When I get around at park , I made up a question to hang out by myself.However , I stuck in it. Question: Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$? Where I stuck in : $1-)$ I know how to write $75$ in the base of $3$ , but I do not know writing $75!$ in the base of $3$. $2-)$ To find the last two digits, we utilize from $(\text{mod }100)$, then which modulus should I use to find its last two digits? Thanks for your helps..
120607
This is a proof from a book on number theory I'm reading. I'm having a hard time following. I think there's a variable here that means two different things at two different times...  Theorem:  If n is a positive integer and p is a prime, then $p^e \vert\vert n!$, where $e=\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$ and r is determined by n by the inequality $p^r \le n &lt;p^{r+1}$  Proof:  For a given integer k, the multiples of $p^k$ that do not exceed  n are $p^k, 2p^k, \cdots, qp^k$ , where q is the largest integer such that $qp^k \le n$. But this says that q is the largest integer not exceeding $n/p^k$, so that  $q= \lfloor n/p^k \rfloor$. Thus, $\lfloor n/p^k\rfloor$ gives the number of positive multiples of $p^k$ that do not exceed n.     Interlude:      I understand the proof so far. We've simply chosen some power of p   and shown that $\lfloor n/p^k \rfloor$ is the largest integer that we can multiply by $p^k$ and stay under n. Here's where I start to get confused.   Now, if $1 \le m \le n$, then $m=qp^k$ with $(q,p) = 1, 0 \le k \le r$, and m contributes precisely k to the total exponent e with which p appears in the canonical representation of n!     Interlude 2:      How exactly does $m=qp^k$ follow from $1 \le m \le n$? Also, this q seems to mean something other than $\lfloor n/p^k\rfloor$ since we seem to be considering the set of all numbers less than or equal to n and greater than or equal to 1 that have the form $m=qp^k$?   Moreover, m is counted precisely k times by the sum  $\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$,   once as a multiple of p, once as a multiple of $p^2$, ... , once as a multiple of $p^k$, and no more. Of course, if $k=0$, then m is not counted in the sum. Therefore, the sum above accounts exactly for the contribution of each m between 1 and n to the exponent e as claimed.     Interlude 3:      What exactly is meant here by counted? I really don't understand the last part of this proof at all...
69873
I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic.    Question:  If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9.    Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$  Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$   therefor  $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$  $9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$  Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself.    The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$.  I think this is fair game here but not 100% confident.  Thanks.
58471
How many $4$ digit even numbers have all $4$ digits distinct ?   This question has been asked if we replace even with positive.   I was solving this Problem like: For first position $1-9$ (No $0$) for next $0-9$ (leaving out the digit which has been choose) for next $0-9$ (leaving out $2$ digits), but on last digit I get stuck. From what I should choose. Will it be $0, 2, 4, 6, 8$, because number needs to be even? How should I solve it?
17911
I am doing this problem mentioned above and I know the answer because I know Euler's Theorem that $$a^{\varphi{(m)}}\equiv{1}\pmod{m}.$$  I used 100 as my modulus and got that the last two digits of $7^{355}$ are $43$.  My question is this.  What prime should I pick to solve this problem using Fermat's theorem $$a^{p-1}\equiv{1}\pmod{p}$$  As a precursor problem, the author (Dudley; Elementary Number Theory) asked to find the last digit of the number above; so using a modulus $5$, $7^4\equiv{1}\pmod{5}$ by Fermat.  Since the answer is $7^{355}\equiv{3}\pmod{5}$, the last digit is either $3$ or $8$.  Since $7$ is odd, the answer will clearly be odd, so $3$ is it.  The book is working linearly, building from the previous chapters and Euler's Theorem isn't mentioned for another two chapters (I remembered Euler from a problem solving class in college).    So again, what prime would be best?  How should I procede?    EDIT: I've seen a bunch of problems on the site and none that I looked at did it using Fermat.  If there is one, that could be posted and I would be happy.  EDIT 2:  it seems that the problem of finding of finding the last two digits using is rather difficult because of the implication that we need something mod 100.  Is there a way around this , like adding a step or complexifying, in order to finish the problem with Fermat?
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 ?
