mathbb{R},\;k \neq 0,\;t>0$

which says that the logarithm of a number $t$ remains unchanged if the base of logarithm, $a$, and the number $t$ are raised to the same power $k$.

     The simultaneous equations in the statement are equivalent to (we take $k=\frac{1}{2}$ in the first term and $k=\frac{1}{3}$ in the second term of each equation)

$\begin{cases}log_{2}\sqrt{x}+log_{2} \sqrt[3]{yz}=2 \\log_{2}\sqrt{y}+ log_{2}\sqrt[3]{xz}=4 \\log_{2} \sqrt{z}+log_{2}\sqrt[3]{xy}=5 \end{cases}$

 and from here we obtain the equivalent equations

$\begin{cases}\sqrt{x}\cdot \sqrt[3]{yz}=2^{2}\\ \sqrt{y}\cdot \sqrt[3]{xz}=2^{4}\\ \sqrt{z} \cdot \sqrt[3]{xy}=2^{5} \end{cases}$                       (1).

     Suppose there are x, y and z for which the last equations are verified (otherwise any conclusion can be deduced). If we multiply the three equations we get

$\sqrt{xyz}\cdot \sqrt[3]{y^{2}z^{2}x^{2}}=2^{11}\Leftrightarrow (xyz)^{\frac{1}{2}+\frac{2}{3}}=2^{11}\Leftrightarrow xyz=2^{\frac{66}{7}}$

where do we get the answer.

      Replace in the first equation of (1) $yz=\frac{2^{\frac{66}{7}}}{x}$ we get

$x^{\frac{1}{2}}\cdot 2^{\frac{22}{7}}\cdot x^{-\frac{1}{3}}=2^{2}$ 

 where we get the value of $x=2^{-\frac{48}{7}}$. The same goes for $y=2^{\frac{36}{7}},\;z=2^{\frac{28}{7}}$ and these numbers check the initial equations.

$\blacksquare$

 

Raspunsul de confirmare a expedierii

Thank you for sending us your solution for one of the problems found
in Crux Mathematicorum.  Our file server has received your submission
and the appropriate editor will be reviewing it in sequence or as
needed.


Below is a receipt confirming the submission you sent us.



Tracking Number: 11004
Received:        Sat Nov 28 16:34:27 2020

From:  Petre Ciobanu
       Scoala Gimnaziala "Samuil Micu" SADU
       Sibiu, Romania
Email: ptr.ciobanu@gmail.com

Type:  Solve a MathemAttic Problem
       (problem MA093)

Files:
  Poblem_MA93_v46n9.pdf
  Sursa_latex_problema_MA93.doc


Comments:

Crux Mathematicorum <crux@cms.math.ca>	sâm., 28 nov., 18:36 (acum 2 zile)
către eu

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

Added May 19, 2021

     Good answer see V447n04, pages 174-175 

          They sum up the equations and get

$log_4{x}+log_4{y}+log_4{z}+log_8{(yz)}+log_8{(zx)}+log_8{(xy)}=11$

 $\Leftrightarrow\;\;log_4{(xyz)}+log_8{x^2y^2z^2)}=11$.

 If we substitute $xyz=2^k$ we get $k\cdot log_4{2}+2k \cdot log_8{2}=11$ and since $log_4{2}=\frac{1}{2},\;\;log_8{2}=\frac{1}{3}$ we find that $\frac{k}{2}+\frac{2k}{3}=11$ hence $k=\frac{66}{7}$.

 =end added=

GH. SIRETCHI - Functii cu PROPRIETATEA DARBOUX

 Univ. BUCURESTI - 1993

Se poate descarca aici

EXERCIȚII FUNCȚIA LINIARĂ

 Este vorba despre functia

$f:\mathbb{R}\rightarrow \mathbb{R},\;\;f(x)=a\cdot x+b,\;\;a\in \mathbb{R},\;b\in \mathbb{R}$

Si la final un MODEL de SIMULARE a EVALAURII NATIONALE

SUCCES !

APICS Mathematics Competitions

 Vezi http://www.math.unb.ca/apics.papers/

APICS Mathematics Contest 1978

 

     Problem 1.  The expression of a positive integer, n in base b is

It is known that the expression of the integer 2n in the same base is

Determine the values of b and n in base 10.

 

ANSWER CiP $b=7$,  $n=480$

      Solution CiP

(1)     $n=1254_{b}=1\cdot b^{3}+2\cdot b^{2}+5\cdot b+4$

$2n=2541_{b}=2\cdot b^{3}+5\cdot b^{2}+4\cdot b+1$

so we have equation

$2(b^{3}+2b^{2}+5b+4)=2b^{3}+5b^{2}+4b+1$

 $\Leftrightarrow b^{2}-6\cdot b-7=0$,

that is, a simple equation of degree 2 whose roots are $b_{1}=-1$ and $b_{2}=7$. But it needs that $b>5$ so $b=7$ and n will be calculated quickly with formula (1), $n=1\cdot 7^{3}+2\cdot7^{2}+5\cdot 7+4=343+98+35+4=480$. It is easily verified that $2n=960=2541_{7}$.

$\blacksquare$