Find last two digits of $33^{100}$.  My try:  So I have to compute $33^{100}\mod 100$  Now by Euler's Function $a^{\phi(n)}\equiv 1\pmod{n}$  So we have $33^{40}\equiv 1 \pmod{100}$  Again by Carmichael Function : $33^{20}\equiv 1 \pmod{100}$  Since $100=2\cdot40+20$ so we have $33^{100}=1\pmod{100}$  So last two digits are $01$  Is it right?
120603
Find the last two digits of $33^{100}$   By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$  So $33^{40}\equiv 1 \pmod{100}$ Then how to proceed?  With the suggestion of  @Lucian:  $33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$  By using the binomial expansion, we have: $33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$ $\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
113972
Are there any simple methods for calculating the digits of $\pi$? Computers are able to calculate billions of digits, so there must be an algorithm for computing them. Is there a simple algorithm that can be computed by hand in order to compute the first few digits?
90091
I bumped across the aforementioned question in my notes while studying today and I have completely forgot how to do this. I remember using CRT to solve a problem like this on one of my tests, too bad they didn't give back my solutions :(.  Since $\gcd(100,2) = 2$, we can't use the usual Euler's theorem to solve via $\pmod {100}$. So $100 = 5^2 2^2$, and applying Euler's on the 25 gives me $2^{20} \equiv 1 \pmod {25}$. However since $\gcd(2,4) = 2$, we can't do the same for $\bmod 4$. Accordingly how do I set up the other modulo congruence so I can apply CRT?
18834
I ran into a cool trick for . This is actually a recurrent relation which states that:  If $D(N)$ denotes the last non zero digit of factorial, then   $$D(N)=4D\left(\left\lfloor{\frac N5}\right\rfloor\right)\cdot D(\mbox{Units digit of $N$})  \qquad \mbox{(If tens digit of $N$ is odd)}$$ $$D(N)=6D\left(\left\lfloor{\frac N5}\right\rfloor\right)\cdot D(\mbox{Units digit of $N$})  \qquad \mbox{(If tens digit of $N$ is even)}$$  Where $\left\lfloor\cdots\right\rfloor$ is greatest integer function.  I was wondering, if anybody could explain why this works?
119485
with -24 $\lt l \le$ 25.  Then I need to conclude that all final 2-digit numbers of the decimal expansion of squares are to be found among those of $0^2, 1^2, 2^2,...., 25^2$.  I'm thinking that I could maybe rewrite the division theorem proof for this but I don't know how/what to do with that. Any ideas?
51740
I am trying to find the last digit of $3^{5^7}$ (or $3^{(5^7)}$, if you will) using modular exponentiation. Here's what I've figured out:  The value I want is $3^{5^7} \mod 10$.  $5^n \mod 10 = 5$ (if $n$ is a positive natural number).  The residues of $3^n \mod 10$ repeat themselves every time $n$ increases by $4$. The pattern is $1, 3, 9, 7, 1, 3, 9, 7, \ldots$ for $n = 0, 1, 2, 3, 4, 5, 6, 7, \ldots$  Now, I don't know how to use these facts to get what I need. I know that the answer $3$ (via WolframAlpha) but I don't know how to connect the dots here. Any help would be appreciated.  Thanks!
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
86119
I know how to find for which $n$ $\phi(n)=n/2$ or $\phi(n)=n/3$, my method for finding those was simply to find primes $p$ that satisfy $\Pi_p$$_|$$_n$$1-1/p$ $  = 1/2$ or $1/3$.  However, I don't know how to find $\Pi_p$$_|$$_n$$1-1/p = n/6$. Intuitively it seems that if I combine results for both $\phi(n) = n/2$ and $\phi(n) = n/3$ I'll get $\phi(n) = n/6$ but it does not work, cause I get number of the form $2^a3^b$ which gives $\phi(n) = n/3$ again.  Is there a way to find $n$ for which $\phi(n)=n/6$? Or do such numbers exist at all?
36916
I heard from a video that there is a theorem that proves that exists a power of two wghose leading or last digits can be any sequence of digits for example your telephone number. Can anyone point me to this theorem please?
