heureka

$$\small{ \begin{array}{llrl} $(A)$ &\text{Find a number }$0\leq x< &14 &$that solves the congruence $9x \equiv 9 \pmod{14}$.\\ $(B)$&\text{Find a number }$0\leq x< &5 &$that solves the congruence $3x \equiv 4 \pmod{5}$.\\ $(C)$&\text{Find a number }$0\leq x< &1000&$that solves the congruence $999x \equiv 998 \pmod{1000}$.\\ $(D)$&\text{Find a number }$0\leq x< &21 &$that solves the congruence $4x \equiv 17 \pmod{21}$. \end{array} }$$

$$\left\{ \begin{array}{l} $(A)$\\ $(B)$\\ $(C)$\\ $(D)$ \end{array} \right\}= \left\{ \begin{array}{lcr} x&=&1\\ x&=&3\\ x&=&2\\ x&=&20 \end{array} \right\}, \small{\text{ because } \left\{ \begin{array}{lcr} 9\cdot 1 &\equiv& 9 \pmod{14}\\ 3\cdot 3 &\equiv& 4 \pmod{5}\\ 999\cdot 2 &\equiv& 998 \pmod{1000}\\ 4\cdot 20 &\equiv& 17 \pmod{21}\\ \end{array} \right\}$$

(A) Convince yourself that 50 is the inverse of 5 modulo 83, and that 12 is the inverse of 7 modulo 83. Find the inverse of 35 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{ The inverse of $35 \pmod{83} = \mathbf{ 19 }$, because $35 \cdot 19 \equiv 1 \pmod {83}$ }}$$

(b) Find the inverse of 49 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{ The inverse of $49 \pmod{83} = \mathbf{ 61 }$, because $49 \cdot 61 \equiv 1 \pmod {83}$ }}$$

What is the inverse of 100 modulo 10001? (Give your answer as a nonnegative integer that is less than 10001.)

$$\small{\text{ The inverse of $100 \pmod{10001} = \mathbf{ 9901 }$, because $100 \cdot 9901 \equiv 1 \pmod {10001}$ }}$$

1.5=1/cos(x) find x

$$\small{\text{$ \begin{array}{rcl} \dfrac{ 1 } { \cos{(x)} } &=& 1.5 \\\\ \dfrac{ 1 } { \cos{(x)} } &=& \dfrac{3}{2} \\\\ \cos{(x)} &=& \dfrac{2}{3} \\\\ \cos{(x)} &=& \dfrac{2}{3} \qquad | \qquad \pm\arccos{()}\\\\ x &=& \pm \arccos{ \left(\dfrac{2}{3} \right) }\\\\ x_{1,2} &=& \pm 48.1896851042\ensurement{^{\circ}}\\\\ \mathbf{x_1} & \mathbf{=} & \mathbf{48.1896851042\ensurement{^{\circ}} } \\\\ x_2 &=& -48.1896851042\ensurement{^{\circ}} \\\\ x_2 &=& -48.1896851042\ensurement{^{\circ}} + 360 \ensurement{^{\circ}} \\\\ \mathbf{x_2} & \mathbf{=} & \mathbf{311.810314896\ensurement{^{\circ}} } \end{array} $}}$$

You can get 28200 if u dont reduce it with 5 % the last day.

You get max 20 * 1410 = 28200.

$$\small{\text{$ \boxed{~~t_n=0.95^n\cdot \left( \dfrac{25547}{0.95}-20\cdot 1410 \right)+20\cdot 1410~~} $}}\\\\ \small{\text{We have $t_n = 27924$ and get n by rearrange: }}\\\\ \small{\text{$ \boxed{ n=\dfrac {\ln{\left( \dfrac{t_n -20\cdot 1410}{ \dfrac{25547}{0.95} -20\cdot 1410} \right)}} {\ln{(0.95)}} } $}}\\\\\\ \small{\text{$ \begin{array}{rcl} n &=& \dfrac {\ln{\left( \dfrac{27924 -20\cdot 1410}{\dfrac{25547}{0.95}-20\cdot 1410} \right)}} {\ln{(0.95)}}\\\\ n &=& \dfrac {\ln{\left( \dfrac{-276}{ -1308.42105263} \right)}} {\ln{(0.95)}}\\\\ n &=& \dfrac {\ln{\left( 0.21094127112\right)}} {\ln{(0.95)}}\\\\ n &=& \dfrac {-1.55617552018} {-0.05129329439}\\\\ \mathbf{n} & \mathbf{=} & \mathbf{30.3387711544} \end{array} $}}$$

