Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1973 Canadian MO Problems/Problem 1"

(Created page with "==Problem== <math>\text{(i)}</math> Solve the simultaneous inequalities, <math>x<\frac{1}{4x}</math> and <math>x<0</math>; i.e. find a single inequality equivalent to the two si...")
 
(→‎Solution)
 
Line 13: Line 13:
  
 
==Solution==
 
==Solution==
−
<math>\text{(i)}</math> <center><math>x<\frac{1}{4x} \Rightarrow 4x^2 < 1 \Rightarrow x^2< \frac{1}{4} \Rightarrow -\frac{1}{2} < x < \frac{1}{2}</math></center>
+
<math>\text{(i)}</math> <center><math>x<\frac{1}{4x} \Rightarrow 4x^2 > 1 \Rightarrow x^2> \frac{1}{4}</math>.
−
Since from the second inequality <math>x<0</math>, our solution is thus: <math>\boxed{-\frac{1}{2} < x < 0}</math>.
+
Since from the second inequality <math>x<0</math>, our solution is <math>x<-1/2</math>.
  
  
Line 34: Line 34:
 
&=& \boxed{\frac{1}{2}} \end{matrix}</math>  
 
&=& \boxed{\frac{1}{2}} \end{matrix}</math>  
 
</center>
 
</center>
−
 
−
 
  
 
==See also==
 
==See also==

Latest revision as of 15:14, 21 July 2012

Problem

$\text{(i)}$ Solve the simultaneous inequalities, $x<\frac{1}{4x}$ and $x<0$; i.e. find a single inequality equivalent to the two simultaneous inequalities.

$\text{(ii)}$ What is the greatest integer that satisfies both inequalities $4x+13 < 0$ and $x^{2}+3x > 16$.

$\text{(iii)}$ Give a rational number between $11/24$ and $6/13$.

$\text{(iv)}$ Express $100000$ as a product of two integers neither of which is an integral multiple of $10$.

$\text{(v)}$ Without the use of logarithm tables evaluate $\frac{1}{\log_{2}36}+\frac{1}{\log_{3}36}$.


Solution

$\text{(i)}$

$x<\frac{1}{4x} \Rightarrow 4x^2 > 1 \Rightarrow x^2> \frac{1}{4}$.

Since from the second inequality $x<0$, our solution is $x<-1/2$.


$\text{(ii)}$ From $4x+13 < 0$, we get that $4x+13 < 0 \Rightarrow x < -\frac{13}{4} \approx -3$. From $x^{2}+3x > 16$, we get:
$x^{2}+3x > 16 \Rightarrow x^{2}+3x - 16 > 0 \Rightarrow x > \frac{-3 + \sqrt{71}}{2} \approx 2 \text{ or } x < \frac{-3-\sqrt{71}}{2}\approx -5$

With these two inequalities, we see that the greatest integer satisfying the requirements is $\boxed{-6}$.


$\text{(iii)}$ $\frac{11}{24} = \frac{286}{624},$ $\frac{6}{13} = \frac{288}{624}$. Thus, a rational number in between $11/24$ and $6/13$ is $\boxed{\frac{287}{624}}.$


$\text{(iv)}$ $100000 = 10 \times 10 \times10 \times10 \times 10 = 2^{5}\times5^{5} = 32\times3125.$ Thus, $100000 = \boxed{32\times3125}.$


$\text{(v)}$

$\begin{matrix} \frac{1}{\log_{2}36}+\frac{1}{\log_{3}36} &=& \log_{36}2+\log_{36}3 \\ &=& \log_{36}{2\times3} \\ &=& \log_{36}6 \\ &=& \boxed{\frac{1}{2}} \end{matrix}$

See also

1973 Canadian MO (Problems)
Preceded by
1973 Canadian MO Problems 		1 • 2 • 3 • 4 • 5 Followed by
Problem 2
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1973_Canadian_MO_Problems/Problem_1&oldid=47692"