115165
Find all integers $n$ between $0\le n &lt; m$ that are relatively prime to $m$, for $m = 4,5,9, 26$. We denote the number of integers $n$ which fulfill the condition by $\phi (m)$, e.g. $\phi (3) = 2$. This function is called "Euler's phi function". What is $\phi (m)$ for $m =4,5,9,26$?
82626
Suppose,I have a number 1256. I want (1256 % 11)! That is 2 . But, here, if i try in this way,  ( (125 % 11 ) * 10 + 6 ) % 11 = 2. That is exactly the same answer 2 as above.  I am confused , how This two process gives the similar answer ??? Even if , I try this method for any number n,m  to find (n mod m) . How this works ?? Can anybody explain ?
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 ?
56139
I know the answer is $198$.  I realise that if $\log _{10}\left( x\right) =y$, the number $x$ has $\lfloor y\rfloor -1$ digits  So I tried $\log ^{\ }_{10}\left( 99^{99}\right) $ = $\log _{10}\left( 100\left( 1-\dfrac {1}{100}\right) \right) $ = $198 + 99\log_{10}\left( 1-\dfrac {1}{100}\right) $ , then I do not know how to proceed. I guess using this method amounts to finding a good approximation to $\log _{10}\left( 99\right)$  I would also be interested to know how one can solve this with the binomial theorem: $\left( 100-1\right) ^{99}$
25880
I was able to work out the cases for $n$ having up to $4$ digits and was wondering if someone could verify my generalization to $m$ digits.  Here I am assuming that when a reversal results in there being a leading $0$ it get ignored (e.g. $76130$ gets reversed to $3167$, $998700$ to $7899$ etc.)  I believe the function can be expressed as $r(n): \mathbb{N} \rightarrow \mathbb{N}$  $$r(n) = \left \lfloor \frac{n}{10^m}\right \rfloor +\sum_{k=0}^{m-1}10^{m-k} \left( \left \lfloor\frac{n}{10^k} \right \rfloor-10 \left \lfloor \frac{\left \lfloor \frac{n}{10^k}\right \rfloor}{10}\right \rfloor \right)$$  where again, $n$ has $m$ digits.  It seems like this combination of the floor function and powers of 10 is the only way to achieve this but is this true?
93985
I'm looking for some solutions to $\varphi\left(n\right)=2^{32}$ where $\varphi$ is the Euler's totient function. I know that if $n=p_{1}^{r_{1}}\cdot\ldots\cdot p_{k}^{r_{k}}$ satisfies $\varphi\left(n\right)=2^{32}$ then \begin{align*}  &amp; 2^{32}=\varphi\left(n\right)=\prod_{i=1}^{k}p^{r_{i}}\left(1-\frac{1}{p_{i}}\right)=n\prod_{i=1}^{k}\left(1-\frac{1}{p_{i}}\right)\\ \Rightarrow\quad &amp; n=\frac{2^{32}}{\prod\left(p_{i}-1\right)}\prod p_{i} \end{align*} So I was looking to compute solutions by finiding primes $p_{i}$ such that $p_{i}-1\mid2^{32}$ and plug them into the last equation. Those are $p_{i}-1\in\left\{ 2^{l}\mid1\leq l\leq32\right\} $  and for example $p_{i}-1=2$ is good because then $p_{i}=3$ is a prime. Plugging it into the equation  gives $$ n=\frac{2^{32}}{\left(3-1\right)}\cdot3=3\cdot2^{31} $$  But then $\varphi\left(3\cdot2^{31}\right)=\varphi\left(3\right)\varphi\left(2^{31}\right)=2\left(2^{31}-2^{30}\right)=2^{31}\boldsymbol{\neq}2^{32}$  What am I missing here?
18547
Let $k\ge 2$ be a natural number.      What is the largest number $n$ with $\lambda(n)=k$, where $\lambda(n)$ is the Carmichael-function. (See :  ) ?   For $k=2$, the maximum seems to be $24$.  For $k=4$, the maximum seems to be $240$.  For $k=6$, the maximum seems to be $504$.  For $k=60$, I found many $n's$, currently I arrived at $227,146,920$. (The program just found a higher number, so the number is not maximal)  I am not sure, but a maximum should exists for every $k$, thus defining the number $n(k)$.     Which growth rate will $n(k)$ probably have ? Exponential ?
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.