heureka

2.
Compute \(\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4}\)

\(\text{AP: $\quad a_n = a_1+(n-1)d,\qquad a_1=2,\ d=6 $} \)

\(\begin{array}{|rcll|} \hline a_n&=&2+(n-1)\cdot6\\\\ a_2 &=& 2+1\cdot 6 = 8 \\ a_3 &=& 2+2\cdot 6 \\ \ldots \\ a_{101} &=& 2+100\cdot 6 =602\\ \hline \end{array}\)

\(\text{GP: $\quad b_n = ar^{n-1},\qquad a=1,\ r=\dfrac{1}{4} $} \)

\(\begin{array}{|rcll|} \hline b_n &=& \left(\dfrac{1}{4}\right)^{n-1} \\\\ b_1 &=& \left(\dfrac{1}{4}\right)^{0} = 1 \\ b_2 &=& \dfrac{1}{4} \\ b_3 &=& \left(\dfrac{1}{4}\right)^{2} \\ \ldots \\ b_{101} &=& \left(\dfrac{1}{4}\right)^{100} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} \\ \hline \end{array}\)

\(\begin{array}{|rclll|} \hline s &=& a_2b_{101}+& a_3b_{100}+a_4b_{99}+\ldots+ a_{101}b_{2} \\ \dfrac{s}{\frac{1}{4}} &=& & a_2b_{100}+ a_3b_{99}+\ldots+ a_{100}b_{2}+a_{101}b_1 \quad | \quad b_1 = 1,\ a_{n+1}-a_n = d \\ \hline s- \dfrac{s}{\frac{1}{4}} &=& a_2b_{101}+& d(b_2+b_3+\ldots + b_{100})-a_{101} \quad | \quad a_2 = 8,\ a_{101} = 602,\ d = 6 \\ -3s &=& 8b_{101}+& 6(\underbrace{b_2+b_3+\ldots + b_{100}}_{=S~ (GP)})-602 \\ -3s &=& 8b_{101}+& 6S-602 \\\\ &&&\begin{array}{|rclll|} \hline S &=& b_2+&b_3+\ldots + b_{100} \\ \dfrac{1}{4}S &=& & b_3+\ldots + b_{100}+b_{101} \\ \hline S - \dfrac{1}{4}S &=& b_2-& b_{101} \\ \dfrac{3}{4}S &=& b_2-& b_{101} \\ S &=& \dfrac{4}{3}b_2-& \dfrac{4}{3}b_{101} \\ \hline \end{array} \\\\ -3s &=& 8b_{101}+& 6\left(\dfrac{4}{3}b_2- \dfrac{4}{3}b_{101} \right)-602 \\ -3s &=& 8b_{101}+& 8b_2- 8b_{101} -602 \\ -3s &=& & 8b_2 -602 \quad | \quad b_2 = \dfrac{1}{4} \\ -3s &=& & 2 -602 \\ -3s &=& & -600 \quad | \quad : (-3) \\ \mathbf{ s} &=& &\mathbf{ 200 } \\ \hline \end{array}\)

\(\mathbf{\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} = 200}\)

1.
The graphs of \(y = x^3 - 3x + 2 \text{ and } x + 4y = 4\) intersect in the points \((x_1,y_1), (x_2,y_2), \text{ and } (x_3,y_3)\).
If \(x_1 + x_2 + x_3 = A \text{ and } y_1 + y_2 + y_3 = B\), compute the ordered pair \((A,B)\).

