Finding largest number with GCD. The question is:
Find the largest number which when divided by 20,25,35 and 40 leaves a remainder of 14,19,29 and 34 respectively.
The solution is to find the difference which in this case is 20-14=6,25-19=6 and similarly for others.
Then finding GCD of all the numbers 20,25,35 and 40 and subtracting 6 from the obtained GCD.
I really did not understand why this solution works, if anyone could explain me the logic behind this in plain english. I am sorry I am weak in mathematics.
 A: If $x$ is the required number, $x+6$ will be divisible by lcm$(20,25,35,40)$
A: First take the LCM of $20,25,35,40$ which is $1400$
Now, Since the $$20-6=14\\25-6=19\\35-6=29\\40-6=34$$So, $1400-6=1394$ will have the remainder of $14,19,29,34$ when divided by $20,25,35,40$
Edit:
Note that if a number is divided by $20$ then it leaves a remainder of $14$ and can be expressed as $20a-6$ or $20a+14$.
Similarly if the number gives remainders of $19,29,34$ when divided by $25,35,40$ respectively it can be expressed as $25b-6$ or $25b+19$ , $35c-6$ or $35c+29$ , $40d-6$ or $40d+34$.
We took $-6$ form because it is common among all the given $4$ divisors.
A: The question is flawed.  There is no largest number like that, as you can add $\operatorname{LCM}(20,25,35,40)=1400$ to any solution and get a larger one.  
The idea only works because you are asked for remainders that are the same amount below the moduli.  If we asked for the smallest number greater than $0$ that was a multiple of $20,25,35,40$ you would answer with $\operatorname{LCM}(20,25,35,40)=1400$.  Now we are asking for a number $6$ less than a multiple of each, so we subtract $6$.  Another way to say the same thing is that $-6$ has the required remainders.  As we said above, you can add $1400$ to any solution and get another, so we add $1400$ to $-6$ and get $1394$.  
This is all based on the Chinese Remainder Theorem.
