5x ≡ 3 (mod 12). The question asked me to find the smallest positive non zero integers that can be plugged in the place of x. I only know how to find the answer by making a list.  5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 0, 5, 10, 3.  It took 12 for 3 to get back to another 3. So the answer is 3, 15(=3+12), 27(=3+12+12). But is there any shortcut or formula to get them more easily?
41147
Problem :   Solve in $\mathbb N$ :   $8n+9\equiv 0\pmod{1163}$  The mathematics give me : $n=435+1163k$ , $k\in\mathbb N$   $1163$ is a prime number   $8n\equiv -9\pmod{1163}$   $8n\equiv 1154\pmod{1163}$   I don't know any ideas to complete my work ?   I have to see your hints
50226
How many integer solutions of $x_1 + x_2 + x_3 + x_4 = 28$ are there with:  (a) $0 ≤ x_i ≤ 12$?  (b) $−10 ≤ x_i ≤ 20$?  (c) $0 ≤ x_i, x_1 ≤ 6, x_2 ≤ 10, x_3 ≤ 15, x_4 ≤ 21$?  \ \  My attempts (I was really only able to solve part (a) but I would appreciate if you could check my work for this one as well):  (a) First I took the total number of integer solutions with no restrictions: There would be ${28+3\choose 3}$ = ${31\choose 3}$ total solutions with no restrictions.   Then I used the inclusion exclusion principle and knew I would have to subtract [P(A) + P(B) + P(C) + P(D)] where A --> $x_1$ is greater than or equal to 13, ..., D--> $x_4$ is greater than or equal to 13. The probabilities of all these cases would equal ${15+3\choose 3}$ = ${18\choose 3}$ because once we assign 13 to one of the $x_i$ there would be $28 - 13 = 15$ remaining to distribute among the $x_i$ values. We would also need to choose which of the $x_i$ to give the 13 to so we would multiply the ${18\choose 3}$ by ${4\choose 1}$ = ${4\choose 1}{18\choose 3}$  We would then have to add back, since we "oversubtracted", the case in which there is an intersection between two of the $x_i$ or in other words two of the $x_i$ receive 13 or more. P$(A\bigcap B)$ = $28-2(13) = 2$ remaining to distribute among the $x_i$ values. So we would also have to choose which of the $x_i$ to give the 13 to so we would have ${4\choose 2}$ ways to choose them. So there would be ${4\choose 2}{2+3\choose 3}$ = ${4\choose 2}{5\choose 3}$ solutions in this case.   Thus the number of solutions would be: ${31\choose 3} - {4\choose 1}{18\choose 3} + {4\choose 2}{5\choose 3}$.  I'm not sure if this is correct and I'm really stuck on b and c so I would appreciate any help or feedback. Thank you!
91566
I recently encountered this on a practice test  Find the smallest positive integer $n$ such that there exists an integer $m$ satisfying $0.33 &lt; \frac{m}{n} &lt; \frac{1}{3}$  My answer:  $$\begin{align}0.33 &lt; \frac{m}{n} &lt; \frac{1}{3}&amp;\implies(\frac{33}{100} &lt; \frac{m}{n})\land(\frac{m}{n}&lt;\frac{1}{3}) \\ &amp;\implies(33n&lt;100m)\land(3m&lt;n) \\\end{align}$$ and thus $n=3m+1$. So $$\begin{align}33n&lt;100m&amp;\implies33(3m+1)&lt;100m \\ &amp;\implies 99m+33&lt;100m\\ &amp;\implies m&gt;33\end{align}$$ Taking $m\geq34$, we find that $n=34\times3+1=103$  After the test, I had the following thoughts, which I do not know how to answer. Questions Will the solution for $n$ always yield the smallest $m$ possible? Does this method (find minimum value of $n$ and substitute) work for all such problems? (where $m,n\in\mathbb{Z}$) Can we generalise $n$ for all possible fraction ranges, and if so, will $n$ always be of a certain form compared to $a,b,c,$ and $d$? (where $\frac{a}{b}&lt;\frac{m}{n}&lt;\frac{c}{d}$).  note: sorry for asking multiple questions in one post (which I know some people frown on) but I feel posting multiple questions with the same 'introduction' (problem+proof) would clutter and be somewhat cumbersome.
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.
89945
Generalize the $3x + 1$ problem as $cx \pm 1$, where $c$ is a positive odd integer and $x$ is a positive integer iterated through the function as far as possible to discover a cycle. If $x$ is even, then you halve it. But if $x$ is odd, you do either $cx + 1$ or $cx - 1$ as the case may be. (If you prefer, $c$ may be negative and you disallow $cx - 1$ for the odd branch; then $|-3x + 1|$ and $3x - 1$ are kind of the same).  With $3x - 1$ and $5x + 1$ it is somewhat well-known that many $x$ don't lead to 1, while with $3x + 1$ the question is unresolved despite intense scrutiny by many professionals and amateurs. For which other $cx \pm 1$ is the question of ultimate arrival at 1 still undetermined despite study by more than a few people? I would appreciate journal articles that look at several different $cx \pm 1$.
