heureka

Both x and y are positive real numbers,

and the point (x,y) lies on or above both of the lines having equations \(2x + 5y = 10\) and \(3x + 4y = 12\).

What is the least possible value of \(8x + 13y\)?

\(\begin{array}{|lrcll|} \hline & 2x + 5y &=& 10 \\ & 3x + 4y &=& 12 \\ & 3x + 4y &=& 12 \\ \hline \text{sum} & 8x + 13y &=& 10+12+12 \\ & \mathbf{8x + 13y} &=& \mathbf{34} \\ \hline \end{array}\)

(b)

In the expansion of \(\left(1 + x\right)^n\), three consecutive coefficients are in the ratio \(1:7:35\).
Find the positive integer \(n\).

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ &&\boxed{ \mathbf{\dbinom{n}{k - 1} = \dfrac{k}{n-k+1} \dbinom{n}{k} }\\ ~\\ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{k}{n-k+1} \dbinom{n}{k} : \dbinom{n}{k} : \dfrac{n-k}{k+1} \dbinom{n}{k} &=& 1 : 7 : 35 \\ \hline \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k}} &=& \dfrac{7}{1} \\\\ \dfrac{n-k+1}{k} &=& 7 \\\\ n-k+1 &=& 7k \\ \mathbf{n} &=& \mathbf{8k -1} \qquad (1) \\ \hline \dfrac{\dfrac{n-k}{k+1} \dbinom{n}{k} } {\dbinom{n}{k}} &=& \dfrac{35}{7} \\\\ \dfrac{n-k}{k+1} &=& 5 \\\\ 5(k+1) &=& n-k \\ 5k+5 &=& n-k \\ 6k &=& n-5 \\ \mathbf{k} &=& \mathbf{ \dfrac{n-5}{6} } \qquad (2) \\ \hline n &=& 8k-1 \quad | \quad k=\dfrac{n-5}{6} \\ n &=& 8 \dfrac{(n-5)}{6}-1 \\ n &=& \dfrac{4}{3}(n-5)-1 \\ & & \dotsb \\ \mathbf{n} &=& \mathbf{23} \\\\ k &=& \dfrac{n-5}{6} \quad | \quad n=23 \\\\ k &=& \dfrac{23-5}{6} \\ & & \dotsb \\ \mathbf{k} &=& \mathbf{3} \\ \hline \dbinom{n}{k - 1} : \dbinom{n}{k} : \dbinom{n}{k + 1} &=& 1 : 7 : 35 \\\\ \dbinom{23}{2} : \dbinom{23}{3} : \dbinom{23}{4} &=& 1 : 7 : 35 \\\\ 253 : 1771 : 8855 &=& 1 : 7 : 35 \\ \hline \end{array}\)

(a)

Simplify \(\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}} \\\\ &=& \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ &=& \dfrac{n-k+1}{k} * \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} } \\\\ &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

Thank you, CPhill !

Let \(\omega\) be a complex number such that \(\omega^5 = 1\) and \(\omega \neq 1\).
\(\dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \omega^4} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3}\).

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \omega^4} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} } \quad | \quad \omega^4 = \dfrac{\omega^5}{\omega}=\dfrac{1}{\omega} \\\\ &=& \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^2}{1 + \dfrac{1}{\omega}} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} \\\\ &=& \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^3}{1 + \omega} + \dfrac{\omega^4}{1 + \omega^3} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^4}{1 + \omega^3} \quad | \quad \omega^3 = \dfrac{\omega^5}{\omega^2}=\dfrac{1}{\omega^2} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^4}{1 + \dfrac{1}{\omega^2}} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega^6}{1 + \omega^2} \quad | \quad \omega^6 = \omega^5 \omega=\omega \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{\omega}{1 + \omega^2} + \dfrac{\omega}{1 + \omega^2} \\\\ &=& \dfrac{2\omega^3}{1 + \omega} + \dfrac{2\omega}{1 + \omega^2} \\\\ &=& 2\left( \dfrac{ \omega^3}{1 + \omega} + \dfrac{ \omega}{1 + \omega^2} \right) \\\\ &=& 2\left( \dfrac{ \omega^3(1 + \omega^2) + (1 + \omega)\omega}{(1 + \omega)(1 + \omega^2) } \right) \\\\ &=& 2\left( \dfrac{ \omega^5 + \omega + \omega^2 + \omega^3 }{ 1 + \omega + \omega^2 + \omega^3 } \right) \quad | \quad \omega^5 = 1 \\\\ &=& 2\left( \dfrac{ 1 + \omega + \omega^2 + \omega^3 }{ 1 + \omega + \omega^2 + \omega^3 } \right) \\\\ &=& \mathbf{2} \\ \hline \end{array}\)

egyptian fractions

can all fractions with a numerator of two be written as the sum of two different unitary fractions with mathematical proof

I assume:
\(\dfrac{2}{n}=\dfrac{1}{a}+\dfrac{1}{b} \)

if \(n\) is even then we set \(n=2k\):

\(\mathbf{\dfrac{2}{n}}=\dfrac{2}{2k}=\dfrac{1}{k}\mathbf{=\dfrac{1}{k+1} + \dfrac{1}{k(k+1)} }\)

if \(n\) is odd then we set \(n=2k-1\):

\(\mathbf{\dfrac{2}{n}}=\dfrac{2}{2k-1}\mathbf{=\dfrac{1}{k} + \dfrac{1}{k(2k-1)} }\)

What's the complexity of the number 15?

15.......{ (1+1)*(1+1)+1 } * (1+1+1)........8

The integer complexity of the number 15 is 8

Thank you, Melody !

Thank you, Melody !

What is the coefficient of \(ab^2c^3\) in \(\left(a + 2b + 3c \right)^6\)?

In general: \(\boxed{\left(a + b + c \right)^n = \sum \limits_{i=0}^{n} \sum \limits_{j=0}^{n-i}\dfrac{n!}{i!j!(n-i-j)!} \cdot a^ib^jc^{n-i-j}}\)

\(\begin{array}{|rlcll|} \hline & n = 6 \\ a^1b^2c^3: & i=1,\ j=2,\ n-i-j = 3 \\\\ & \left(a + 2b + 3c \right)^6 = \dotsb + \dfrac{6!}{1!2!3!} \cdot a^1(2b)^2(3c)^3 + \dotsb \\\\ & \left(a + 2b + 3c \right)^6 = \dotsb + \dfrac{4\cdot 5 \cdot 6}{2} \cdot 2^2 \cdot 3^3 \cdot a^1 b^2 c^3 + \dotsb \\\\ & \left(a + 2b + 3c \right)^6 = \dotsb + 4\cdot 5 \cdot 6 \cdot 2 \cdot 27 \cdot a^1 b^2 c^3 + \dotsb \\\\ & \left(a + 2b + 3c \right)^6 = \dotsb + \mathbf{6480} \cdot a^1 b^2 c^3 + \dotsb \\\\ \hline \end{array} \)

The coefficient of \(\mathbf{ab^2c^3}\) is \(\mathbf{6480}\)