\(\mathbf{\text{Vieta:}} \)

\(\begin{array}{rcll} x^3+ a_2x^2+a_1x+a_0 &=& 0 \qquad \mathbf{ a_2 =-(x_1+x_2+x_3)} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline x + 4y &=& 4 \quad |\quad y = x^3 - 3x + 2 \\ x + 4(x^3 - 3x + 2) &=& 4 \\ 4x^3-11x+4 &=& 0 \quad |\quad : 4 \\ x^3-\dfrac{11}{4}x+1 &=& 0 \quad |\quad a_2 = 0~ !\\\\ \mathbf{x_1+x_2+x_3 } &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x + 4y &=& 4 \\ x &=& 4-4y \\ \mathbf{x} &=& \mathbf{4(1-y)} \\\\ y &=& x^3 - 3x + 2 \quad |\quad \mathbf{x = 4(1-y)}\\ y &=& 64(1-y)^3 -3\cdot 4(1-y)+2 \\ \ldots \\ 64y^3-129y^2+181y-54 &=& 0 \quad |\quad : 64 \\ y^3-3y^2+\dfrac{181}{64}y-\dfrac{54}{64} &=& 0 \quad |\quad a_2 = -3~ !\\\\ -(y_1+y_2+y_3) &=& a_2 \\ -(y_1+y_2+y_3) &=& -3 \\ \mathbf{y_1+y_2+y_3 } &=& \mathbf{3} \\ \hline \end{array} \)

\(\mathbf{(A,B) = (0,3)}\)

A geometric series \(b_1+b_2+b_3+\cdots+b_{10}\) has a sum of 180.
Assuming that the common ratio of that series is \(\dfrac{7}{4}\),
find the sum of the series \(b_2+b_4+b_6+b_8+b_{10}\).

Bestimme den Betrag der Summenkraft FΣ und den Winkel zwischen FΣ und der x-Achse

\(\begin{array}{|rcll|} \hline \vec{F_1} &=& \dbinom{50N}{0} \\ \vec{F_2} &=& \dbinom{60N}{0} \\ \vec{F_3} &=& \dbinom{-40N\cos(40^\circ)}{40N\sin(40^\circ)} \\ \vec{F_4} &=& \dbinom{-30N\cos(30^\circ)}{-30N\sin(30^\circ)} \\ \mathbf{\vec{F_{\Sigma}}} &=& \mathbf{\vec{F_1}+\vec{F_2}+\vec{F_3}+\vec{F_4}} \\ \vec{F_{\Sigma}} &=& \dbinom{50N}{0}+\dbinom{60N}{0}+\dbinom{-40N\cos(40^\circ)}{40N\sin(40^\circ)}+\dbinom{-30N\cos(30^\circ)}{-30N\sin(30^\circ)} \\ \vec{F_{\Sigma}} &=& \dbinom{50N+60N-40N\cos(40^\circ)-30N\cos(30^\circ)} {0+0+40N\sin(40^\circ)-30N\sin(30^\circ)} \\ \mathbf{\vec{F_{\Sigma}}} &=& \mathbf{\dbinom{53.3774601617N} {10.7115043875N}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline |\vec{F_{\Sigma}}| &=& \sqrt{53.3774601617^2N^2+10.7115043875^2N^2} \\ \mathbf{|\vec{F_{\Sigma}}|} &=& \mathbf{54.4416162467N} \quad | \quad \text{Betrag der Summenkraft} \\\\ \alpha &=& \arctan\left(\dfrac{10.7115043875N}{53.3774601617N}\right) \\ \alpha &=& \arctan(0.20067467345) \\ \mathbf{\alpha} &=& \mathbf{11.3470968211^\circ} \quad | \quad \text{Winkel zwischen $\vec{F_{\Sigma}}$ und der $x$-Achse} \\ \hline \end{array}\)

The solutions to the equation

\(6x^2 + 10x = 4 - 10x - 6x^2 \text{ can be written in the form }\\ x=\frac{P\pm \sqrt Q}{R}, \text{ where P and R}\\ \text{are relatively prime integers and } R>0. \\ \text{What is the product PQR} ? \)