31441
How can I find all integers $x≠3$ such that $x−3|x^3−3$?  I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
78226
Use Euclid's algorithm to find a multiplicative inverse of 15 mod 88, hence solve the linear congruence $15x\equiv 20 \pmod{88}$  So far I have:   $88=5\cdot15+13$  $15=1\cdot13+2$  Backwards substitution gives  $15v+18w=1$   $1=15-1\cdot13$   $=15-1(88-5\cdot15)$   $=15\cdot15-1\cdot88$   now very stuck what to do from here:(
69195
Prove that for any set of $2n+3$ integers from the interval $[-2n-1,2n+1]$ there is a triple $(x,y,z)$ such that $x+y+z=0$.    Example  : Choose $5$ number from $\{-3,-2,-1,0,1,2,3\}$ there are $x,y,z$ with $x+y+z=0$.
21541
For any nonzero integers $ a $ and $ b $, there exist integers $ s $ and $ t $ such that $ \gcd(a,b) = as + bt $. Moreover, $ \gcd(a,b) $ is the smallest positive integer of the form $ as + bt $.  I know of one proof of this question, in which we consider $$ S = \{ am + bn ~|~ \text{$ m,n \in \mathbb{Z} $ and $ am + bn &gt; 0 $} \}, $$ but I didn’t manage to get the proof. Can someone please explain this theorem?
58186
Help me understand problems of this type a bit more intuitively.   The solution $C(3+11−1,11)$ seems simple enough, but I got stuck thinking about how many integers you are choosing from within $x_1$, $x_2$, and $x_3$ and I was trying to get an $n$ - as in $C(n+r-1,r)$ - this way. Also, I though the $r$ was 3 since you have $x_1$, $x_2$, $x_3$ choices. Basically, I went wrong at every step. Help!
37981
I want to solve for:     Number of integers solutions to the equation $x_1+x_2+x_3=15$ with $x_1,x_2\leq 5$ and $x_3\leq 7$ for non negative integers $x_1,x_2,x_3$   How can this be done using the principle of inclusion/exclusion?
64693
I actually got this part and I got $16$ choose $2$, which would be $120$. The part I didn't get which wouldn't fit into the title was in how many of these solutions is $x_1 \geq 1$, $x_2 \geq 2$,and $x_3 \geq 3$? I'm not very good at combinatorics so I don't really know. I think the way to approach it would be saying how many solutions where $x &lt; 1 +$ solutions where $x &lt; 2 +$ solutions where $x &lt; 3$ and subtract that from $120$, but I don't know how to do that.
105135
Find the order of $8/9, 14/5,48/28$ in the additive group of $\mathbb{Q}/\mathbb{Z}$    JUNK  Guessing asking for $8/9 * k = 0 $ (being a whole integer) for smallest integer so for this one is 9 . another guess would be the smallest element of $Q/Z$ being something like $9/8$  $14/5$ would be 5  $48/28=12/7$ would be 7?
110629
$$ \begin{cases} x\equiv 2 \pmod{3}\\ x\equiv 1 \pmod{4}\\ x\equiv 3 \pmod{5} \end{cases} $$  $ n_1=3\\ n_2=4\\ n_3=5\\ N=n_1 * n_2 * n_3 =60\\ m_1 = 60/3 = 20\\ m_2 = 60/4 = 15\\ m_3 = 60/5 = 12\\ gcd(20,3)=1=-20*1+3*7\\ gcd(15,4)=1=-15*1+4*4\\ gcd(12,5)=1=12*3-5*7\\ x=-20*2-15*1+12*3\equiv -19 \equiv 41 \pmod{60}\\ $  The above is what I've tried so far. Can someone tell me where I'm going wrong? It's my first time doing this and looking at the explanation and examples for the Chinese Remainder Theorem makes my head want to explode.
3760
How many ways can I write a positive integer $n$ as a sum of $k$ nonnegative integers up to commutativity?  For example, I can write $4$ as $0+0+4$, $0+1+3$, $0+2+2$, and $1+1+2$.    I know how to find the number of noncommutative ways to form the sum: Imagine a line of $n+k-1$ positions, where each position can contain either a cat or a divider. If you have $n$ (nameless) cats and $k-1$ dividers, you can split the cats in to $k$ groups by choosing positions for the dividers: $\binom{n+k-1}{k-1}$. The size of each group of cats corresponds to one of the nonnegative integers in the sum.
39993
I need to find the number of possibilities for which the following equation exists:  $$x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70$$  Each variable is a non-negative integer.  I tried simplifying the question to the point of finding the number of possible solutions to each equation seperatly:  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 70,$$ which is ${79}\choose{10}$.  $$x_1 + x_2 + x_3 \cdots + x_{10} = 69,$$ which is ${78}\choose{10}$.  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 68,$$ which is ${77}\choose{10}$, and so forth.  Then I thought about adding all of these together. This seems a bit too much work considering that the textbook solution is: ${80}\choose{10}$, and I can't seem to figure out the idiomatic way of approaching these kind of questions.   Any help is greatly appreciated.
115622
How many nonnegative integer solutions are there to the pair of equations  \begin{align}x_1+x_2+\dots +x_6&amp;=20  \\ x_1+x_2+x_3&amp;=7\end{align}  How do you find  non-negative integer solutions?