Find $x$ given remainder conditions The problem: Find the smallest positive integer $x$ such that
$x$ divided by $4$ has remainder $1$
$x$ divided by $5$ has reminder $2$
$x$ divided by $6$ has remainder $3$
Now, my first idea was to add to each divisor its the remainder and multiply the quantities obtained.
$x=(4+1)(5+2)(6+3)=315$ 
But $315$ does not satisfy all the conditions above, and I don't know how to get the smallest integer that satisfies the conditions. Any help? 
 A: If $x \equiv 1\pmod{4}$, then $$x \equiv 1,5,9,13,17\pmod{20}$$ Of these, only $17$ is also $2 \pmod{5}$. Therefore 
$$x \equiv 17 \pmod{20}$$
Next, the LCM of $20$ and $6$ is $60$.
If $x \equiv 17 \pmod{20}$, then $$x \equiv 17, 37, 57 \pmod{60}$$
Of these, only $57$ is $3\pmod{6}$. This means
$$x \equiv 57\pmod{60}$$
This means $\boxed{57}$ is the smallest $x$ that satisfies your conditions.
A: You may have solved problems of the kind: Find x such that it leaves a remainder 3 when divided by 4, 5 or 6.
In the above scenario you first find out a number which divisible by each of 4, 5 and 6 (lcm or it's multiple) and then you add the remainder 3 to that number. You can find infinitely many such numbers as there are infinite number of multiples of $lcm(4,5,6)$.
Here notice that $4-1 = 3$, $5-2 = 3$ and $6-3 = 3$. So basically we are solving the exactly same problem here, we have just changed the remainder from $3$ to $-3$.
As $lcm(4,5,6) = 60$, such numbers are of form $60n - 3$, for any integer $n$.
