4.5 Multiplication and Division in Base Systems

Step 1: Let’s apply the process demonstrated and outlined above to find the base 7 multiplication table. It should be clear that 0 multiplied by anything, regardless of base, will give 0, and that 1 multiplied by anything, regardless of base, will be the value of “anything.” So, we start with the table below:

Step 2: Notice $2\times 2=4$ is there. But we didn’t hit a problematic number there (4 works fine in both base 10 and base 6). The same is true for $2\times 3$ and $3\times 2$, which equal 6. But what is $2\times 4$? If we use the repeated addition concept, $2\times 4=2+2+2+2=6+2$. According to the base 7 addition table in the solution for Example 4.29, $6+2=11$. So, we add that to our table:

Step 3: Next, we need to fill in $2\times 5$. Using repeated addition, $2\times 5=2+2+2+2+2=11+2=13$ if we use our base 7 addition rules. So, we add that to our table:

Step 4: Finally, $2\times 6=2+2+2+2+2+2+2=13+2=15$. And so we add that to our table:

Step 5: A similar analysis will give us the remainder of the entries. Here is $4_7\times 5_7$ demonstrated:

This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that $4_7+4_7={11}_7$ and then applying the addition processes we’ve always known, but with the base 7 table in the solution for Example 4.29. Using those addition rules, the rest of the table is given below:

Let’s apply the repeated addition to base 12. Here is $7_{12}\times 9_{12}$ demonstrated:

This is done by using the addition rules from Addition and Subtraction in Base Systems, namely that $7_{12}+7_{12}={12}_{12}$ and then applying the addition processes we’ve always known, but with the base 12 table in the solution for Example 4.31. Using those addition rules, the rest of the table is given below:

1. Step 1: Use the base 6 multiplication table (Table 4.8) and, when necessary, the base 6 addition table (Table 4.4).

   Set up this calculation using columns:

   	4	5
   x	2	4

   Step 2: Multiply the 1s digits, 5 and 4, using the base 6 multiplication table (Table 4.8). There we see the result is 32₆. So, we enter the 2 and carry the 3.

   	3
   	4	5
   x	2	4
   		2

   Step 3: So, now we multiply the 4 and the 4, then add the 3 (just as you would do if multiplying two base 10 numbers!). $4_6\times 4_6={24}_6$ (from the base 6 table [Table 4.8]), then ${24}_6+3_6={31}_6$. So, we enter the 31.

   	3
   	4	5
   x	2	4
   3	1	2

   Step 4: Now we move on to the 2 in the “tens” place in the bottom value. We multiply the 2₆ and the 5₆, and we get 14₆. So, we enter the 4 and carry the 1.

   Step 5: Next up, we multiply the 2 and the 4, and then add 1. This gives us ${12}_6+1_6={13}_6$. We enter those on that second line.

   		1
   		4	5
   	x	2	4
   	3	1	2
   1	3	4	0

   Step 6: Now we add down the columns.

   		1
   		4	5
   	x	2	4
   1	3	1	2
   1	3	4	0
   2	0	5	2

   Step 7: The 3 and the 3 add to 10 in base 6, so we enter the 0 and carry the 1. We now have the result:

   ${45}_6\times {24}_6={2052}_6$.
   
2. Step 1: Use the base 2 multiplication table (Table 4.10) and, when necessary, the base 2 addition table in the solution for Example 4.29. Set up this calculation using columns:

   	1	0	1
   x	1	1	0

   Step 2: Using the pattern established above, and the processes from multiplication from base 10, we find the following:

   			1	0	1
   x			1	1	0
   			0	0	0
   		1	0	1
   	1	0	1

   Step 3: Adding down the columns results in the following:

   			1	0	1
   x			1	1	0
   			0	0	0
   		1	0	1
   	1	0	1
   	1	1	1	1	0

   So, ${101}_2\times {110}_2={11110}_2$.

Step 1: Use the base 12 multiplication table in the solution for Example 4.39 and, when necessary, the base 12 addition table in the solution for Table 4.7. Set up this calculation using columns:

Step 2: First, the 4 is multiplied by 3A, resulting in the first line.

Step 3: Now we move on to the 7 in the “tens” place in the bottom value.

Step 4: Now we add down the columns.

Step 5: The 3 and the A add to 11 in base 12, so we enter the 1 and carry the 1.

We now have the result: $3{\mathrm{A}}_{12}\times {74}_{12}={2414}_{12}$.

1. Looking at the multiplication table for base 6 (Table 4.8), we see that $5_6\times 2_6={14}_6$. Using that, we know that ${14}_6÷5_6=2_6$.
2. Looking at the multiplication table for base 12 in the solution for Example 4.39, we see that $7_{12}\times {\mathrm{A}}_{12}=5{\mathrm{A}}_{12}$. Using that, we know that $5{\mathrm{A}}_{12}÷7_{12}={\mathrm{A}}_{12}$.

Since the problem is in base 6, the symbol set available is 0, 1, 2, 3, 4, and 5. The 8 in the answer is clearly not a legal symbol for base 6. Looking back to the base 6 multiplication table (Table 4.8), we see that $4_6\times 2_6={12}_6$.

If this problem was a base 10 problem, this would be the correct answer. However, in base 12, $8_{12}\times 7_{12}$ is not 56, but is instead 48. To correct this error, carefully use the multiplication table for base 12 (Table 4.9). If properly used, the correct answer would be ${18}_{12}\times 7_{12}=\mathrm{B}{8}_{12}$.