heureka

When three standard dice are tossed, the numbers x, y, and z are obtained.

Find the probability that xyz=180.

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)?

$\text{Let $ABC = x$ and $DEF = y$ }$

$\begin{array}{|rcll|} \hline 4(ABC,DEF) &=& 4(1000x+y) \\\\ 9(DEF,ABC) &=& 9(1000y+x) \\\\ \hline \\ 4(1000x+y) &=& 9(1000y+x) \\\\ 4000x+4y &=& 9000y+9x \\\\ 3991x &=& 8996y \\\\ 13*307x &=& 13*692y \\\\ 307x &=& 692y \\ \hline \mathbf{ x } &=& \mathbf{692} \\\\ \mathbf{ y } &=& \mathbf{307} \\ \hline \end{array}$

$4(692,307) = 9(307,692)$

Find positive integers (a,b) so that sqrt(37 + 12*sqrt(7)) = a + b*sqrt(7).

$\begin{array}{|rcll|} \hline \sqrt{37 + 12*\sqrt{7}} &=& \sqrt{9+28 + 12*\sqrt{7}} \\ &=& \sqrt{9+ 12*\sqrt{7} + 28 } \\ &=& \sqrt{9+ 2*6*\sqrt{7} + 28 } \\ &=& \sqrt{\left(3+2\sqrt{7}\right)^2 } \\ \mathbf{\sqrt{37 + 12*\sqrt{7}} } &=& \mathbf{3+2\sqrt{7} } \\ \hline \end{array}$

In a geometric sequence of real numbers,

the sum of the first 2 terms is 7,

and the sum of the first 6 terms is 91.

What is the sum of the first 4 terms?

$\begin{array}{|rcll|} \hline s_2 &=& a_1 + a_2 \\ \mathbf{s_2} &=& \mathbf{a_1 + a_1r} \\ \hline \end{array} \begin{array}{|rcll|} \hline s_4 &=& a_1 + a_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_1r^2 + a_1r^3 \\ s_4 &=& s_2 + (a_1+a_1r)r^2 \\ s_4 &=& s_2 + s_2r^2 \\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\ \mathbf{s_4} &=& \mathbf{ 7(1+r^2) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s_6 &=& a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_1r^4 + a_1r^5 \\ s_6 &=& s_4 + (a_1+a_1r)r^4 \\ s_6 &=& s_4 + s_2r^4 \quad | \quad s_6 = 91,~s_2=7\\ 91 &=& s_4 +7r_4 \\ \mathbf{s_4} &=& \mathbf{ 91-7r_4 } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s_4 = 7(1+r^2) &=& 91-7r_4 \\ 7(1+r^2) &=& 91-7r_4 \quad | \quad : 7 \\ 1+r^2 &=& 13-r^4 \\ \mathbf{r^4+r^2 -12} &=& \mathbf{ 0 } \\ \hline r^2 &=& \dfrac{ -1\pm\sqrt{1+4*12} }{2} \\\\ r^2 &=& \dfrac{ -1\pm 7 }{2} \\\\ r^2 &=& \dfrac{ -1 {\color{red}+} 7 }{2} \\\\ \mathbf{r^2} &=& \mathbf{3} \\ \hline s_4 &=& 7(1+r^2) \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}$

Define

$A=\dfrac{1}{1^2}+\dfrac{1}{5^2}-\dfrac{1}{7^2}-\dfrac{1}{11^2}+\dfrac{1}{13^2}+\dfrac{1}{17^2}-\dfrac{1}{19^2}-\dfrac{1}{23^2} + \dotsb,$

which omits all terms of the form $\dfrac{1}{n^2}$where  is an odd multiple of 3, and

$B=\dfrac{1}{3^2}-\dfrac{1}{9^2}+\dfrac{1}{15^2}-\dfrac{1}{21^2}+\dfrac{1}{27^2}-\dfrac{1}{33^2}+\dfrac{1}{39^2}-\dfrac{1}{45^2} + \dotsb,$

which includes only terms of the form $\dfrac{1}{n^2}$ where  is an odd multiple of 3.