\(\begin{array}{|rcll|} \hline 6x^2 + 10x &=& 4 - 10x - 6x^2 \\ 12x^2 +20x -4 &=& 0 \\\\ x &=& \dfrac{-20\pm\sqrt{20^2-4\cdot 12\cdot(-4) } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{400+4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4\cdot 100 +4\cdot 48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm\sqrt{4}\sqrt{100+48 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{148 } } {2\cdot 12} \\ x &=& \dfrac{-20\pm2\sqrt{4\cdot 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm2\sqrt{4}\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-20\pm 4\sqrt{ 37 } } {4\cdot 6} \\ x &=& \dfrac{-4\cdot 5\pm 4\sqrt{37 } } {4\cdot 6} \\ x &=& \dfrac{ 4\cdot (-5\pm \sqrt{37}) } {4\cdot 6} \\ \mathbf{x} &=& \mathbf{\dfrac{ -5\pm \sqrt{37} } {6} } \quad | \quad P=-5,\ Q=37,\ R = 6 \\\\ PQR&=& -5\cdot 37 \cdot 6 \\ \mathbf{PQR} &=& \mathbf{-1110 } \\ \hline \end{array}\)

Thank you Alan!

Thank you CPhill!

Hallo, kann mir bitte jemand Helfen, wie man auf die Gewichtungsfaktoren von w1, w2 und w3 kommt ? 

\(\begin{array}{|rcll|} \hline (1-3,0858)\cdot w_1 + 4\cdot w_2+5\cdot w_3 &=& 0 \\ \dfrac{1}{4}\cdot w_1 + (1-3,0858)\cdot w_2+3\cdot w_3 &=& 0 \\ \dfrac{1}{5}\cdot w_1 + \dfrac{1}{3}\cdot w_2+(1-3,0858)\cdot w_3 &=& 0 \\ w_1 + w_2 + w_3 &=& 1 \\ \hline \end{array} \)

\((1-3,0858) = -2,0858 \)

\(\begin{array}{|rcll|} \hline -2,0858\cdot w_1 + 4\cdot w_2+5\cdot w_3 &=& 0 \\ \dfrac{1}{4}\cdot w_1 -2,0858\cdot w_2+3\cdot w_3 &=& 0 \\ \dfrac{1}{5}\cdot w_1 + \dfrac{1}{3}\cdot w_2-2,0858\cdot w_3 &=& 0 \\ w_1 + w_2 + w_3 &=& 1 \\ \hline \end{array} \)

\(\begin{array}{rcll} w_1 &=& 1-(w_2+w_3) \\ \mathbf{w_1} &=& \mathbf{1-w_2-w_3} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline -2,0858\cdot (1-w_2-w_3) + 4\cdot w_2+5\cdot w_3 &=& 0 \\ -2,0858+2,0858w_2+2,0858w_3 + 4\cdot w_2+5\cdot w_3 &=& 0 \\ 6,0858w_2+7,0858w_3 &=& 2,0858 \\ w_2 &=& \dfrac{2,0858-7,0858w_3}{6,0858} \qquad (1) \\ \hline \dfrac{1}{4}\cdot (1-w_2-w_3) -2,0858\cdot w_2+3\cdot w_3 &=& 0 \\ 0,25 -0,25w_2-0,25w_3 -2,0858\cdot w_2+3\cdot w_3 &=& 0 \\ -2,3358w_2+2,75w_3 &=& -0,25 \\ w_2 &=& \dfrac{0,25+2,75w_3}{2,3358} \qquad (2) \\ \hline w_2 = \dfrac{2,0858-7,0858w_3}{6,0858} &=& \dfrac{0,25+2,75w_3}{2,3358} \\ \dfrac{2,0858-7,0858w_3}{6,0858} &=& \dfrac{0,25+2,75w_3}{2,3358} \\ w_3 &=& \dfrac{2,3358\cdot 2,0858-6,0858 \cdot 0,25 }{6,0858\cdot 2,75 +2,3358\cdot 7,0858 } \\ \mathbf{w_3} &=& \mathbf{0,10065687810} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline w_2 &=& \dfrac{0,25+2,75w_3}{2,3358} \quad | \quad \mathbf{w_3=0,10065687810} \\ w_2 &=& \dfrac{0,25+2,75\cdot 0,10065687810}{2,3358} \\ \mathbf{w_2} &=& \mathbf{0,22553575424} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline w_1 &=& 1 - w_2 - w_3 \\ w_1 &=& 1 -0,22553575424-0,10065687810 \\ \mathbf{w_1} &=& \mathbf{0,67380736766} \\ \hline \end{array}\)

