Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$ Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$
 A: You have $x\equiv 0\bmod 7, 2\bmod 4, 1 \bmod 3$.
By the Chinese Remainder Theorem there is a unique solution modulo $7 \times 4 \times 3=84$. Note that this works because $7, 4 , 3$ are mutually coprime.
We can immediately see that there will be $\lfloor \frac {999}{84}\rfloor= 11$ or $\lceil \frac {999}{84}\rceil=12$ such integers, and this will depend on whether the lowest positive solution is greater than $999-11\times 84 = 75$.
Now we can solve this in various ways, including the full machinery of CRT for the general case. Here we know we have a multiple of $7$ and from the second equation we know it must be even, hence a multiple of $14$.
Since $14 \equiv 2 \bmod 4$ we know that we need $14+28k$ to get the right answer $\bmod 4$. Testing the options give $70$ as the least positive solution.
I've left some gaps for you to fill in, since this will help you understand the way of working so that you can solve other similar problems.
A: We want to find integers $x$ that satisfy$$\begin{align}x&=7k\\x&=4l+2\\x&=3m+1\end{align}$$
Substitute $x=7k$ into the second equation and obtain
$$7k = 4l+2\\\implies 8k-k=4l+2\\\implies -k=4\heartsuit+2\\\implies k = 4n+2 $$
So we have $k=4n+2$, where $n$ is an integer. Then
$$x=7(4n+2) = 28n+14$$
For $x$ to satisfy the last equation, we must have
$$28n+14=3m+1\\\implies n+27n+14=3m+1\\\implies n = 3\spadesuit -1\\\implies n = 3s+2$$
This yields
$$x=28(3s+2)+14 = 84s+70$$
All in all, you want to find the integers $x=84s+70$ such that
$$\color{blue}{0\lt 84s+70\lt 1000}$$
A: I assume you are not familiar with modular arithmetic. (If you want to dig deeper, you should try to look into that).
So, it goes like this:
Let's set $x_1 = 7k$ for some $k$. The first constraint is fulfilled. We want to find $k$, such that the second constraint is also fulfilled. You'll find, that $k=2$ works, because $x_1 = 7\cdot 2 = 4\cdot 3 + 2$. Let's set $x_2 = 7\cdot 2 + 7\cdot 4\cdot n = 14 + 28n$ for some $n$. $x_2$ fulfills the first and second constraint. If you set $n = 2$, then $x_2 = 14+28\cdot 2 = 3\cdot 23 + 1$, hence $14+28\cdot 2 = 70$ fulfills all three constraints, and so do $70 + 3\cdot 4\cdot 7\cdot y = 70 + 84\cdot y$ for all $y$. Now choose $y=1,2,\dots$ to get all solutions.
