heureka

What is the remainder when 1009^109 is divided by 101? Thanks for help.

$\begin{array}{|rcll|} \hline && 1009^{109} \pmod {101} \qquad &| \qquad 1009 \pmod {101} = -1 \\ &\equiv & (-1)^{109} \pmod {101} \qquad &| \qquad (-1)^{109}=-1\\ &\equiv & -1 \pmod {101} \\ &\equiv & 100 \pmod {101} \\ \hline \end{array}$

What is the remainder when $5^{137}$ is divided by 8?

$\begin{array}{|rcll|} \hline \text{Because the } gcd(5,8) = 1 \\ 5^{\varphi(8)} \equiv 1 \pmod 8 \\ \hline \end{array}$

$\varphi(n) \text{ is the Euler's totient function}$

$\begin{array}{|rcll|} \hline 8 &=& 2^3 \\ \varphi(8) &=& 8 \cdot \left( 1-\frac12 \right) \\ \varphi(8) &=& 4 \\ 5^{4} &\equiv& 1 \pmod 8 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && 5^{137} \pmod 8 \\ &\equiv& 5^{4\cdot 34 + 1 }\pmod 8 \\ &\equiv& 5^{4\cdot 34}\cdot 5 \pmod 8 \\ &\equiv& (5^{4})^{34}\cdot 5 \pmod 8 \qquad | \qquad 5^{4} \equiv 1 \pmod 8 \\ &\equiv& 1^{34}\cdot 5 \pmod 8 \\ &\equiv& 1 \cdot 5 \pmod 8 \\ &\equiv& 5 \pmod 8 \\ \hline \end{array}$

The remainder is 5

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1. Your response should have a factorial.

$\begin{array}{|rcll|} \hline n &=& (n+1) - 1 \\ \hline n\cdot n! &=& [~(n+1) - 1~]\cdot n! \\ n\cdot n! &=& (n+1)\cdot n! - n! \\ \mathbf{n\cdot n!} &\mathbf{=}& \mathbf{(n+1)! - n!} \\ \hline \end{array}$

$\begin{array}{|r|r|c|r|} \hline n & n\cdot n! &=& (n+1)! - n! \\ \hline 1 & 1* 1! &=& \not \! 2! - 1! \\ 2 & 2* 2! &=& \not \! 3! - \not \! 2! \\ 3 & 3* 3! &=& \not \! 4! - \not \! 3! \\ 4 & 4* 4! &=& \not \! 5! - \not \! 4! \\ 5 & 5* 5! &=& \not \! 6! - \not \! 5! \\ \dots & \dots & & \dots \\ n & n\cdot n! &=& (n+1)! - \not \! n! \\ \hline \Sigma & 1*1!+2*2!+...+n*n! &=& (n+1)! - 1! \\ \hline \end{array}$

Find a closed form for

S_n=1*1!+2*2!+...+n*n!.

for integers n>=1.

Your response should have a factorial.

Example for $n=4$:

$\begin{array}{|rcll|} \hline S_4 &=& 1*1!+2*2!+3*3!+4*4! \quad & | \quad 4! = 3!*4 \\ S_4 &=& 1*1!+2*2!+3*3!+4*3!*4 \\ S_4 &=& 1*1!+2*2!+3!*(3+4*4) \\ S_4 &=& 1*1!+2*2!+3!*(3+4^2) \quad & | \quad 3! = 2!*3 \\ S_4 &=& 1*1!+2*2!+ 2!*3*(3+4^2) \\ S_4 &=& 1*1!+2!*[~2+ 3*(3+4^2)~] \quad & | \quad 2! = 1!*2 \\ S_4 &=& 1*1!+1!*2*[~2+ 3*(3+4^2)~] \\ S_4 &=& 1!*\{~1+2*[~2+ 3*(3+4^2)~]~\}\quad & | \quad 1! = 1 \\ S_4 &=& 1*\{~1+2*[~2+ 3*( \underbrace{3+4^2}_{ \underbrace{=4-1+4^2}_{ \underbrace{=4*(4+1)-1}_{=4*5-1} } })~]~\} \\ S_4 &=& 1*\{~1+2*[~2+ 3*( 4*5-1 )~]~\} \\ S_4 &=& 1*[~1+2*(~2+ 3*4*5-3 ~)~] \\ S_4 &=& 1*[~1+2*(~3*4*5-1 ~)~] \\ S_4 &=& 1*(~1+2*3*4*5-2 ~) \\ S_4 &=& 1*(~ 2*3*4*5-1 ~) \\ S_4 &=& 1*2*3*4*5-1\\ S_4 &=& 5!-1\\ 1*1!+2*2!+3*3!+4*4! &=& 5!-1\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline S_n &=& 1*1!+2*2!+...+n*n! \\ S_n &=& (n+1)!-1 \\ \hline \end{array}$

Kmax= (6.63*10^-34J*s) (7.09*10¹⁴s) - 2.17*10^-19J

Im not sure if the 10¹⁴s is suppose to be 10¹⁴s^-1 or not.

