help..  Find the last 3 digits of $$2003^{2002^{2001}}$$  I try..  I need find $$2003^{2002^{2001}}\equiv x(mod\ 1000)$$ ??  How?
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 $k$.
122886
I think i know how to solve it but is that the best way? Is there a better way (using number theory). What i do is: knowing that  1st power last digit: 3 2nd power last digit: 9 3rd power last digit: 7 4rh power last digit: 1 5th power last digit: 3    $3^{347} = 3^{5\cdot69+2} = (3^5)^{69} \cdot3^2 = 3\cdot3^2=3^3=27 $ so the result is $7$.
18430
Find Remainder when $N=2009^{2009}-1982^{2009}-1972^{2009}+1945^{2009}$ is divided by 1998  I have written the given number $N$ as  $$N=(1998+11)^{2009}-(1998-16)^{2009}-(1998-26)^{2009}+(1998-53)^{2009}$$ and by binomial theorem the remainder when $N$ divided by $1998$ is  $$R=11^{2009}+16^{2009}+26^{2009}-53^{2009}$$  but how to find remainder when $R$ is divided by $1998$
51253
$6$ and $210$ share the property that both are the products of both two and three consecutive numbers. $6$ is $2\times3$ and $1\times2\times3$ and $210$ is $14\times15$ and $5\times6\times7$. It was easy enough to write a program to search for more numbers with this property, I found that there were no more up to at least $1{,}000{,}000{,}000{,}000$ ($1$ Trillion). But it is beyond me to prove that there are either no more numbers like this or to find the next one. Any ideas?
1672
I have had a go at this for a while now and can't seem to get anywhere.  $$2^{(2^{517})} \pmod {23}$$  I see that Fermat's Little Theorem must come into play but I can't see where to pull $22$ from. Any help would be appreciated.  Here is my attempt:  $$2^{(2^{517} \!\!\mod {22})} \equiv  2^{(2^{517})} \pmod {23}$$ Then proceed to do Chinese remainder theorem to find the exponent?
62212
I've been trying to figure out if a number is divisible by $13$.  As I'm saying this in first person, I think I'm supposed to take the rightmost digit of the number, for example, $39$, multiply it by 9, and finally subtract the digit from the rest of the digits put in order to the left of that rightmost digit.  The final result has to be divisible by $13$.  So, in $39$, $9*9=81-3=78$, and $78$ is divisible by $13$.  I think this is it.  Am I on the right digit track?  I just want some answers with words that will satisfy me.
5610
What will be the ones digit of the remainder in: $$\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}$$
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 ?
105719
I wish to explain to a grade 5 kid, the answer to the question - 'When can we know that the observing the last digit of a number will not be sufficient to determine its divisibility by a given N?'  For example - It doesn't work for 3, because we get all sorts of digits, i.e. from 0 to 9, at the units place. The same reason works for 7 and 9.   But what would be a convincing argument for 4 and 8? Even though the last digits form a pattern - 0, 2, 4, 6 and 8 - but we're still not sure about its divisibility.
106487
In many places it is said that the last digits of the powers of the numbers from 1 to 9 have certain cycles. For example the last digits of powers of 2 repeat in a cycle of $4, 8, 6, 2$, and the last digits of powers of 9 repeat in a cycle of $1, 9$.  It seems like this works for bigger numbers as well. The last digits of any number's powers seem follow the cycle of the number's last digit's cycle. For example, the cycle of the last digits of powers of 7 is $9, 3, 1, 7$, and the cycle of the last digits of powers of 1097 are $9, 3, 1, 7$. I've been experimenting with my calculator and I haven't found a single counterexample, so my guess is that it's true for all numbers. That is, the last digits of powers of a number $n$ follow the same cycle as the last digits of the number $n$'s last digit's powers. Could someone show me a proof of this?
124256
Why are the ten digits in base 10 noted $0, 1, 2 ,3 ,4 ,5 ,6 ,7 ,8 ,9$ ? I heard it has to do with angles, but since I can't see how $9$ has $9$ angles, this may not be the historical reason ?
79525
-> 683  -> 678  ???  (Doesn't really bother me too much but figured I'd report the bug...)
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?
116853
There is a number with 6 different digits, if we pick the last digit of that number and place before that number we got $5$ times our number.  How to find such a number?
99761
How many numbers of 6 digits, that can be formed with digits 2,3,9. And also divided by 3?  $$$$ I was trying to to add 2^6 (when there is no 2)+ C(6,2)2^2 (when 2 can be in two places)+C(6,3)2^3 (when 2 can be in three places)...
15969
Is there any way I can search based on the last three digits of a file name?  I have a file with many invoices that are easy to reference with the last three numbers. Is there any way I can search something like *765 to search for ends with 765  Another example.. sometimes files are prefixed with a 0, sometimes not.. 093241251 vs 93241251 and if I just search 93241251 it will not find the 0 prefix version
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