 Please help! Do not know hot to solve

\(\begin{array}{|rcl|rcl|} \hline g(x) &&& f(x)&=&g^{-1}(x) \\ \hline g(-15)&=&0 & f(0)&=&-15 \\ g(0)&=&3 & f(3) &=& 0 \\ g(3)&=&9 & f(9) &=& 3 \\ g(9)&=&20 & f(20) &=& 9 \\ \hline \end{array} \)

\(\mathbf{f(f(9))} = f(3) \mathbf{= 0} \)

the water park is a popular field trip destination. falls high filled 2 vans and 9 buses with 383 students.

rainy high filled 8 vans and 3 buses with 179 students. find the number of students in each van and each bus

\(\text{Let $\mathbf{v} =$ students in a $\mathbf{\text{van} }$ } \\ \text{Let $\mathbf{b} =$ students in a $\mathbf{\text{bus} }$ } \)

\(\begin{array}{|lrcll|} \hline \text{rainy high}: & 8v + 3b &=& 179 \quad | \quad \cdot 3 \\ & 24v + 9b &=& 537 \quad (1) \\\\ \text{falls high}: & 2v + 9b &=& 383 \quad (2) \\ \hline (1)-(2): & 24v + 9b - ( 2v + 9b ) &=& 537 - 383 \\ & 24v + 9b - 2v + 9b &=& 154 \\ & 22v &=& 154 \\ & v &=& \frac{154}{22} \\ & \mathbf{ v } &=& \mathbf{7} \\\\ & 2v + 9b &=& 383 \\ & 2\cdot 7 + 9b &=& 383 \\ & 14+ 9b &=& 383 \\ & 9b &=& 369 \\ & b &=& \frac{369}{9} \\ & \mathbf{ b } &=& \mathbf{41} \\ \hline \end{array}\)

 In each van are 7 students.
 In each bus are 41 students.

Suppose that x is an integer that satisfies the following congruences:
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
What is the remainder when x is divided by 105?

\(\begin{array}{|lrcll|} \hline & \mathbf{3+x} &\mathbf{ \equiv } &\mathbf{ 2^2 \pmod{3^3} } \\ & x &\equiv & 2^2-3 \pmod{3^3} \\ (1) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 1 \pmod{27} }\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{5+x} &\mathbf{ \equiv } &\mathbf{ 3^2 \pmod{5^3} } \\ & x &\equiv & 3^2-5 \pmod{5^3} \\ (2) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 4 \pmod{125} }\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{7+x} &\mathbf{ \equiv } &\mathbf{ 5^2 \pmod{7^3} } \\ & x &\equiv & 5^2-7 \pmod{7^3} \\ (3) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 18 \pmod{343} }\\ \hline \end{array}\)

The congruences are now:

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 1 \pmod{27} } \\ (2) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 4 \pmod{125} } \\ (3) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 18 \pmod{343} } \\ \hline \end{array} \)