82160
If we reverse the digits of $12$ we will get $21$. $12^{2}=144$. If we reverse its digits we will get $441$ which is $21^{2}$.  Here is the puzzle. How many such two digit numbers are there? Digits must be different.  We got an algebraic method to solve it if the square is a three digit number.It given here :  Let $a&gt;b$  $(10a+b)^{2}=100x+10y+z$ and $(10b+a)^{2}=100z+10y+x$  Difference between these two equations will lead to $a^{2}-b^{2}=x-z$  $x-z \leq 8$.Then  $a^{2}-b^{2} \leq 8$.  Since squares are of three digits $a &lt;4$.But since $a&gt;b ,a \neq 1$.  $b=1 \Longrightarrow a^{2} \leq 9$.Then $a=1,2,3$ but only possibilities are $2 $ and $3$.  $b=2 \Longrightarrow a^{2} \leq 12 \Longrightarrow a= 1,2,3$. But possibility is $a=3$. So next such pair will be $13,31$.  Higher values of $b$ will not work since $a$ cannot exceed 3. This is an explanation we obtained from a mathematics group.It works well if the square is a three digit number.What will happen if square is a 4 digit number... Or if we try to extend this problem for numbers having more than two digits ?
3207
Since $16 \div{-3} = -5.\overline{3}$, I thought I could also express this as $16 \div{-3} = -5\:R\:1$ or in other words $16\mod{-3} = 1$. My calculator tells my it is in fact $-2$.  Along the same lines, I see that $-16\mod{3} = 2$. So, while the true answer evaluates to $-5.\overline{3}$ again, the "remained notation" is $-6\:R\:2$ here, and $-6\:R\:-2$ in the first example.  Why do we go "past" $15$ to $18$ (or in the second example, past $-15$ to $-18$) and calculate the remainder from there?  Further warping my brain, $-16\mod{-3} = -1$, so it looks like we "stop" at $-15$ in this example.
8931
Many years ago,  I noticed that $987654321/123456789 = 8.0000000729\ldots$.  I sent it in to Martin Gardner at Scientific American and he published it in his column!!!  My life has gone downhill since then:)  My questions are:   Why is this so? What happens beyond the "$729$"? What happens in bases other than $10$?
73819
I'm trying to find the last two digits of ${2012}^{2012}$. I know you can use (mod 100) to find them, but I'm not quite sure how to apply this. Can someone please explain it?
95063
Find the value of:$$\frac{\sqrt{45} + \sqrt{18}} {\sqrt{7+2\sqrt{10}}}$$      I need help solving this question. Every reply is appreciated.  Thanks!
71773
Find The Last 3 digits of the number $2003^{2002^{2001}}$  BY number theory or otherwise,   Also i would like to ask is there a property observed in the numbers of the form $k^n$, where for some $k, n$ is varied then the digits of $k^n$ are periodic,  for example,  $2^n$,  its last digit is periodic with period 4, its second last digit is periodic $4\cdot 5 = 20$ its third last digit is periodic with periodic with period $20\cdot 5 =100$  I have observed this property with other numbers as well, though period might vary,for different values of $k$.
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?
120624
Base $10$ uses these digits: $\{0,1,2,3,4,5,6,7,8,9\};\;$ base $2$ uses: $\{0,1\};\;$ but what would base $1$ be?    Let's say we define Base $1$ to use: $\{0\}$. Because $10_2$ is equal to $010_2$, would all numbers be equal?  The way I have thought Base 1 might be represented is tally marks, $0_{10}$ would be represented by nothing. So, $5$ in Base 1 would be represented by $00000$? Or we could define Base 1 to use: $\{$|$\}$ and $5$ would be |||||?
121952
My question is this : how many distinct two digit numbers can be produced from numbers $4, 3, 3, 1$? When applying the formula $$\frac{4!}{(4-2)!2!}$$ you come up with $6$, yet when doing the problem manually, I come up with $7$ numbers, namely $41, 43, 14, 13, 33, 31, 34$. Can anybody explain this discrepancy?  Thanks.
50585
This is my question: Find the last 2 digits of $123^{562}$  I don't even understand where to begin on this problem. I know I'm supposed to use Euler's theorem but I have no idea how or why. Any help? Thanks
42887
$$ \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? $$   I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.  Thanks in advance!
106448
What is the remainder when $2^{2016}$ is divided by $47$?   I have done Fermat's little theorem and I have now this:  $2^{2016} \equiv 2^{38} \pmod{47}$  My issue is that $2^{38}$ is too large of a number to work with, how do I continue?