cours/kangourou 2024-07-15 problèmes du jour.md

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[!question] Question Find numbers from 0 to 100, but only using 4 times the digit "4" (and any operator)

Trouver les nombres de 100, mais seulement en utilisant 4 fois le chiffre 4 (et des opérateurs)

Ideas / Idées

You can always use up a 4 using : "$\sqrt{ \sqrt{ x^{4} } } = x$" (does nothing). This allows formulas with only 3 or 2 times the 4. On peut toujours utiliser un 4 avec : "$\sqrt{ \sqrt{ x^{4} } } = x$" (ne fait rien). Cela permet des formules avec seulement 2 ou 3 fois le 4.

Solutions

n = 0

• 4 - 4 + 4 - 4
• 44 - 44

n=1

n = 41

• 41 = 44 - \sqrt{ 4 } - (\lim\limits_{ n \to \infty } \sqrt[n]{4})

n = 71

• 71 = (\Gamma(4) + \Gamma(\sqrt{ 4 })) \cdot (\Gamma(\sqrt{ 4 }) \cdot \Gamma(\sqrt{ 4 }))
• 71 = (4! + (\sqrt{ 4 })!) \cdot ((\sqrt{ 4 })! \cdot (\sqrt{ 4 })!)

General solution / Solution générale

1

\displaystyle\int _{-4}^{4} \dfrac{\underbrace{(x \times x)+(x \times x)+\cdots + (x \times x)}_{6n \text{ times } x}}{4^{4}} \, dx = n

2

\displaystyle\sqrt{ 4 } \cdot\frac{\ln \left( \frac{\overbrace{\sqrt{ \sqrt{ \sqrt{ \sqrt{ \cdots 4 } } } }}^{n \text{ times }} }{\ln(4)} \right)}{\ln(4)}

[!definition] Logarithm / Logarithme \log_{n}(n^{x}) = x

[!definition] Natural logarithm / Logarithme naturel \ln(x) = \log_{e}(x) So : / Alors : \ln(e^{x}) = x

3

0 = 4 - 4 1 = (\sqrt{ 4 })! 2 = \sqrt{ 4 } 3 = \dfrac{4!}{\sqrt{ 4 }} 4 =

[!definition] Factorial / Factorielle n! = 1 \times 2 \times 3 \times \cdots \times n

[!definition] Double factorial / Double factorielle n!! = 1 \cancel{\times 2} \times 3 \cancel{\times 4} \times \cdots \times n if n is odd / si n est impair n!! = \cancel{1 \times} 2 \times \cancel{3 \times} 4\times \cdots \times n if n is even / si n est pair

[!definition] Concatenation New operator / Nouvel opérateur : 4 \cdot 4 = 44 71 \cdot 92 = 7192

2

\displaystyle\log_{\log_{4}(\sqrt{ 4 })} \left( \log_{4} \left( \underbrace{\sqrt{ \sqrt{ \sqrt{ \sqrt{ \dots \sqrt{ 4 } } } } }}_{n \text{ times}} \right) \right) = n

Why ? / Pourquoi ?

\log_{4}(\sqrt{ 4 }) = \frac{1}{2}

\log_{4}(4) = 1 \log_{4}(4^{2}) = 2 \vdots \log_{4}(4^{p}) = p

\log_{4}(\sqrt{ 4 }) = \log_{4}\left( 4^{\frac{1}{2}} \right) = \frac{1}{2} \log_{4}(\sqrt{ \sqrt{ 4 } }) = \log_{4}\left( 4^{\frac{1}{4}} \right) = \frac{1}{4} \log_{4}(\sqrt{ \sqrt{ \sqrt{ 4 } } }) = \log_{4}\left( 4^{\frac{1}{8}} \right) = \frac{1}{8} \log_{4}(\sqrt{ \sqrt{ \sqrt{ \sqrt{ 4 } } } }) = \log_{4}\left( 4^{\frac{1}{16}} \right) = \frac{1}{16} \vdots \displaystyle\log_{4}\left( \underbrace{\sqrt{ \sqrt{ \sqrt{ \dots \sqrt{ 4 } } } }}_{n \text{ times}} \right) = \log_{4} \left( 4^{\frac{1}{2^{n}}} \right) = \frac{1}{2^{n}}

\displaystyle\log_{\frac{1}{2}} \left( \frac{1}{2 \times n} \right) = n

!

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