\(\begin{array}{|rclrclrcl|} \hline (1) \quad \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 1 \pmod{27} } \\ x &=& 1 + 27 m, \ m\in \mathbb{Z} \\\\ (2)\quad \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 4 \pmod{125} } \\ x &=& 4 + 125n, \ n\in \mathbb{Z} \\ \hline 1+27m &=& 4+125n \\ 27m &=& 3+125n \\ m &=& \dfrac{3+125n}{27} \\ m &=& \dfrac{135n+3-10n}{27} \\ m &=& 5n+ \underbrace{ \dfrac{3-10n}{27} }_{=a} \\ \mathbf{m} &=& \mathbf{ 5n+a } & a &=& \dfrac{3-10n}{27} \\ & & & 27a &=& 3-10n \\ & & & 10n &=& 3-27a \\ & & & n &=& \dfrac{3-27a}{10} \\ & & & n &=& \dfrac{-30a+3+3a}{10} \\ & & & n &=& \dfrac{-30a+3+3a}{10} \\ & & & n &=& -3a+ \underbrace{ \dfrac{3+3a}{10} }_{=b} \\ & & & \mathbf{n} &=& \mathbf{ -3a+b} & b &=& \dfrac{3+3a}{10} \\ & & & & & & 10b &=& 3+3a \\ & & & & & & 3a &=& 10b - 3\\ & & & & & & a &=& \dfrac{10b - 3}{3} \\ & & & & & & a &=& \dfrac{9b - 3 + b }{3} \\ & & & & & & a &=& 3b-1+ \underbrace{ \dfrac{ b }{3} }_{=c} \\ & & & & & & \mathbf{a} &=& \mathbf{3b-1+c} \qquad \mathbf{b=3c}\\ & & & & & & a &=& 3(3c)-1+c \\ & & & & & & \mathbf{a} &=& \mathbf{ 10c-1 } \\ & & & n &=& -3(10c-1) +3c \\ & & & \mathbf{n} &=& \mathbf{ 3-27c } \\ m &=& 5(3-27c )+10c-1 \\ \mathbf{ m} &=& \mathbf{ 14-125c} \\\\ x &=& 1+27m \\ x &=& 1+27(14-125c) \\\\ (4) \quad \mathbf{x} &=& \mathbf{ 379-3375c} \\ \text{or} \mathbf{ x }&=& \mathbf{ 379 \pmod{3375}} \\ \hline \end{array}\)

The congruences are now:

\(\begin{array}{|lrcll|} \hline (4) & \mathbf{ x }&=& \mathbf{ 379 \pmod{3375}} \\ (3) & \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 18 \pmod{343} } \\ \hline \end{array}\)

\(\small{ \begin{array}{|rclrclrcl|} \hline (4) \quad \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 379 \pmod{3375} } \\ x &=& 379 -3375 m, \ m\in \mathbb{Z} \\\\ (3)\quad \mathbf{x} &\mathbf{ \equiv } & \mathbf{ 18 \pmod{343} } \\ x &=& 18 + 343n, \ n\in \mathbb{Z} \\ \hline 379 -3375 m &=& 18 + 343n \\ 343n &=& 361 -3375 m \\ n &=& \dfrac{361 -3375m}{343} \\ n &=& \dfrac{-3430m+343 +55m +18 }{343} \\ n &=& -10m+1+\underbrace{\dfrac{ 55m+18 }{343}}_{=a} \\ \mathbf{n} &=& \mathbf{-10m+1+a } & a &=& \dfrac{55m+18 }{343} \\ & & & 343a &=& 55m+18 \\ & & & 55m &=& 343a-18 \\ & & & m &=& \dfrac{343a-18}{55} \\ & & & m &=& \dfrac{330a+13a-18}{55} \\ & & & m &=& 6a+\underbrace{\dfrac{13a-18}{55}}_{=b} \\ & & & m &=& 6a+b & b &=& \dfrac{13a-18}{55} \\ & & & & & & 55b &=& 13a-18 \\ & & & & & & 13a &=& 55b+18\\ & & & & & & a &=& \dfrac{55b+18}{13} \\ & & & & & & a &=& \dfrac{52b+13+5+3b}{13} \\ & & & & & & a &=& 4b+1+\underbrace{\dfrac{5+3b}{13} }_{=c} \\ & & & & & & a &=& 4b+1+c & c &=& \dfrac{5+3b}{13} \\ & & & & & & & & & 13c &=& 5+3b \\ & & & & & & & & & 3b &=& 13c -5 \\ & & & & & & & & & b &=& \dfrac{13c -5}{3} \\ & & & & & & & & & b &=& \dfrac{12c-3+c-2}{3} \\ & & & & & & & & & b &=& 4c-1+\underbrace{\dfrac{c-2}{3}}_{=d} \\ & & & & & & & & & \mathbf{b} &=& \mathbf{4c-1+d} \qquad \mathbf{c=2+3d}\\ & & & & & & & & & b &=& 4(2+3d)-1+d \\ & & & & & & & & & \mathbf{b} &=& \mathbf{13d+7} \\ & & & & & & a &=& 4(13d+7)+1+2+3d \\ & & & & & & \mathbf{a} &=& \mathbf{31+55d} \\ & & & m &=& 6(31+55d)+13d+7 \\ & & & \mathbf{m} &=& \mathbf{343d+193}\\ n &=& -10(343d+193) +1+31+55d \\ \mathbf{n} &=& \mathbf{ -3375d-1898 } \\\\ x &=& 18 + 343n \\ x &=& 18 + 343(-3375d-1898 ) \\ x &=& 18-343\cdot 1898- 343\cdot 3375d \\ x &=& -650996 -1 157 625 d \\\\ x &\equiv & -650996 \pmod{1 157 625} \\ x &\equiv & -650996+1 157 625 \pmod{1 157 625} \\ x &\equiv & 506629 \pmod{1 157 625} \\\\ x &=& 506629 + 1 157 625z,\ z\in\mathbb{Z} \quad | \quad 1 157 625 = 3^35^37^3 \\ x &=& 506629 + 3^35^37^3z \\ \hline \end{array} }\)

