PROBLEM MA127 CRUX MATHEMATICORUM V47 No 6

 Pag 275 - En

Pag 276 - Fr

ANSWER CiP 

$\log_5 12=\frac{2\cdot a+b}{1-a}$

     Solution CiP

          We have $\log_5 12 =\log_5(2^2 \cdot 3)=\log_5 2^2+\log_5 3$, so

$\log_5 12 =2\cdot \log_5 2 + \log_5 3. \tag{1}$

          First, $\frac{1}{a}=\log_2 10=\log_2 (2\cdot 5)=\log_2 2 +\log_2 5=1+\frac{1}{\log_5 2}$ so

$\frac{1}{\log_5 2}=\frac {1}{a}-1$, whence it results

$\log_5 2 =\frac {a}{1-a}. \tag{2}$

           Secondly, $\frac{1}{b}=\log_3 10 =\log_3 2 +\log_3 5=\frac{\log_{10} 2}{\log_{10} 3}+\frac{1}{\log_5 3}$, so $\frac{1}{\log_5 3}=\frac {1}{b}-\frac {a}{b}$, whence it results

$\log_5 3=\frac{b}{1-a}. \tag{3}$

          Replacing formulas (2) and (3) in formula (1) we get the answer.

$\blacksquare$

          

REMARK CiP

          More briefly $\log_5 12=\frac{\log_{10} 12}{\log_{10} 5}=\frac{\log_{10} 2^2 \cdot 3}{\log_{10} \frac{10}{2}}=\frac{\log_{10} 2^2 +\log_{10} 3}{\log_{10} 10-\log_{10} 2}=\frac{2\cdot \log_{10} 2 +\log_{10} 3}{1-\log_{10} 2}$ and we get the same answer.

$\blacksquare \blacksquare$

 

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

Type:  Solve a MathemAttic Problem
       (problem MA127)

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Crux Mathematicorum <crux@cms.math.ca>	10:12 (acum 22 de minute)
către eu

PROBLEM OC531 - CRUX MATHEMATICORUM, V47n5

 Pag 238 - En

Pag 239 - Fr

ANSWER CiP

If $7 \mid k$ then solutions are $(k,0),\;(k,k),\;(\frac{3k}{7},-\frac{k}{7})$,

and, if any, the pairs $(x_1,y_1),\;(k-x_1+y_1,y_1).$

If $7 \nmid k$ then the solution $(\frac{3k}{7},-\frac{k}{7})$(although rational),

 is not made up of integers; the others remain valid.

Examples

 If $k=-4$, the equation has solutions

$(-4,0),\;(-4,-4),\;(-1,-3),\;(-6,-3).$

If $k=7$, the equation has solutions

$(7,0),\;(7,7),\;(3,-1),\;(3,6),\;(10,6).$ 

 

 

Solution CiP

          There are two pairs $(x,y)$ that obviously check the equation: $(k,0)$ and $(k,k)$.

          The given equation is equivalent to

$$x^2-xy+2y^2-kx-ky=0\; ;\; x+y \neq 0. \tag{1} \label{eq1}$$

           Let $(x_1,y_1)$ be a solution of equation (1), with the condition $x_1+y_1 \neq 0$. This means that the equation with unknow $x$

$$x^2-(k+y_1)x+2y_1^2-ky_1=0 \tag{2}$$

has the solution $x=x_1$. Being of the second degree for  $x$, she has another solution $x_2$ which, according to Viete's Relations, she checks $x_1+x_2=k+y_1$, so $x_2=k-x_1+y_1.$

     For the pair $(x_2,y_1)$ to fit the given problem it must be $x_2+y_1 \neq 0$. But $x_2+y_1=0 \; \Leftrightarrow \;k-x_1+2y_1=0 \; \Leftrightarrow \; x_1=k+2y_1$, and writing that $k+2y_1$ check equation (2) we immediately get $y_1=0$, then $x_1=k$.

          So for any solution $(x_1,y_1) \neq (k,0)$ of the equation (1) we get another solution by the transformation

$$(x_1,y_1)\; \mapsto \; (k-x_1+y_1,y_1). \tag{3}$$

       We have $(k,k) \overset{(3)}{\mapsto} (k-k+k,k)=(k,k)$, so by transforming (3) we do not get a NEW solution. Let's look for all the solutions that remain invariant through this transformation, so $k-x_1+y_1=x_1$, whence $y_1=2x_1-k$. If we write that $(x_1,2x_1-k)$ check equation (1)

$$x_1^2-x_1(2x_1-k)+2(2x_1-k)^2-kx_1-k(2x_1-k)=0$$

$\Leftrightarrow \;7x_1^2-10kx_1+3k^2=0$ we get $x_1=k$ or $x_1=\frac{3k}{7}$. In the secon case $y_1=-\frac{k}{7}$. So the only invariant pairs by transformation (3) are $(k,k)$ and $(\frac{3k}{7}, -\frac{k}{7})$.

          So, if $k \;\vdots \; 7$ (means "k is divisible by 7"), then the equation (1) has three solutions $(k,0),\; (k,k),\;(\frac{3k}{7},-\frac{k}{7})$ to which, if there are others $(x_1,y_1)$, an even number of solutions $(x_1,y_1)$ and $(k-x_1+y_1,y_1)$ is added. In total an odd number.

     Otherwise ($7 \nmid k$) the solution $(\frac{3k}{7},-\frac{k}{7})$ does not consits of integers, and the number of solutions is even.

$\blacksquare$

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Tracking Number: 12749
Received:        Wed Jul  7 08:52:45 2021

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

Type:  Solve an Olympiad Corner Problem
       (problem OC531)

Files:
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I could not get a PDF of this document

Crux Mathematicorum <crux@cms.math.ca>	11:53 (acum 7 ore)
către eu