On how to do a division in a different base. I was wondering if there was a quick way to compute multiplication and division in a base different from base $10$.
Say for example we are in base $12$ then $3*16=40$ the way I do this is by noticing that in base $10$ we have that $3*16=48$ and that $48= 4*12$ so in base twelve $4*16 = 48 - 4*2 = 40$ where $2$ is given by $12-10$.
Is this the fastest approach? 
Also, in the same way, I can notice that when $12*4<ab<12*6$ (where $a,b$ belong to the number sistem in base 12) the last digit of the number in base $12$ will be the same as that in base $10$.
What about division? in particular is there a quick way to figure out the last digit of the result of a division by two numbers?

Faster approach is digit by digit.  Just you would in base $10$.
$3*16$.  You multiply the the $3*6$.  As six is half of 12 (just like 5 is half of ten) $3*6 = 12 + 6 = 16$.  We write the $6$ down and carry the one.  $3*1 = 3$ and we add the one we carried.  So we get $3*16 = 46$.
Notice: $3_{10}*16_{10}=48_{10} = 40_{12} \ne 3_{12}*16_{12} = 46_{12}=54_{10}=3_{10}*18_{10}$.
Keep in mind $16_{12} = 18_{10}$.
