2014000+x represents the 7-digit number.
To be divisible by 4, 5 and 6 we can try to solve:
2014000+x=4a=5b=6c where a, b and c are integers.
4 and 5 divide into 2014000, but 6 divides into it with a remainder of 4.
If 4 and 5 divide into a number, then the number is also divisible by 4×5=20.
So we can forget about 2014000 and just consider x, which must be divisible by 20, so x=20m where m is an integer; and 4+x must be divisible by 6.
We can write 4+x=6n where n is an integer, or, in other words, x=6n-4.
So we have two equations for x: x=20m and x=6n-4, therefore:
20m=6n-4 or 6n=20m+4, or 3n=10m+2, so n=(10m+2)/3.
We want 10m+2 to be divisible by 3, so the smallest positive value we can choose for m is 1, making n=12/3=4. The next smallest value would be 4 (n=42/3=14), then 7 (n=72/3=24) and so on.
Sticking with m=1 and n=4, we can work out x=20m=20.
The smallest 7-digit number is therefore 2014020, divisible by 4, 5, and 6.
But we need the largest value of x and x<1000.
m=1, 4, 7, ..., 3y+1 where y is an integer. The largest value of m is when 20m<1000, that is, 60y+20<1000, 3y+1<50, 3y<49, y<16⅓. So y=16 and m=3×16+1=49. x=20m=980 and the 7-digit number is 2014980, divisible by 4, 5 and 6.