I cant figure out the multiples of 6,7,8,and 12
Are you just trying to find the set of all numbers that 6,7,8, and 12 all divide into without a remainder?
Start with 12. Anything that is divisible by will also be divisible by 6. So we have:
$\displaystyle 12n$
Now, consider 8. 8 will go into even multiples of 12. So we have:
$\displaystyle 24n$
Now, consider 7. 7 only goes into seven multiples of 12 (7*12, 14*12, 21*12, etc.). 7*12 = 84, but 8 only goes into even multiples of 12. So we have:
$\displaystyle 168n$