What is the remainder when x is divided by 105?

\(105 = 3*5*7\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ x \pmod{3*5*7}} \\ &\equiv & 506629 + 3^35^37^3z \pmod{3*5*7} \\ & \equiv & 506629\pmod{3*5*7} + 3^35^37^3z \pmod{3*5*7} \\ & \equiv & 4+ 0 \pmod{3*5*7} \\ & \mathbf{\equiv} & \mathbf{ 4 \pmod{3*5*7} } \\ & \mathbf{\equiv} & \mathbf{ 4 \pmod{105} } \\ \hline \end{array}\)

The remainder is 4 when x is divided by 105 

There are two positive integers c for which the equation
\(5x^2 + 11x+c=0\) has rational solutions.

What is the product of those two values of \(c\)?

\(\begin{array}{|rcll|} \hline \mathbf{ 5x^2 + 11x+c} &=& \mathbf{0} \\\\ x &=& \dfrac{-11\pm \sqrt{121-4\cdot 5c} }{2\cdot 5} \\ \hline 121-4\cdot 5c &>& 0 \\ 121-20c&>& 0 \\ 121&>&20c \\ \mathbf{\dfrac{121}{20}} &>& \mathbf{c} \\ \hline \end{array}\)

The possible values of positive integers \(c\) are \(\{1,2,3,4,5,6\}\)

\(\begin{array}{|c|l|c|} \hline c & \sqrt{121-20c} & \text{rational solutions} \\ \hline \mathbf{6} & \sqrt{121-20\cdot 6}=\sqrt{1}= 1 & \checkmark \\ \hline 5 & \sqrt{121-20\cdot 5}=\sqrt{21} & \\ \hline 4 & \sqrt{121-20\cdot 4}=\sqrt{41} & \\ \hline 3 & \sqrt{121-20\cdot 3}=\sqrt{61} & \\ \hline \mathbf{2} & \sqrt{121-20\cdot 2} =\sqrt{81} = 9 & \checkmark\\ \hline 1 & \sqrt{121-20\cdot 1}=\sqrt{101} & \\ \hline \end{array}\)

So \(c=6\) and \(c=2\) and \(2\cdot 6 = 12\)