It will take to get 27924 about 30.3387711544 days

day 30 = 27919.16 

day 31 = 27933.20


$$\small{\text{$ 
 \begin{array}{lr}
 \small{\text{day }} 1 & 26957.00 \\
 \small{\text{day }} 2 & 27019.15 \\
 \small{\text{day }} 3 & 27078.19 \\
 \small{\text{day }} 4 & 27134.28 \\
 \small{\text{day }} 5 & 27187.57 \\
 \small{\text{day }} 6 & 27238.19 \\
 \small{\text{day }} 7 & 27286.28 \\
 \small{\text{day }} 8 & 27331.97 \\ 
 \small{\text{day }} 9 & 27375.37 \\
 \small{\text{day }} 10 & 27416.60 \\
 \small{\text{day }} 11 & 27455.77 \\
 \small{\text{day }} 12 & 27492.98 \\
 \small{\text{day }} 13 & 27528.33 \\
 \small{\text{day }} 14 & 27561.92 \\
 \small{\text{day }} 15 & 27593.82 \\
 \small{\text{day }} 16 & 27624.13 \\
 \small{\text{day }} 17 & 27652.92 \\
 \small{\text{day }} 18 & 27680.28 \\
 \small{\text{day }} 19 & 27706.26 \\
 \small{\text{day }} 20 & 27730.95 \\
 \small{\text{day }} 21 & 27754.40 \\
 \small{\text{day }} 22 & 27776.68 \\
 \small{\text{day }} 23 & 27797.85 \\
 \small{\text{day }} 24 & 27817.96 \\
 \small{\text{day }} 25 & 27837.06 \\
 \small{\text{day }} 26 & 27855.20 \\
 \small{\text{day }} 27 & 27872.44 \\
 \small{\text{day }} 28 & 27888.82 \\
 \small{\text{day }} 29 & 27904.38 \\
 \small{\text{day }} 30 & 27919.16 \\
 \small{\text{day }} 31 & 27933.20
 \end{array}
 $}}$$

if i have 25547 and add 1410 to that sum i get = 26957. if i reduce that new sum with 5 % i get = 25609,15. if i repeat this process each day, how many days will it take to get to 27924?

If I have properly understood: You get max 19 * 1410 = 26790

(19 = 0.95/0.05)

You never reach 27924!

$$\small{\text{$ \begin{array}{rcl} \dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } \\ &=& \dfrac{ 3^{2n+2n}-3^{2n} } { 3^{2n+n} + 3^{2n} } \\\\ &=& \dfrac{ 3^{2n}3^{2n}-3^{2n} } { 3^{2n}3^{n} + 3^{2n} } \\\\ &=& \dfrac{ 3^{2n} (3^{2n}-1) } { 3^{2n} (3^{n} + 1) } \\\\ &=& \dfrac{ 3^{2n}-1 } { 3^{n}+1 } \\\\ &=& \dfrac{ (3^{n}-1)(3^{n}+1) } { 3^{n}+1 } \\\\ \mathbf{ \dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } }&\mathbf{=} & \mathbf{3^{n}-1} \end{array} $}}$$

$$3.\\\\ \small{\text{$ \dfrac{ \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right) * \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) } { \dfrac{ x } { x^2+2xy+y^2 } } $}}$$

$$\small{\text{$ \begin{array}{rcl} &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x^2-xy }{ x^4-y^4 } \right) * \left( \dfrac { x^2+2xy+y^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x(x-y) }{ x^4-y^4 } \right) * \left( \dfrac { (x+y)^2 } { x } \right) \\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)^2 }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x-y)(x+y)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ x^4-y^4 } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ (x^2-y^2)(x+y) }{ (x^2-y^2)(x^2+y^2) } \right)\\\\ &=& \left( \dfrac{x}{x+y}+\dfrac{y}{x-y}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{x(x-y)+y(x+y)}{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &=& \left( \dfrac{ x^2+y^2 }{(x-y)(x+y)}\right)* \left( \dfrac{ x+y }{ x^2+y^2 } \right)\\\\ &\mathbf{=} & \mathbf{ \dfrac{ 1 }{x-y} } \end{array} $}}$$

