Integral solution to an equation For what largest value of $X$ would $$(4X-1)/5 ,(4X-2)/6 ,(4X-3)/7 $$give integers such that $$4x+2$$ is a four digit number ?
My approach:
In somewhat convoluted way ,let a=4x-1 etc or a is a number which leaves remainder 1 when divided by 4, b leaves remainder 3 when divided by 5 , c leaves 5 when divided by 7.
Lcm of 4,5,7 should be multiplied by largest possible number till we get a four digit number, then subtract 2 from it and we are done.
But obviously, its more like hit and trail and not mathematics.
Any other better solution will be appreciated. 
 A: The problem is solved using the Chinese Remainder Theorem (CRT), but since that requires coprime moduli, we have to begin by replacing $4x$ with $y$ and later looking at which solutions for $y$ are multiples of $4$.
First, let's restate the divisibility properties in terms of modular arithmetic: $$5\mid (4x-1) \Rightarrow y\equiv 1 \bmod 5$$
$$6\mid (4x-2) \Rightarrow y\equiv 2 \bmod 6$$
$$7\mid (4x-3) \Rightarrow y\equiv 3 \bmod 7$$
By CRT, there is one number smaller than $5\cdot 6\cdot 7=210$ that satisfies those conditions, and that number is $206$. We can find that number by first considering numbers which are $\equiv 1 \bmod 5$. Those are $6,11,16,21,26,\dots$. Note that they occur every five apart. Next identify the members of that list that are $\equiv 2 \bmod 6$. Those are $26,56,86,116,\dots$. They occur every $5\cdot 6=30$ apart. Finally from that list identify the number that is $\equiv 3 \bmod 7$. It is $206$. Numbers that meet all three conditions will be spaced $5\cdot 6\cdot 7=210$ apart. They will have the form $y=206+210k$.
Neither $206$ nor $210$ is divisible by $4$. But the sum $y=206+210k$ will be divisible by $4$ when $k$ is odd. So we can say that $4x=416+420n$. You wish to identify the largest occurrence of this number less than $9999$.
$416+420n<9999 \Rightarrow n<\frac{9999-416}{420}=22.82$. So for integer value, $n=22$, $4x=9656$, $x=2414$, and $4x+2=9658$.