Someone told me it should but it doesnt say on the problem. 

$\begin{array}{|rcll|} \hline K_{max} &=& ( 6.63*10^{-34}J*s) ( 7.09*10^{14}s^{-1} ) - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-34}*10^{14} J*s*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-34+14} J*s*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^1*s^{-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^{1-1} - 2.17*10^{-19}J \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*s^{0} - 2.17*10^{-19}J \quad | \quad s^{0} = 1 \\\\ K_{max} &=& 6.63*7.09*10^{-20} J*1 - 2.17*10^{-19}J \\\\ K_{max} &=& 47.0067*10^{-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10*10^{-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10^{1-20} J - 2.17*10^{-19}J \\\\ K_{max} &=& 4.70067*10^{-19} \ J - 2.17*10^{-19} \ J \\\\ K_{max} &=& (4.70067- 2.17)*10^{-19} \ J \\\\ K_{max} &=& 2.53067*10^{-19}\ J \\ \hline \end{array}$

e= (1.7*10^3J)-(3.3*10^2J)                  I got 1.4*10^-3J but i dont think it is right. 

            (1.7*10^3J)

$\begin{array}{|rcll|} \hline e &=& \dfrac{(1.7*10^3J)-(3.3*10^2J)}{(1.7*10^3J)} \\\\ e &=& \dfrac{1.7*10^3J}{1.7*10^3J} -\dfrac{3.3*10^2J}{1.7*10^3J} \\\\ e &=& 1 -\dfrac{3.3*10^2{\color{red}\not} J }{1.7*10^3{\color{red}\not} J } \\\\ e &=& 1 -\dfrac{3.3*10^2 }{1.7*10^3 } \\\\ e &=& 1 -\dfrac{3.3*10^{2-3} }{1.7} \\\\ e &=& 1 -\dfrac{3.3*10^{-1} }{1.7} \\\\ e &=& 1 -\dfrac{0.33}{1.7} \\\\ e &=& 0.80588235294 \\ e &=& 8.0588235294 *10^{-1}\\ \hline \end{array}$

Find the sum of all possible positive integer values of b

$\begin{array}{|rcll|} \hline b & < & \frac{25}{8} \\ b & < & 3.125 \\\\ &\text{positive integer:}& b = 1 \qquad b = 2 \qquad b = 3\\\\ &\text{rational roots:}& b = 2 \qquad b = 3\\\\ && 2+3 =5\\ \hline \end{array}$

K= 1/2(6.6*10^2kg)(2.11*10^4m/s)^2

Solve for K

K= 1/2(6.6*10 ^2 kg)(2.11*10^4m/s)^2

$\begin{array}{|rcll|} \hline K &=& \frac12*(6.6*10 ^2 kg)*(2.11*10^4\frac{m}{s})^2 \\ K &=& \frac12*6.6*10 ^2 kg * (2.11*10^4\frac{m}{s})^2 \\ K &=& \frac12*6.6*10 ^2 kg * 2.11^2*10^8\frac{m^2}{s^2} \\ K &=& \frac12*6.6*10 ^2 * 2.11^2*10^8\frac{ kg * m^2}{s^2} \\ K &=& \frac12*6.6 * 2.11^2*10^{10} \frac{ kg * m^2}{s^2} \\ K &=& 14.69193 *10^{10} \frac{ kg * m^2}{s^2} \\ K &=& 1.469193 *10^{11} \frac{ kg * m}{s^2} \cdot m \qquad | \qquad 1~N = 1~\frac{kg*m}{s^2} \\ \mathbf{K} &\mathbf{=}& \mathbf{1.469193 *10^{11} N \cdot m} \\ \hline \end{array}$

Find the sum of all possible positive integer values of b such that the quadratic equation

$2x^2 + 5x + b = 0$

has rational roots.

$\begin{array}{|rcll|} \hline && 2x^2 + 5x + b = 0 \\\\ x& =& \frac{-5\pm\sqrt{5^2-4\cdot 2\cdot b}}{2\cdot 2} \\\\ x &=& \frac{-5\pm\sqrt{25-8\cdot b}}{4} \\\\ 25-8\cdot b &\ge& 0 \\ 25 &\ge& 8\cdot b \\ 8\cdot b &\mathbf{<}& 25\\ \mathbf{b} &\mathbf{<}& \mathbf{\frac{25}{8}} \\ \hline \end{array}$