Find The age for a rock for which you determine that 65% as the original uranium – 238 remainswhile the other 35% has decayed into lead

I assume: uranium-lead decay route $${}^{238}U$$ to $${}^{206}{Pb}$$

(i)

$$\boxed{~N_t = N_0\cdot e^{-k\cdot t} ~~(1)~} \quad \small{\text{k, is called the decay constant.}}\\ \small{\text{$N_0$ is the initial number of nuclei (at time zero), and}} \\ \small{\text{$N_t$ is the number remaining after the time interval.}} \\ \small{\text{We can also write our equation:~}} \boxed{~\ln{ \left( \dfrac{N_t}{N_0} } \right) = -k\cdot t ~~(2)~}$$

(ii)

$$\small{\text{Relationship between the decay constant, k, and half-life, $t_{1/2}$.}}\\ \small{\text{We set $N_{t_{1/2}} = \dfrac{N_0}{2} $ in (1)~so we have $ \dfrac{N_0}{2}=N_0\cdot e^{-k\cdot t_{1/2}} $ and get }}\\ \boxed{~ k = \dfrac{ \ln{(2)} }{ t_{1/2} } ~~(3)~}$$

(iii)

$$\small{\text{We get t, if we rearrange (2). We have~~}} \boxed{~ t = -\dfrac{1}{k} \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(4)~}\\ \small{\text{We set k in (3) into (4). We have~~}} \boxed{~ t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{N_t}{N_0} } \right)~~(5)~}\\$$

(iv)

$$\small{\text{$N_0 = {}^{238}U + {}^{206}Pb\cdot \dfrac{238}{206} $. And we have $35\%$ of ${}^{238}U$ is ${}^{206}Pb$ so }}\\ \small{\text{$N_0 = {}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}. $We obtain the latter quantity of ${}^{206}Pb$ by}}\\ \small{\text{multiplying the present mass of lead-206 by the ratio of the atomic mass of uranium}}\\ \small{\text{to that of lead, into which it has decayed. $N_t ={}^{238}U$. }}$$

$$\small{\text{ Finally $\dfrac{N_t}{N_0} = \dfrac{ {}^{238}U } {{}^{238}U + ({}^{238}U\cdot 35\%) \cdot \dfrac{238}{206}} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } $ }}\\ \boxed{ \dfrac{N_t}{N_0} = \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } }$$

(v)

$$\small{\text{If we regard the half-life $t_{1/2}$ for the decay of}}\\ \small{\text{uranium-238 $({}^{238}U)$ to lead-206 $({}^{206}Pb)$ is $\mathbf{4.468\cdot 10^9~ yr }$, }}\\ \small{\text{we can calculate how old the rock is.}}\\ \small{\text{We now have $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right) $ }}\\ \small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot\ln{ \left( \dfrac{1}{ 1 + 35\% \cdot \dfrac{238}{206} } } \right) $ }} \\ }}\\ \small{\text{or $t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot \left[ \ln{ (1) } - \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) \right] $ }} \\ }}\\ \small{\text{Because $\ln{(1)}= 0$ }} \\ \small{\text{$t = -\dfrac{t_{1/2}}{ \ln(2) } \cdot \left[ 0 - \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) \right] $ }} \\\\ \small{\text{and finally}}\\\\ \boxed{t = t_{1/2} \cdot \dfrac { \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) } { \ln(2) }}$$

(vi)

$$\small{\text{We calculate:}}\\ \small{\text{$ \begin{array}{rcl} t &=& t_{1/2} \cdot \dfrac { \ln{ \left( 1 + 35\% \cdot \dfrac{238}{206} } \right) } { \ln(2) }\\ t &=& 4.468\cdot 10^9~ yr \cdot \dfrac { \ln{ \left( 1.40436893204} \right) } { \ln(2) }\\\\ t &=& 4.468\cdot 10^9~ yr \cdot \dfrac { 0.33958804319 } {0.69314718056 }\\\\ t &=& 4.468\cdot 10^9~ yr \cdot 0.48992198586 \\\\ \mathbf{t} &\mathbf{=}& \mathbf{ 2.18897143282\cdot 10^9~ yr } \end{array} $}}$$

The rock is 2 188 971 432.82 years old 

Hallo sabi,

your way is not wrong.