Multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively 
I want the multiple of $7$ that has remainder $1,2,3$ divided by $2,3,4$ respectively.

Now I have had a search of similar questions, but I may have missed it since I am on a phone.
This question should be trivial to me, but I am having trouble, so that is depressing enough, please just give me a hint.

It would appear that I am dealing with:
$$\frac{7x}2\equiv 1\pmod 7$$
$$\frac{7x}{3} \equiv 2\pmod 7 $$
$$\frac{7x}{4} \equiv 3 \pmod 7$$
and then I find inverses for $2,3,4$ and multiply both sides by these, then apply Chinese remainder theorem. But having the different remainders under the same modulus makes no sense, so this leads me to believe I have formulated the question wrong, and chosen the wrong modulus. Thanks
 A: Let $n$ be the number you want, then $n+1$ is a multiple of $2,3,4$ that is congruent to $1\bmod 7$. So a multiple of $12$ that is $1$ mod $7$. since $12$ is $5$ mod $7$ and the inverse of $5\bmod 7$ is $3$ we get $12\cdot3=36$ is a valid solution for $n+1$. thus $35$ is a valid value for $n$
Notice a much more general way to solve congruence systems is via the chinese remainder theorem.
A: Solution:
First recognise that $x \equiv 3 \pmod 4 \implies x \equiv 1 \pmod 2$. The latter criterion is redundant.
So you can simply solve this system:
$$x \equiv 3 \pmod 4$$
$$x \equiv 2 \pmod 3$$
From the second equation, $x = 3s+2$.
Subbing into second equation, $3s+2 \equiv 3 \pmod 4$, giving $3s \equiv 1 \implies s \equiv 3^{-1} = 3 \pmod 4$. Hence $s = 4t + 3$.
So $x = 3(4t+3) + 2 = 12t + 11$ or $x \equiv -1 \pmod{12}$, which is the solution of all three congruences.
Finally, you have to incorporate the divisibility by seven criterion:
$x = 7y \implies 7y \equiv -1 \pmod{12} \implies y \equiv -1.7^{-1} = -7 \equiv 5 \pmod{12}$.
Hence the required solution is $x = 7(12v + 5)$, also expressible as $x \equiv 35 \pmod{84}$. Some possible particular solutions are: $35, 119, 203,..$ etc.
