SOLUTION: I have work for this question but i couldn't get the proper answer, please help me solve it. Multiply 11321*12 base 4

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Question 979038: I have work for this question but i couldn't get the proper answer, please help me solve it.
Multiply 11321*12 base 4

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
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%2B1=1
0%2B2=2
0%2B3=3
1%2B1=2
1%2B2=3
0%2A1=0
0%2A2=0
0%2A3=0
1%2A1=1
1%2A2=2
1%2A3=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%2B3=10=four
2%2B2=10=four
2%2B3=11=five
3%2B3=12=six
2%2A2=10=four
2%2A3=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%2A12 you start by multiplying 11321%2A2=23302 .
(It turns out that in base 4, 11321%2A2=23302 ) .
Then you multiply 11321 times the 1 from 12 .
(It is 11321%2A1=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%2B3=12 for the fourth digit from the right,
you still have to add the 1 carried over to get 12%2B1=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%2B1=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%2A2=23302
The far right digit (the ones digit) is 1%2A2=2 .
The digit to the left of that comes from 2%2A2=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%2A2=12=six ,
but you also have to add the 1 carried over,
so you have 12%2B1=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%2A2=2 ,
but you also have to add the 1 carried over,
so you have 2%2B1=3 ,
so you write the 3 as the fourth digit from the right.
The fifth digit from the right) comes from 1%2A2=2 ,
and since you have nothing carried over,
you just write the 3 as the fifth digit from the right.