What is the value of $187^2 \mod 100$?
33620
I was wondering if there is a simpler way to find the last to digits of a power such as $3^{45}$. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result.  Here's my method:  Using the fact that $3^4=81\equiv -19 \pmod{100}$ and $19^5=2,476,099\equiv -1 \pmod{100}$  $$ 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} $$  I just felt like this was sort of a messy way to find out that the last two digits are 43.
97258
Given $n\in\Bbb N$, what is the least $a&gt;1$ with $a^{2^n}\equiv1\bmod2015$?   Is there a solution not using Fermat's little theorem or the Chinese remainder theorem, any ideia?
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?
98817
Let $c$ is complex-number satisfying :  $(c+\frac{1}{c}+1)(c+\frac{1}{c}) = 1$  So, how could i get $(3c^{100}+\frac{2}{c^{100}}+1)(c^{100}+\frac{2}{c^{100}}+3)$ ?
106926
Find the last $2$ digits of $$\large7^{7^{7^{10217}}}$$   So far I have: $17\cdot 601=10217$ and $7=7 \pmod{10}$. Any help greatly appreciated.
101081
I feel the answer should be $nC3 \times 3! \times 10^{n-3}$.
21441
Some one please help me with this question. I am getting -10.5 as the wrong final answer.I found the formula in the following slide.Since the result was coming out in complex number i got its modulus.That is how i ended up with a 10.5 (2 * 5.25).
72373
Without strenuous arithmetic.   Is there a program I can download to do so?  What are the quadratic residues modulo $5^4$ or $5^5$?  Thanks!
30116
What is the calculation way to find out the last $2$ digits of $7^{7^{7^7}}$? WolframAlpha shows  $...43$.
38241
So I'm supposed to express:  $$1.24\overline{123}=1.24\;123\;123\;123\;\ldots$$ as the ratio of two integers.  So I got $$1+\frac{24}{100}+\frac{123}{10^5}+\cdots$$  I don't know if this is correct, and if it is, I also don't know how I would get the common ratio from this. Would the common ratio just be $\frac{1}{10^5}$? Am I doing this wrong?
13240
Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.
58304
Here is some information we got : We know the value of $n$, with size $1043$. We know the value of $p$ (size $20$) and $q$ (size $1023$) as the factors. $e = 65537.$ $\varphi(n)$ = $(q-1)(p-1)$ When I calculated $\gcd$ and $\text{modinv}$, I got : $\gcd(e,\varphi(n)) = 65537$ $modinv(e,\varphi(n)) = 1 $ So we can tell that they are not relatively prime. So, how to compute the d, and get the value of m? I'm not that good with math, so I cant understanding well the theory. so can anyone please make an example implementation or write a clear formula?
11316
How can I solve $x^2 \equiv 19 \pmod {59}$?  I know that we can just try squaring numbers from 1 to 58 , but this is a very slow method, is not their a quicker one?
111922
I came across the following problem and do not know how to tackle it.       Find the last two digits of $ 7^{81} ?$     Can someone point me in the right direction? Thanks in advance for your time.
115222
For example $151x - 294\equiv 44\pmod 7 $. How would I go about solving that?  The answer says to simplify it into the $ax\equiv b\pmod c $ form first, but I have no clue on how to get rid of the $294$... help?
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?
124071
Find the number of integer pairs 0 ≤ a, b ≤ 100 such that $a^{20}$ ≡ $b^{50} \pmod {101}$.  Here is the solution: "Since is prime, there exists a primitive root g in modulo 101. For some integers x and y, with 1 ≤ x, y ≤ 100, let a ≡ $g^x$ and b ≡ $g^y$. Hence, $g^{20x}$ ≡ $g^{50y}$ (mod 101). Therefore, $g^{20x−50y}$ ≡ 1 (mod 101), and 100|20x − 50y, or alternatively, 10|2x − 5y. Since 2|2x − 5y, we have 2|y. Also, 5|2x − 5y, so 5|x. Hence, the congruence holds for all x and y such that x is a multiple of 5 and y is a multiple of 2. There are 20 choices for x and 50 choices for y, so our answer is 20 · 50 = 1000."  In general, I'm very confused on the logical reasoning behind these steps. I'm a beginner in learning primitve roots and perhaps don't have enough background yet.  I'm confused why you can let a ≡ $g^x$ and b ≡ $g^y$.  Wouldn't that mean both $g^{x}$ and $g^{y}$ are both one Modulo 101? Finally, why does 100 have to divide 20x-50y. In fact, shouldn't it be 20x-50y dividing 100? I thought ord(a) of x always has to divide totient of x, and therefore 20x-50y|totient 101 which is 100.