+118658 Hi HSC
That is because this integral does not converge.
\(\int_{2}^{\infty} \frac{1}{2n-1}dn\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\left[\frac{ln(2n-1)}{2}\right]_2^\infty\\ =\infty-\frac{ln3}{2}\\ =\infty\)
back up by wolfram|alpha
https://www.wolframalpha.com/input/?i=integra+of+1%2F(2x-1)dx+from+x%3D2+to+x%3Dinfty
I was surprised by this because here is the graph...
The area of the green section represents the integral.... it looks like it converges doesn't it....
Maybe Heureka or Alan or some other person might like to comment ://
+26388 What is the remainder of: 13^1031 mod 599 =? Thanks for help.
Fermat's little theorem states that if p is a prime number, then for any integer a,
\({\displaystyle a^{p}\equiv a{\pmod {p}}}\)
If a is not divisible by p, Fermat's little theorem is equivalent
\( {\displaystyle a^{p-1}\equiv 1{\pmod {p}}}\)
see: https://en.wikipedia.org/wiki/Fermat%27s_little_theorem
Let p = 599 (prime number)
Let a = 13 (prime number)
gcd(13,599) = 1 ! so 13 and 599 are relatively prime, we can use Fermat's little theorem.
\(\begin{array}{|rcll|} \hline a^{p-1} &\equiv& 1{\pmod {p}} \\ 13^{599-1} &\equiv& 1{\pmod {599}} \\ 13^{598} &\equiv& 1{\pmod {599}} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline && 13^{1031} \pmod{599} \\ &\equiv & 13^{598+433} \pmod{599} \\ &\equiv & 13^{598}\cdot 13^{433} \pmod{599} \quad & | \quad 13^{598} \pmod{599} = 1 \\ &\equiv & 1\cdot 13^{433} \pmod{599} \\ &\equiv & 13^{433} \pmod{599} \\ &\equiv & 13^{8\cdot 54 + 1} \pmod{599} \\ &\equiv & (13^{8})^{54}\cdot 13 \pmod{599} \quad & | \quad 13^8 \pmod{599} = 541 \\ &\equiv & 541^{54}\cdot 13 \pmod{599} \\ &\equiv & 541^{3\cdot 18 }\cdot 13 \pmod{599} \\ &\equiv & (541^{3})^{18}\cdot 13 \pmod{599} \quad & | \quad 541^3 \pmod{599} = 162 \\ &\equiv & 162^{18}\cdot 13 \pmod{599} \\ &\equiv & 162^{3\cdot 6}\cdot 13 \pmod{599} \\ &\equiv & (162^{3})^{6}\cdot 13 \pmod{599} \quad & | \quad 162^3 \pmod{599} = 425 \\ &\equiv & 425^{6}\cdot 13 \pmod{599} \\ &\equiv & 425^{3\cdot 2}\cdot 13 \pmod{599}\\ &\equiv & (425^{3})^{2}\cdot 13 \pmod{599} \quad & | \quad 425^3 \pmod{599} = 181 \\ &\equiv & 181^{2}\cdot 13 \pmod{599} \\ &\equiv & 425893 \pmod{599} \\ &\equiv & 4 \pmod{599} \\ \hline \end{array}\)
