SOLUTION: find the greatest 4-digit number which when divided by 6,12,18,24 and 30 leaves5,11,17,23 and 29 as remainder respectively.

Theorem:

Suppose k and n are positive integers.

Then when kn-1 is divided by n the quotient is k-1 and the remainder is n-1.

Proof:
 
        quotient
divisor)dividend
        quotient×divisor
               remainder 

Or:

dividend - (quotient)×(divisor) = remainder
             (kn-1) - (k-1)×(n) = n-1 
     

Therefore 1 less than any common multiple of 6,12,18,24 and 30 will leave
a remainder of 1 less than those, which are 5,11,17,23 and 29 respectively.  

The least common multiple of 6,12,18,24 and 30 is 360, so 1 less or 359 
would be an answer if we didn't have the requirement that it be the greatest
4-digit answer.

However ANY multiple of 360 is also a multiple of 6,12,18,24 and 30. So
we need only find the greatest 4-digit multiple of 360 and subtract 1.

We divide the greatest 4 digit number 9999 by 360:

      27
360)9999
    720
    2799
    2520
     279

So 27×360 = 9720 is the greatest 4-digit multiple of 6,12,18,24 and 30, and
if we subtract 1, we get the greatest 4-digit number which leaves remainder
5,11,17,23 and 29 respectively when divided by 6,12,18,24 and 30, respectively.

Answer: 9720-1 = 9719

Edwin