Answer to March 2012 Problem of the Month

Solution to the Problem:

The answer is 2520.

To be divisible by 2, you need a 2 in the answer.
To be divisible by 3, you need a 3.
To be divisible by 4, you need an additional 2 (You don't need a 4 because there is already a 2 earlier).
To be divisible by 5, you need a 5.
To be divisible by 6, you don't need any additional numbers (there is already a 2 and a 3).
To be divisible by 7, you need a 7.
To be divisible by 8, you need an additional 2.
To be divisible by 9, you need an additional 3.
Now multiply 2 x 3 x 2 x 5 x 7 x 2 x 3 = 2,520.

However, delahoydep@kpmath.com and Dakota Rees sent in the answer of 0, and 0 is divisible by each of the numbers 2, 3, 4, ..., 9!!!

Correctly solved by: