heureka

Let \(f(x)=3x+2\) and \(g(x)=ax+b\), for some constants \(a\) and \(b\).
If \(ab=20\) and \(f(g(x))=g(f(x))\) for \(x=0,1,2,\ldots ,9\), find the sum of all possible values of \(a\).

Let \(f(x)=3x+2\) and \(g(x)=ax+b\), for some constants \(a\) and b.
If \(ab=20\) and \(f(g(x))=g(f(x))\) for \(x=0,1,2,\ldots ,9\), find the sum of all possible values of \(a\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{3x+2} \quad &| \quad x=g(x) \\\\ f(g(x)) &=& 3g(x)+2 \quad &| \quad g(x)=ax+b \\ f(g(x)) &=& 3(ax+b)+2 \\ f(g(x)) &=& 3ax+3b+2 \quad &| \quad b=\dfrac{20}{a} \\ f(g(x)) &=& 3ax+3*\dfrac{20}{a}+2 \\ \mathbf{f(g(x))} &=& \mathbf{3ax+ \dfrac{60}{a}+2} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{g(x)} &=& \mathbf{ax+b} \quad &| \quad x=f(x) \\\\ g(f(x)) &=& af(x)+b \quad &| \quad f(x)=3x+2 \\ g(f(x)) &=& a(3x+2)+b \\ g(f(x)) &=& 3ax+2a+b \quad &| \quad b=\dfrac{20}{a} \\ \mathbf{g(f(x))} &=& \mathbf{3ax+2a+\dfrac{20}{a}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(g(x)) } &=& \mathbf{g(f(x))} \\\\ 3ax+ \dfrac{60}{a}+2 &=& 3ax+2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}+2 &=& 2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}-\dfrac{20}{a}+2 &=& 2a \\\\ \dfrac{40}{a} +2 &=& 2a \quad &| \quad * a \\\\ 40 +2a &=& 2a^2 \quad &| \quad : 2 \\ 20 + a &=& a^2 \\ \mathbf{ a^2-a-20} &=& \mathbf{0} \\\\ a _{1,2} &=& \dfrac{1 \pm \sqrt{1-4(-20)} }{2} \\ a _{1,2} &=& \dfrac{1 \pm \sqrt{81} }{2} \\\\ a _{1,2} &=& \dfrac{1 \pm 9 }{2} \\\\ a_1 = \dfrac{1}{2} + \dfrac{9}{2} && a_2 = \dfrac{1}{2} - \dfrac{9}{2} \\ a_1+a_2 &=& 2\times \dfrac{1}{2} \\ \mathbf{a_1+a_2} &=& \mathbf{1} \\ \hline \end{array}\)

What is the units digit of 3 to the 2004th power?

\(\begin{array}{|rcll|} \hline && \mathbf{3^{2004} \pmod{10}} \\ &\equiv & 3^{2*1002} \pmod{10} \\ &\equiv & \left(\mathbf{3^2}\right)^{1002} \pmod{10} \quad & | \quad \mathbf{3^2 \equiv -1 \pmod{10}} \\ &\equiv & \left(-1\right)^{1002} \pmod{10} \\ &\equiv & \mathbf{1 \pmod{10} } \\ \hline \end{array}\)

The units digit of 3 to the 2004th power is 1

Thank you, CPhill !

You have linear functions p(x) and q(x).
You know p(2)=3, and p(q(x))=4x+7 for all x.
Find q(-1).

My attempt:

linear functions

\(\text{Let $p(x)=ax+b$} \\ \text{Let $q(x)=ux+v$}\)

\(\begin{array}{|rcll|} \hline && \boxed{p(x)=ax+b \\ p(2) = a*2 + b \quad | \quad p(2) = 3 \\ 3 = 2a + b \\ \mathbf{b = 3-2a} } \\ p(x) &=& ax+b \quad &| \quad x=q(x) \\ p\Big(q(x)\Big) &=& aq(x)+b \quad &| \quad q(x)=ux+v \\ p\Big(q(x)\Big) &=& a(ux+v)+b \\ p\Big(q(x)\Big) &=& aux+av+b \quad &| \quad p\Big(q(x)\Big) = 4x+7 \\ 4x+7 &=& (au)x+(av+b) \quad &| \quad \text{compare} \\ && \boxed{\mathbf{au = 4},\\ ~\\ \mathbf{av+b = 7} \quad | \quad b = 3-2a \\ av+3-2a = 7 \\ av-2a = 4\\ a(v-2) = 4 \quad | \quad 4 = au \\ a(v-2)=au \\ \mathbf{v-2 = u} } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline q(x) &=& ux+v \quad &| \quad u = v-2 \\ q(x) &=& (v-2)x+v \quad &| \quad x=-1 \\ q(-1) &=& (v-2)(-1)+v \\ q(-1) &=& -v+2 +v \\ \mathbf{q(-1)} &=& \mathbf{2} \\ \hline \end{array} \)

The solutions to \(2x^2 - 10x + 13 = 0\) are \(a+bi\) and \(a-bi\), where a and b are positive.
What is \(a*b\)?

\(\begin{array}{|lrcll|} \hline (1) & (a+bi)+(a-bi) &=& \dfrac{10}{2} \\ & (a+bi)+(a-bi) &=& 5 \\ & 2a &=& 5 \\ & \mathbf{a} &=& \mathbf{2.5} \\ \hline (2) & (a+bi)(a-bi) &=& \dfrac{13}{2} \\ & a^2-(bi)^2 &=& 6.5 \\ & a^2-b^2i^2 &=& 6.5 \quad | \quad i^2 = -1 \\ & \mathbf{a^2+b^2} &=& \mathbf{6.5} \quad | \quad a=2.5 \\ & 2.5^2+b^2 &=& 6.5 \\ & b^2 &=& 0.25 \\ & \mathbf{b} &=& \mathbf{0.5} \\ \hline & a*b &=& 2.5*0.5 \\ & \mathbf{a*b} &=& \mathbf{1.25} \\ \hline \end{array}\)

Points \(A\) and \(B\) are on side \(\overline{YZ}\) of rectangle \(WXYZ\) such that \(\overline{WA}\) and \(\overline{WB}\) trisect \(\angle ZWX\).
If \(BY = 3\) and \(AZ = 6\), then what is the area of rectangle \(WXYZ\)?

\(\begin{array}{|lrcll|} \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ (2) & \tan(2A) &=& \dfrac{y}{6} \\\\ \hline (1)\times (2): & \tan(A)\times \tan(2A) &=& \dfrac{6}{y}\times \dfrac{y}{6} \\\\ & \tan(A)\times \tan(2A) &=& 1 \quad | \quad \tan(2A)=\dfrac{2\tan(A)} {1-\tan^2(A)} \\ & \tan(A)\times\dfrac{2\tan(A)} {1-\tan^2(A)} &=& 1 \\\\ & \dfrac{2\tan^2(A)} {1-\tan^2(A)} &=& 1 \\\\ & 2\tan^2(A) &=& 1-\tan^2(A) \\ & 3\tan^2(A) &=& 1 \\\\ & \tan^2(A) &=& \dfrac{1}{3} \\\\ & \tan(A) &=& \mathbf{ \dfrac{1}{\sqrt{3}} } \\ \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ & \dfrac{1}{\sqrt{3}} &=& \dfrac{6}{y} \\\\ & \sqrt{3} &=& \dfrac{y}{6} \\\\ & \mathbf{y} &=& \mathbf{6\sqrt{3}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3) & \tan(A) &=& \mathbf{\dfrac{y}{x-3}} \\\\ & x-3 &=& \dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{6\sqrt{3}}{\dfrac{1}{\sqrt{3}}} \\\\ & x &=& 3+ 6\sqrt{3}\sqrt{3} \\\\ & x &=& 3+ 18 \\\\ & \mathbf{x} &=& \mathbf{21} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The area of rectangle }~ WXYZ &=& xy \\ &=& 21*6\sqrt{3} \\ &=& \mathbf{126\sqrt{3}} \\ &=& \mathbf{218.238401754} \\ \hline \end{array}\)

Points \(A\), \(B\), and \(C\) are on sides\( \overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\), \(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + 14^2 \\ x^2 &=& (3*6)^2 + 14^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}\)

\(\mathbf{AB \approx 22.8}\)

On Tuesday, I worked t+1 hours and earned 3t-3 dollars per hour.
My friend Andrew worked 3t-5 hours but only earned t+2 dollars an hour.
At the end of the day, I had earned two dollars more than he had. What is the value of t?

\(\begin{array}{|rcll|} \hline (3t-3)\dfrac{\text{dollars}}{\text{hours}}\times (t+1)\text{hours} &=& $2+ (t+2)\dfrac{\text{dollars}}{\text{hours}}\times (3t-5)\text{hours} \\\\ (3t-3)(t+1) \text{dollars} &=& $2+ (t+2)(3t-5) \text{dollars} \\\\ (3t-3)(t+1)&=& 2+ (t+2)(3t-5)\\ 3(t-1)(t+1)&=& 2+ (t+2)(3t-5)\\ 3(t^2-1)&=& 2+ (t+2)(3t-5)\\ 3t^2-3 &=& 2+ (t+2)(3t-5)\\ 3t^2-3&=& 2+ 3t^2-5t+6t-10 \\ -3&=& 2+ -5t+6t-10 \\ -3&=& 2+t-10 \\ -3&=& t-8 \\ t &=& 8-3 \\ \mathbf{t} &=& \mathbf {5} \\ \hline \end{array}\)

In an election, 1200 total votes were cast for the 3 candidates.
The second-place candidate received 125 fewer votes than the winner and 200 votes more than the third-place candidate.
How many votes did the winner receive?

\(\text{Let winner candidate = c_1 } \\ \text{Let second-place candidate = c_2 } \\ \text{Let third-place candidate = c_3 } \)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{c_2} &=& \mathbf{c_1 - 125} \\ \hline (2) & c_2 &=& c_3 + 200 \\ & c_3 &=& c_2 - 200 \quad | \quad \mathbf{c_2=c_1 - 125} \\ & c_3 &=& c_1 - 125 - 200 \\ & \mathbf{c_3} &=& \mathbf{c_1 - 325} \\ \hline (3) &c_1+ c_2+c_3 &=& 1200 \quad | \quad \mathbf{c_2=c_1 - 125},\ \mathbf{c_3=c_1 - 325} \\ &c_1+ (c_1 - 125) + (c_1 - 325) &=& 1200 \\ &3c_1- 125 - 325 &=& 1200 \\ &3c_1- 450 &=& 1200 \\ &3c_1 &=& 1200 + 450 \\ &3c_1 &=& 1650 \quad | \quad : 3 \\ & \mathbf{c_1} &=& \mathbf{550} \\ \hline \end{array}\)

The winner did receive 550 votes?