I am supposed to find the last digit of the number $3^{459}$. Wolfram|Alpha gives me $9969099171305981944912884263593843734515811805621702621829350243852275145577745\\3002132202129141323227530694911974823395497057366360402382950449104721755086093\\572099218479513977932448616356300654729978057481366551670706\color{red}{\mathbf{7}}$  Surely there's some sort of numerical trick to doing this. I thought maybe modular arithmetic was involved? Any ideas on how to approach this problem?
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 $7$.
17609
I am trying to find the smallest number if we are given the sum of its digits. Suppose that sum of digits  is 9 then it should be 9 instead of 18,36,63 and similarly if sum of digits is  11 then desired answer is 29 not 92 or any other number bigger than 29.I tried to write sum of all numbers upto 53 and got this but i am not able to come up with a general formula.                         from 1 to 9 it is just 9.FROM 10 to 18 it is 19,29,39,...... and for 19(1+9) it is 199  that is increase by 100. from 19 to 27 it is 199,299,399,499.... and for 28 it is 1999 that is increase by 1000. from 29 to 36 it is 29999,39999,49999,..... for 37 it is 19999 (increse by 10000). for 38 to 45 it is it is 29999,39999,49999,........999999 and for 46 it is 199999 (increased by 100000)
118577
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$ Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.  How to calculate the missing digits? I know that large factorials can be estimated using : $$15! \approx \sqrt{2\pi\cdot 15} \cdot \left(\frac{15}{e}\right)^{15}$$ which is not feasible to calculate by hand. The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11: $1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $ $-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $ The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
46053
I'm not sure if it's correct, but what I have so far is; $$21n^5 + 10n^3 + 14n ≡ (1 + 0 - 1) ≡ 0 \mod 5$$ but I'm having trouble solving it in $\bmod 3$. I have: $$21n^5 + 10n^3 + 14n ≡ (0 + (?) + 2),$$ I'm not sure how to solve the $10n^3$ part of the congruence in $\mod 3$. Also, if I have any mistakes please let me know.
120507
What is the correct answer to this expression:    $26^{32}  \pmod  {12}$    When I tried in Wolfram Alpha the answer is $4$, this is also my answer using Fermat's little theorem, but in a calculator the answer is different, $0.$
108470
I tried remainder theorem but that does not simplify it.   I tried factorizing $1446$ as $2\cdot3\cdot241$ and got remainders when numerator is divided by $2,3$ and $241$ individually but then I did not know what to do next? Stuck!
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.
55292
For how many positive integer $k$ does the ordinary decimal representation of the integer $k\text { ! }$ end in exactly $99$ zeros ?  By inspection I found that $400\text{ !}$ end in exactly $99$ zeros , but $399\text{ !}$ does NOT end in $99$ zeros ; using the formula $$\text{ number of zeros at the end }=\sum_{n=1}^{\infty}\left[\frac{k}{5^n}\right]$$where , $[x]$ denotes the greatest integer part not exceeding $x$. I also found that, for $k=401,402,403,404$ the number of zeros is same, but for $k=405$  the number of zeros increase by $1$ ; as $405$ is divisible by $5$ again , after $400$. Thus I got that there are only $5$ integers satisfying the condition which are $400,401,402,403,404$. The question is possible duplicate of  or  but my question is different from these two questions.  Does there exist any other rule  or easy formula from where I can get how many integers are there
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?
43525
Let $A = 4444^{4444}$;  Then sum of digits of $A = B$; Then sum of digits of $B = C$; Then sum of digits of $C = D$;  Find $D$.  What should be the approach here?
115516
i have looked at/practiced several methods for solving ex: $7^{9999}$. i have looked at techniques using a)modulas/congruence b) binomial theorem c) totient/congruence d) cyclicity.   my actual desire would be a start to finish approach using totient/congruence. i have figured out how to work the individual steps but not how to combine them and in what order to be able to solve any "end digits" questions.
79525
-> 683  -> 678  ???  (Doesn't really bother me too much but figured I'd report the bug...)
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
83643
I need to find the last digit of $103^{103^{103^{103^{103}}}}$ so the value in $\mod10$. I know  \begin{align} 103^{103^{103^{103^{103}}}}&amp;=(100+3)^{103^{103^{103^{103}}}}\\ &amp;=100\cdot(stuff)+3^{103^{103^{103^{103}}}}\\ &amp;=3^{103^{103^{103^{103}}}} \mod10 \end{align}  And then I am stuck. I have observed that the last digit of a power of 3 is cyclic: $$3 \to 9 \to 7 \to 1 \to 3 \to 9 \to 7 \to\cdots$$  This means it depends on $103^{103^{103^{103}}}\mod4$.  Is this the right method? do I just need to keep doing this or is there a simpler "trick"?
108341
How to solve and see resolution of $13^{53} \pmod 7$ using Fermat little Theorem? Using , I know it gives me 6 as an answer to this problem..., but why? How is the resolution? Thanks,
114494
Find  $$\frac{1}{3}+\frac{1}{10}+\frac{1}{21}+\frac{1}{45}+....\infty$$ I tried to rewrite it as $$S=\frac{1}{1\times3}+\frac{1}{2\times5}+\frac{1}{3\times7}+\frac{1}{5\times9}+.....$$  To identify the pattern but i could not go on
49397
How to solve $ 13x \equiv 1 ~ (\text{mod} ~ 17) $?   Please give me some ideas. Thank you.