SOLUTION: Find the sum of all 4 digit natural numbers of the form 4AB8 which are divisible by 2,3,4,6,8, and 9

To be divisible by 2,3,4,6,8, and 9
we list all the like factors in
columns to get the least common
multiple of all those: 

2=     2
3=           3
4=     2·2 
6=     2    ·3 
8=     2·2·2 
9=           3·3
----------------
       2·2·2·3·3 = 72 

So we find all the multiples of 72
that are 4-digits, that begin with a
4 and end with an 8.

The smallest feasible number is
4008, which divided by 72 gives 55.66,
a decimal, not a whole number.

The largest feasible number is
4998, which divided by 72 gives 69.42,
also a decimal, not a whole number.

Of course neither of those qualify,
but they tell us the bounds to look
for numbers to multiply by 72. Since 
72 ends in 2, to get a number
ending in 8, we know that a number
ending in 4 will give an answer
ending in 8, since 2·4 = 8.  Also we
know that a number ending in 9 
multiplied by 72 will give an answer
ending in 8 since 2·9 = 18.  

The only two digit numbers in that
range which end in 4 or 9 are
59, 64, and 69.  Multiplying them by
72 gives the answers

72·59 = 4248
72·64 = 4608
72·69 = 4968

Add them up and get 13824

Edwin