1973 Canadian MO Problems/Problem 1

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

$\text{(i)}$

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

Since from the second inequality $x<0$ , our solution is thus: $\boxed{-\frac{1}{2} < x < 0}$ .

$\text{(ii)}$ 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, $100000 = \boxed{32\times3125}.$

$\text{(v)}$

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

1973 Canadian MO (Problems)
Preceded by 1973 Canadian MO Problems	1 • 2 • 3 • 4 • 5	Followed by Problem 2

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1973 Canadian MO Problems/Problem 1

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