Question based on finding the dividend 
The least number which on division by $35$ leaves the remainder 25 and on division by 45 leaves the remainder $35$ and on division by $55$ leaves the remainder $45$ is _______

I understood the difference between divisor and remainder is 10 in all the three cases. 
I want to know why the dividend will be LCM of (35, 45 and 55)-10
This is a gmat exam question.
 A: Hint:
$$
\begin{align}
\operatorname{LCM}(35,45,55)&\equiv0\pmod{35}\\
\operatorname{LCM}(35,45,55)&\equiv0\pmod{45}\\
\operatorname{LCM}(35,45,55)&\equiv0\pmod{55}
\end{align}
$$
and
$$
\begin{align}
x&\equiv25\equiv-10\pmod{35}\\
x&\equiv35\equiv-10\pmod{45}\\
x&\equiv45\equiv-10\pmod{55}
\end{align}
$$
A: Leaving aside the requirement to find the least number, there will be solutions that recur every $\operatorname{LCM}(35,45,55)=k$.  This is because if you have one solution and add $k$ to it, you have added a multiple of each of $35,45,55$.  Then note that $-10$ is a solution if you do not restrict yourself to positive numbers because you can write $-10=-1 \cdot 35 +25$ and similarly for the others.  The next solution up will then be $-10+\operatorname{LCM}(35,45,55)$  
For an example with smaller numbers, you could look for the smallest number that leaves $3$ when divided by $4$ and $5$.  Here we have $\operatorname{LCM}(4,6)=12$ and a base solution $-1$, so the smallest positive solution is $11$  You can let a spreadsheet compute =MOD(n,4) and =MOD(n,6)  for each number from $-1$ to $24$ to see the pattern that the remainders repeat every $12$.
A: Note:
$$X=35m+25=45n+35=55p+45\Rightarrow$$
$$X+10=35(m+1)=45(n+1)=55(p+1)\Rightarrow$$
$$min(m,n,p)\iff LCM\{35,45,55\} \Rightarrow$$
$$X=35(m+1)-10=45(n+1)-10=55(p+1)-10.$$
