ogeometrie: noiembrie 2020

sâmbătă, 28 noiembrie 2020

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

W ( $a\in \mathbb{R},\;a>0,\;a\neq1$ )

$log_{a^{k}}t^{k}=log_{a}t$ , 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.

$\blacksquare$

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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=

marți, 24 noiembrie 2020

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), $n=1\cdot 7^{3}+2\cdot7^{2}+5\cdot 7+4=343+98+35+4=480$ . I $2n=960=2541_{7}$ .