Suppose the solution is N.

In order to leave a remainder of 1 when divided by 4, N must be 1 more
than a multiple of 4. So there exists an integer a such that

N = 4a+1

In order to leave a remainder of 2 when divided by 5, N must be 2 more
than a multiple of 5. So there exists an integer b such that

N = 5b+2

In order to leave a remainder of 3 when divided by 6, N must be 3 more
than a multiple of 6. So there exists an integer c such that

N = 6c+3

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So we have:

 N  =  4a+1  =  5b+2  =  6c+3

Add 3 to each of those

N+3 =  4a+4  =  5b+5  =  6c+6

Factor the gcf out of the last three:

N+3 = 4(a+1) = 5(b+1) = 6(c+1) 

This shows that N+3 must be a multiple of 4,5, and 6.
The smallest multiple of those is their LCM = 60

Therefore N+3 = 60
            N = 57

Checking:

    14       11        9
  4)57     5)57     6)57
    4        5        54 
    17        7        3 
    16        5
     1        2

Answer: 57

Edwin