heureka

Find the constant term in the expansion of \(\left(2z - \frac{1}{\sqrt{z}}\right)^9\).

\(\begin{array}{|rcll|} \hline && \dbinom{9}{6}*(2z)^3*\left(\frac{1}{\sqrt{z}}\right)^6 \\\\ &=& \dbinom{9}{6}*8z^3*\dfrac{1}{z^3} \\\\ &=& \dbinom{9}{6}*8\\\\ &=& \dbinom{9}{9-6}*8\\\\ &=& \dbinom{9}{3}*8\\\\ &=& \dfrac{9}{3}*\dfrac{8}{2}*\dfrac{7}{1}*8\\\\ &=& 3*4*7*8 \\\\ &=& 672 \\ \hline \end{array}\)

Question:

\(\text{Let $CA=b$} \\ \text{Let $CB=a$} \\ \text{Let $AB=c$} \\ \text{Let $CM=l$} \\ \text{Let $\angle AMC = \varphi$} \\ \text{Let $\angle CMB = 180^\circ-\varphi$} \)

\(cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ AMC} \\ \hline b^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) \qquad (1) \\ \hline \end{array}\\ cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ CMB} \\ \hline a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(180^\circ-\varphi) \\ a^2 &=& l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (1)+(2)& b^2 + a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) +l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \\ & b^2 + a^2 &=& l^2 + \frac{c^2}{4}+l^2 + \frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + 2\frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + \frac{c^2}{2} \\ & CA^2 + CB^2 &=& 2CM^2 + \frac{AB^2}{2} \\ \hline \end{array}\)

Question:

My attempt:

\(\begin{array}{|rcll|} \hline \vec{M} &=& \frac{\vec{A}+\vec{C}}{2} \\ \vec{N} &=& \frac{\vec{B}+\vec{D}}{2} \\ \hline \vec{M}\vec{N} &=& \frac{(\vec{A}+\vec{C})}{2}* \frac{(\vec{B}+\vec{D})}{2} \\ 4\vec{M}\vec{N} &=& (\vec{A}+\vec{C}) (\vec{B}+\vec{D}) \\ 4\vec{M}\vec{N} &=& \vec{A}\vec{B} +\vec{A}\vec{D}+\vec{C}\vec{B}+\vec{C}\vec{D} \\ \hline \end{array}\)

GS

\(\sqrt{2*5^0},\sqrt{2*5^1},\sqrt{2*5^2},\sqrt{2*5^3},\\ \sqrt{2*5^4}, \sqrt{2*5^5}, \sqrt{2*5^6}=5^3\sqrt{2}\)

The letters A, B, C, D, E, and F represent digits and

\(ABC,DEF\) represents a positive six-digit integer.

What is the number ABC,DEF if \(4(ABC,DEF) = 9(DEF,ABC)\)?

Typo:

\(\begin{array}{|rcll|} \hline ABCD &=& A^2B^2 \\ ABCD &=& 7^211^2 \\ \mathbf{ABCD} &=& \mathbf{5929} \\ \hline \end{array}\)

Four positive integers \(A\), \(B\), \(C\) and \(D\) have a sum of \(36\).
If \(A+2 = B-2 = C+2 = D-2\),
what is the value of the product \(ABCD\)?

\(\begin{array}{|rcll|} \hline A+2 &=& C+2 \\ \mathbf{A} &=& \mathbf{C} \\ \hline \end{array} \begin{array}{|rcll|} \hline B-2 &=& D-2 \\ \mathbf{B} &=& \mathbf{D} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A+B+C+D &=& 36 \\ A+B+A+B &=& 36 \\ 2A+2B &=& 36 \quad | \quad : 2 \\ A+B &=& 18 \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline A+2 &=&B-2 \\ A-B &=&-4 \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)+(2): & 2A &=& 18-4 \\ & 2A &=& 14 \quad | \quad : 2 \\ &\mathbf{A} &=& \mathbf{7} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (1)-(2): & 2B &=& 18+4 \\ & 2B &=& 22 \quad | \quad : 2 \\ &\mathbf{B} &=& \mathbf{11} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline ABCD &=& A^2B^2 \\ ABCD &=& 7^21^2 \\ \mathbf{ABCD} &=& \mathbf{5929} \\ \hline \end{array}\)

Find the number in the form x3579x that is divisible by 42.

335790 / 42 = 7995

Five friends live on the same street.

Their houses are at points A, B, C, D, and E, with the distances shown.

The five friends decide to meet at point P so that the total walking distance

for all five friends is minimized.
What is AP?

\(\begin{array}{|ccccc|c|c|r|} \hline A & B&C&D&E &\text{Meet at} & \text{walking distance} &AP \\ & & & & & & \text{for all} \\ & & & & & & \text{five friends} \\ \hline 0&5&9&11&14&A&39&0\\ \hline 5&0&4&6&9&B&24&5 \\ \hline 9&4&0&2&5&C&\color{red}20&9 \\ \hline 11&6&2&0&3&D&22&11 \\ \hline 14&9&5&3&0&E&31&14 \\ \hline \end{array}\)

The total walking distance
for all five friends is minimized at Point C and AP = 9

What is the smallest positive integer \(n\) such that
\(3n\) is a perfect square and
\(2n\) is a perfect cube?

My attempt:

\( \begin{array}{|rcll|} \hline 3n &=& 3*&3*2*3*2*3 =2^2(3^2)^2=324=18^2 \\ 2n &=& 2*&3*2*3*2*3 =2^33^3=216=6^3 \\ \hline n &=& & 3*2*3*2*3 = 108 \\ \hline \end{array}\)