In the case I have two numbers large enough to justify using scientific notation twice $A \times 10^{B \times 10^C}$ or $Ae+Be+C$ How would I calculate Modulo without taking the numbers or any part of them out of scientific notation?
3352
I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:   $13^{100} \bmod 7$ $7^{100} \bmod 13$   I've also heard of this formula:  $$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n $$  But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)  Any hints or pointers in the right direction would be great.
17890
I saw the answer to  and it is the same problem, but i didn't get how to use the tip S.C.B gave. This was the tip: $"$First, assume that it is not $a$ primitive root $(\text{mod m})$. Then we have that there exists such $r&lt;\phi(m)$ such that $$a^r\equiv 1(\text{mod m})$$ Now use that, if $n=mk$ $$ϕ(mk)=ϕ(m)ϕ(k)\frac{d}{ϕ(d)}≥ϕ(m)ϕ(k)&gt;rϕ(k)$$ where $d=gcd(m,k)"$ and i saw the answer to this question using group theory, but i want an answer using elementary number theory, if you have a different answer, or tip, it would be good as well. What i tried was this: $n=mk$, then $$a^{\phi(n)}\equiv 1(\text{mod n})\Rightarrow a^{\phi(mk)}\equiv1(\text{mod mk}) \Rightarrow a^{\phi(mk)}\equiv1(\text{mod m})$$ But i don't know how to follow from here
54300
Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.
61554
If we multiply 2 number and take the mod with prime this is equivalent to first taking mod with the prime of individual number and then multiplying the result and again taking mod.  $$ ab\bmod p = ((a\bmod p)(b\bmod p))\bmod p$$  does there exist any proof for this?  Does it work for composite moduli too? Then I can use the Chinese remainder Theorem to calculate the result does there any other way apart from Chinese remainder Theorem to solve the problem?
10899
I am working on a problem I am pretty close to solving but I can't figure out the last part. I used some algebraic manipluation to break the problem down. The problem is:     Show that the following congruence is true      a^b = ((a mod n)^b) (mod n)      for any positive integers a, b, and n.      Hint: Number a can be written as a = q*n + r, where q = floor brackets a/n, and r = a mod n.   My attempt it:     (q*n +r)^b = ((a mod n)^b) * (mod n)      (q*n)^b + (a mod n)^b = ((a mod n) ^b) * (mod n)      (a)^b -(r)^b = ((a mod n) ^b) * (mod n)      (a)^b = modn  ----> divided by ((a mod n)^b)   Am I doing this correctly and if so how can I prove this is true? Is there a trick I am missing? I think that by simplifying it further I would just keep going in a loop.  Thanks!
79879
I would like to factorise $496241$. I know the answer is $677 \times 733$. But I don't know how to get there.   Here is the full question:   "A message has been encoded using RSA with a modulus of $m = p q = 496241$ (with $p$ and $q$ being primes) and encoding exponent of $218821$.  You are advised that $\{13631, 142703\}$ is a valid encoding–decoding pair for the same modulus, $m$.  (a) Use this information to determine $\phi(m)$ for this modulus. (Using software to directly factorise $m$ is not a valid option for doing this part.)  (b) Verify your answer by determining the primes $p$ and $q$. Show how these combine to give both $m$ and $\phi(m)$."  I need it so I can solve (a), and it says I cannot use software to directly factorise m. So I guess pen and paper. But there are not tutorial online. Thanks
90050
I want to calculate   $ 8^{-1} \bmod 77 $   I can deduce  $ 8^{-1} \bmod 77$ to $  8^{59} \bmod 77 $ using Euler's Theorem.  But how to move further now. Should i calculate $ 8^{59} $  and then divide it by $ 77 $ or is there any other theorem i can use ?
14039
Show that  $$a^2 \mathrm{\ mod \ } b = (a \mathrm{\ mod \ } b)^2 \mathrm{\ mod\ } b$$ for $ a, b \in \mathbb{Z}^+ $.  this was derived from an Informatics olympiad question.
68160
How do I simplify this expression?     $276^{246} \mod 323$    Thank you so much.
24393
I have the following equation: $$ n \equiv M^k\mod(b) $$ where n, a, b are integers, and M is unknown. How do I solve this equation to find the M value? Those a and b are public keys of the RSA algorithm and n is the result of encryption. I need to find the encrypted message.
99833
Currently working on this problem and I'm having trouble seeing how it goes from one line to the next.    $45x \equiv 63\mod 11$ goes to $x \equiv 8\mod 11$     Any help would be awesome thanks.  Please try to explain how it works to help me see thanks.
22739
Is this purely personal style, or are the generally accepted guidelines? Clearly, when dealing with numbers either $2 \cdot 2$ or $2 \times 2$ is needed, but within formulae where an absence of an operator implies multiplication, when should a \cdot be used? I tend to use it only sparingly if it aids in grouping. (Of course, barring the cases where it's required, e.g. a dot product.)
118796
For example, in RSA, we use this for encryption: $ciphertext = (m^e \mod n)$  and for decryption.  If our message is "hello world", then what number do we have to put as $m$ in the RSA formula?
29031
I would like to know how to find multiplicative inverses in $\mathrm{GF}(2^8)$. I know how to multiply two elements of $\mathrm{GF}(2^8)$ (for example, I know that 0x1b*0xaa equals 0x8c).  I follow , but I still don't understand the process. More specifically: How exactly (step-by-step) are logarithm and antilogarithm tables for AES' finite field formed?
111114
I'm told to find:  $105 308^{7125} \pmod {11}$  I'm not exactly sure how to go about calculating this. I know that I could split the exponent into multiples of it, for instance.  $7125 = 7 * 10 * 10 * 10 * \cdots $ whatever else  As such: $105308^{7(10)(etc..)}$  But even then the base number is too big to be multiplied to get an actual answer on a calculator, any ideas?
44298
From the original data, we know that they are nearly 0.087 rather that 0.087 x 10^-2. Why is there scientific notation (10^-2)? It looks ridiculous. Please help me figure it out. Thanks very much!   MWE:  \documentclass[border=5mm]{standalone} \usepackage{pgfplots} \pgfplotsset{compat=newest} \begin{document} \definecolor{mycolor1}{rgb}{1,0,1}  \pgfplotstableread{ X           Y1 1    0.0873514639102964  2    0.0873656441069913  3    0.0873509320240032  4    0.0873490368814644  5    0.0873430960617957  6    0.0873334626756388  7    0.0873430596012905  8    0.0873485872405474  9    0.0873519301953318  10  0.0873474824394306 11  0.08735183859007     12  0.0873408783047386 13  0.0873439016123011 14  0.0873424452271129 15  0.0873452848079798 16  0.0873467228778811 17  0.087348210873026    18  0.0873464310394191 19  0.0873469131127228 20  0.0873449430894595 }\datatable  \pgfplotsset{     width=9cm,     height=6cm,     scale only axis,     xmin=0.5, xmax=20.5,     tick align = outside,     yticklabel style={/pgf/number format/fixed},   } \begin{tikzpicture}  \begin{axis}[     xtick={1,...,20},     ymin=0.08715,     ymax=0.08740,     ytick={0.08715,0.08720,0.08725,0.08730,0.08735,0.08740},     yticklabels={0.08715,0.08720,0.08725,0.08730,0.08735,0.08740},     xlabel={$x$},     ylabel={$y$},     y axis line style={mycolor1},     ytick style={mycolor1},     yticklabel style=mycolor1,     yticklabel style={/pgf/number format/fixed},     axis x line*=bottom,     axis y line*=left] \addplot [     color=mycolor1,     solid,     line width=1.5pt,     mark size=2.7pt,     mark=square*,     mark options={solid,fill=mycolor1}]     table [x=X, y=Y1]{\datatable}; \label{plot1} % So we can recall the plot style later on \end{axis} \end{tikzpicture}% \end{document}
59304
I am finding a 4-digit number $abcd$ (in base-10 representation) satisfying the following property.  $$abcd=a^b c^d$$  I have been running my mind around this problem since a long time, but got no success.  I wrote a python program for a number with this property, and got the answer to be $2592$ since   $$2592=2^5\times 9^2$$  But I am looking for a purely mathematical way to solve this problem.   Thanks!