SOLUTION: Find greatest number of six digits which on being divided by 6,7,8,9,10 leaves 4,5,6,7,8 as remainders respectively.

Let the required number be N

then there exist positive integers A,B,C,D,E such that

N = 6A+4 = 7B+5 = 8C+6 = 9D+7 = 10E+8

Add 2 to each one:

N+2 = 6A+6 = 7B+7 = 8C+8 = 9D+9 = 10E+10

N+2 = 6(A+1) = 7(B+1) = 8(C+1) = 9(D+1) = 10(E+1)

So N+2 must be divisible by 6,7,8,9, and 10

Therefore N+2 must be a multiple of the least common multiple
of 6,7,8,9,10

6=2×3
7=7
8=2×2×2
9=3×3
10=2×5

So the least common multiple of those is 2×2×2×3×3×5×7 = 2520

So N+2 must be a multiple of 2520

Therefore 

N+2 = 2520k

N = 2520k-2

N is a 6 digit number, which means

100000 <= N <= 999999

100000 ≦ 2520k-2 ≦ 999999

100002 ≦ 2520k ≦ 1000001

We divide through by 2520

39.68333333... ≦ k ≦ 396.8257937...

So  40 ≦ k ≦ 396

Since we want N to be large as possible,

we take k = 396

So N = 2520(396)-2 = 997918 

Answer: 997918

Edwin