Determine $\dfrac{A}{B}$.

$\begin{array}{|rcll|} \hline A&=&\dfrac{1}{1^2}+\dfrac{1}{5^2}-\dfrac{1}{7^2}-\dfrac{1}{11^2}+\dfrac{1}{13^2}+\dfrac{1}{17^2}-\dfrac{1}{19^2}-\dfrac{1}{23^2} + \dotsb \\ \hline B&=&\dfrac{1}{3^2}-\dfrac{1}{9^2}+\dfrac{1}{15^2}-\dfrac{1}{21^2}+\dfrac{1}{27^2}-\dfrac{1}{33^2}+\dfrac{1}{39^2}-\dfrac{1}{45^2} + \dotsb \\\\ 9B&=&\dfrac{1}{1^2}-\dfrac{1}{3^2}+\dfrac{1}{5^2}-\dfrac{1}{7^2} +\dfrac{1}{9^2}-\dfrac{1}{11^2} +\dfrac{1}{13^2}-\dfrac{1}{15^2} +\dfrac{1}{17^2}-\dfrac{1}{19^2}+\dfrac{1}{21^2} + \dotsb \\\\ 9B&=&A-\dfrac{1}{3^2}+\dfrac{1}{9^2}-\dfrac{1}{15^2}+\dfrac{1}{21^2}-\dotsb \\\\ 9B&=&A-\left( \underbrace{ \dfrac{1}{3^2}-\dfrac{1}{9^2}+\dfrac{1}{15^2}-\dfrac{1}{21^2}+\dotsb }_{=B}\right) \\\\ 9B&=&A-B \\ 10B &=& A \\ \mathbf{ \dfrac{A}{B} } &=& \mathbf{10} \\ \hline \end{array}$

There are 7 students in a class: 2 boys and 5 girls.
If the teacher picks a group of 3 at random,
what is the probability that everyone in the group is a girl?

$\begin{array}{|rcll|} \hline \dfrac{ \binom{2}{0}_{\text{Boys}}\binom{5}{3}_{\text{Girls} } } {\binom{7}{3}_{\text{Boys and Girls} } } &=& \dfrac{2}{7}\\ \text{or} \\ \dfrac{5}{7}*\dfrac{4}{6}*\dfrac{3}{5}&=& \dfrac{2}{7}\\ \hline \end{array}$

The probability is $\approx 28.57 \%$

For a certain value of k, the system has no solutions.

$\begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}$

Real numbers x and y satisfy
x + xy = 250y
x - xy = -240y

My attempt:

$\begin{array}{|lrcll|} \hline (1): & x + xy &=& 250y \\ (2): & x - xy &=& -240y \\ \hline (1)+(2):& 2x &=& 10y \quad | \quad :2 \\ &x &=& 5y \quad \text{or} \quad y=\dfrac{x}{5} \\\\ & \dfrac{x}{y} &=& 5 \\ \hline (1):& x + xy &=& 250y \quad|\quad :y \\ & \dfrac{x}{y} + x &=& 250 \\\\ & 5+x &=& 250 \\ & \mathbf{x} &=& \mathbf{245} \\ \hline &y &=& \dfrac{x}{5} \\\\ &y &=& \dfrac{245}{5} \\ & \mathbf{y} &=& \mathbf{49} \\ \hline \end{array}$

The function $f(n)$ satisfies $f(1)=1$ and $f(2n+1)=f(n)+2$ for $n \ge 0$.
Find $f(15)$.

$\begin{array}{|lrcll|} \hline n=1:& f(2*1+1) &=& f(1) + 2 \quad | \quad f(1)=1 \\ &f(3) &=& 1+2 \\ &f(3)&=& 3 \\ \hline n=3:& f(2*3+1) &=& f(3) + 2 \quad | \quad f(3)=3 \\ &f(7) &=& 3+2 \\ &f(7)&=& 5 \\ \hline n=7:& f(2*7+1) &=& f(7) + 2 \quad | \quad f(7)=5 \\ &f(15) &=& 5+2 \\ &\mathbf{ f(15) } &=& \mathbf{7} \\ \hline \end{array}$