Question 317488
What is the sum of the digits of the smallest positive integer that is divisible by 2, 3, 4, 6, and 7? 
<pre><b>
To be divisible by 2 it must have factor 2

So write down what we have so far, which is only a 2 factor: 

<font size = 5>2</font>

To be divisible by 3 it must have factor 3 

So write *3 by the 2. So far we have:

<font size = 5>2*3</font>

To be divisible by 4=2*2 it must have factor 2 twice. It already has factor 2
once so we just need to write *2 beside 2*3 to have the factor 2 twice:

So we write *2 by the 2*3, and so far we have:

<font size = 5>2*3*2</font> 

To be divisible by 6=2*3 it must have factors 2 and 3. It already has those
once so we don't need to write anything else and so far we we still have:

<font size = 5>2*3*2</font>

To be divisible by 7 it must have factor 7 

So write *7 by the 2*3*2.  And we end up with

<font size = 5>2*3*2*7 = 84</font>

The sum of the digits of 84 is 8+4 or 12, choice D)

Edwin</pre>