What will be the reminder of (1n +2n +3n +4n)/5? Simply (1n +2n +3n +4n) mod 5=? Where n&lt;=10100000. My idea: The reminder may be 0 or 1 or 2 or 3 or 4 when a number is divided by 5. We can easily calculate it when n is too small but when n  is too big like  n&lt;=10100000 how can we calculate it?
43674
Calculate for every positive integer $n$ and for every prime $p$ the expression  $$1^n +2^n + \cdots +(p-1)^n \mod p$$    I need your help for this. I don't know what to do, but I'll show you what I know.   Wilson's theorem The identity $X^p-X = \prod_{a \in \mathbb{F}_p-1}(X-a)$ $\exists a \in \mathbb{Z}, \ a^2 \equiv -1 \mod p \qquad \iff \qquad p \equiv 1 \mod 4$   Now can you please provide me a hint?
118556
I'll describe the problem with an example.  Find integers $n$ and $m$ such that $n\cdot13 + m\cdot101 = 1$. This is the same as solving the equation $n\cdot13 \equiv 1 \pmod {101}$  I'm revising for a maths exam in a few days and it seems like a lot of the questions and examples rely on an ability to solve this type of problem. I can't seem to find a method in my notes, the solutions just get plucked out of thin air! What is a neat way of doing it on pen and paper with a calculator?
52847
I found two examples that I'd like to understand:  Example 1: Find $112358132134 \mod 9$.  Solution is: $1+1+2+3+5+8+1+3+2+1+3+4 \mod 9 = 7 \mod 9 = 7$  Example 2: Find $62831853 \mod 11$.  Solution is: $(-6)+2+(-8)+3+(-1)+8+(-5)+3 \mod 11 = -4 \mod 11 = 7$  My question is: Based on what property we sum the digits of the number ( as Example 1) or subtract (as Example 2)?
25377
I did the following base case $n = 5$ $$\begin{align*}     4(5) &amp;\lt 2 ^5\\     20 &amp;\lt 32 \end{align*}$$ So true.  $$\begin{align*}     4n &amp;\lt 2^n\\ n &amp;\lt 2^{n-2}\\ \log_2(n)+2 &amp;\lt n \end{align*}$$ But I don't think this is right. Where do I add in the $(n+1)$.
20105
$T(n) = 2$ if $n=0$  $T(n) = 9T(n-1)-56n +63$ if $n&gt;=1$  Repeated substitution  $k=1$  $T(n) = 9T(n-1)-56n+63 $  $k =2$  $T(n) = 81T(n-2) -560n + 1134$   $k =3$  $T(n) = 729T(n-3) -5096n + 15309$  I cant find the pattern for the n term and the integer  For now i just have  T(n) = $9^k(n-k)$
117732
Question 1: Denote $a \mod b$ as $a \% b$, where $a$ and $b$ are some integers  Evaluate 12^32475 % 21    The following is what I tried:    12^32475 % 21 = (12^3 % 21)^10825 % 21 = 6^10825 % 21 = ... = 6^433 % 21   But how to continue ?  Question 2:  I don't understand this: n-1= -1 mod n. Where is this from? And does this line become the next line?    Context: my class is discrete math for computer science. So far for related topics, it covers modular inverse, (extended) GCD algorithm, Fermat's Little Theorem and Chinese Remainder Theorem (CRT).
29747
To find last two digits  of $2^{100}$  I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about  $a=b \pmod n$. I have also learned about multiplication and addition of classes.  Can someone explain to me step by step on how to apply modular arithmetic to this problem?  Thanks
87175
So I start with:  $$a_n = \left(\frac23\right)^n + n$$ I know $a_1=\frac53, a_2=\frac{22}9, a_3=\frac{340}{81}, a_4=\frac{1247}{243} $ Then I do: $$a_{n-1} = \left(\frac23\right)^{n-1} + (n-1)$$    Do I now do $a_n - a_{n-1}$ or $a_n\over a_{n-1}$ ?
26520
I am not looking for an answer on this. Just need to clarify why my approach is failing -  $N_1 + N_2 + N_3$, i.e. single digit, double digit, 3 digit  single $= 2, 4, 6, 8$, i.e 4   double = X non-zero $= 8 \cdot 4 = 32$ X zero $= 9 \cdot 1 = 9$  Now the confusing part three digit, breaking into 4 cases   X zero zero $= 9$   X nz nz $= 7 \cdot 8 \cdot 4 = 224$   X z nz $= 8 \cdot 1 \cdot 4 = 32$   X nz z $= 8 \cdot 9 \cdot 1 = 72$  Three digit total comes to $= 9 + 224 + 32 + 72 = 337$. This answer is wrong and it should be $328$. What am I missing in the logic? Please suggest.
61995
What's the $n$-th digit in the sequence $S$ of numbers formed by the natural numbers, i.e.,  $n$-th digit in the sequence 1 2 3 4 5 6 7 8 9 10 11 12... ? For example 11th digit in the sequence is 0.
10400
Suppose I have $$ \large\sum_{i=1}^{10} 2. $$ Do I just add $2$ to itself $10$ times? I have worked on more complex ones with $n$ and such in the place where the $2$ is, but I have never done it when there is just a number there.
114348
Problem Statement:- The number of zeroes at the end of the integer   $$100!-101!+\ldots-109!+110!$$   I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.   And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.  Also, I dont know whether I am using the right tag feel free to correct it if its wrong.
78572
How do i prove that $17^n−12^n−24^n+19^n≡0(\mod35)$ for every possitive integer n. Can anyone give me a hint of how to start?
124252
Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.  So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod 1000$. I  don't know how to proceed since 1000 seems to be a bit high.
78178
With given number N, how to determine first number after N with same amount of odd and even divisors? For example if we have N=1, then next number we are searching for is :   2 because divisors: odd : 1 even : 2   I figured out that this special number can't be odd and obviously it can't be prime. I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number. Cheers
81866
Since last week I have been working on a way, how to sum $1$ and $2$ to have $133$.  So for instance we can have $133$ $1s$ or $61$ $s$2 and one and so on. Looking back to the example: if we sum: $1 + 1... + 1 = 133$ there is only one way. But for the second one there will be $131$ possible ways. And I have to do this for every possible combination. I am stuck with it and I have no idea whatsoever on how to begin.  Any ideas or methods I could use guys?
35607
I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.