Difference between revisions of "1973 Canadian MO Problems"

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==Problem 1==
 
==Problem 1==
  
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<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 simultaneous inequalities.
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<math>\text{(ii)}</math> What is the greatest integer that satisfies both inequalities <math>4x+13 < 0</math> and <math>x^{2}+3x > 16</math>.
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<math>\text{(iii)}</math> Give a rational number between <math>11/24</math> and <math>6/13</math>.
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<math>\text{(iv)}</math> Express <math>100000</math> as a product of two integers neither of which is an integral multiple of <math>10</math>.
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<math>\text{(v)}</math> Without the use of logarithm tables evaluate <math>\frac{1}{\log_{2}36}+\frac{1}{\log_{3}36}</math>.
  
  

Revision as of 20:27, 16 December 2011

Problem 1

$\text{(i)}$ 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

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Resources

1973 Canadian MO