Question 979038
Base 4 arithmetic is easier than base 10 arithmetic,
because there are less "math facts" that your elementary teachers would make you memorize.
Some are just like for base 10 arithmetic:
{{{0+1=1}}}
{{{0+2=2}}}
{{{0+3=3}}}
{{{1+1=2}}}
{{{1+2=3}}}
{{{0*1=0}}}
{{{0*2=0}}}
{{{0*3=0}}}
{{{1*1=1}}}
{{{1*2=2}}}
{{{1*3=3}}}
Some other "math facts" look weird, different.
However, you do not need to memorize them,
because you have already memorized more than enough.
Just remember that
when you would write a number in base 10 with a digit other that 0, 1, 2, or 3,
the same number would look different in base 4.
The number after 3 is written as 4 in base 10,
but you write it as 10 in base 4.
The number after that is written at 5 in base 10,
but you write it as 11 in base 4.
And so on, so the counting numbers in base 4 are:
1, 2, 3, 10, 11, 12, 13, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, ...
So the strange math facts are:
{{{1+3=10=four}}}
{{{2+2=10=four}}}
{{{2+3=11=five}}}
{{{3+3=12=six}}}
{{{2*2=10=four}}}
{{{2*3=12=six}}} .
Other than that, you use the same algorithms (procedures) for adding and multiplying.
You work with columns and "carry the 1" the same way, for example.
To multiply {{{11321*12}}} you start by multiplying {{{11321*2=23302}}} .
(It turns out that in base 4, {{{11321*2=23302}}} ) .
Then you multiply {{{11321}}} times the {{{1}}} from {{{12}}} .
(It is {{{11321*1=11321}}} , as it would be for base 10).
You write that with a {{{0}}} at the end,
as {{{113210}}} ,
right below {{{23302}}} ,
and you add them, carrying the 1's as needed.
. 23302
+113210
--------
 203112 , because


when you add 3+2=11 for the third digit from the right,
you write 1 and carry the other 1;


when you add {{{3+3=12}}} for the fourth digit from the right,
you still have to add the 1 carried over to get {{{12+1=13}}} ,
and you have to write 3 as the fourth digit from the right,
and carry the 1,

and for the fifth digit from the right,
{{{2+1=3}}} ,
but when you add the 1 carried over {{{3=1=10}}} ,
and you have to write the {{{0}}} as the fifth digit from the right,
and carry the 1.


It turns out that in base 4, {{{11321*2=23302}}}
The far right digit (the ones digit) is {{{1*2=2}}} .
The digit to the left of that comes from {{{2*2=10}}},
so that is a {{{0}}} , and you "carry the 1".
The digit to the left of that (the third digit from the right) comes from {{{3*2=12=six}}} ,
but you also have to add the {{{1}}} carried over,
so you have {{{12+1=13}}} ,
so you write the {{{3}}}} as the third digit from the right,
and "carry the 1".
The fourth digit from the right) comes from {{{1*2=2}}} ,
but you also have to add the {{{1}}} carried over,
so you have {{{2+1=3}}} ,
so you write the {{{3}}} as the fourth digit from the right.
The fifth digit from the right) comes from {{{1*2=2}}} ,
and since you have nothing carried over,
you just write the {{{3}}} as the fifth digit from the right.