3.5 Division of Whole Numbers

With physcial blocks, or even sets, division can be thought of as removing some fixed number of elements several times and counting the remainder. This problem is asking to start with $10$ and removed groups of $2$ until we're finished. Lets use

$$\{a,b,c,d,e,f,g,h,i,j\}$$

to represent 10. Now we'll removed $2$ at a time.

$$\{a,b\}\quad \{c,d,e,f,g,h,i,j\}$$

After our first subtraction we have $2+8=10$ or $10-2=8$. Repeating this process

$$\{a,b\},\quad \{c,d\},\quad \{e,f\}, \quad \{g,h\},\quad \{i,j\}$$

we find there are 5 sets containing to elements.

We end up with $10-5(2)=0$, which means $5(2)=10$

1. We're going to start by subtracting $10$'s from $111$ until we get as close to zero as possible.

   subtraction	process	remainder
   	111
   1	111-10(1)	101
   2	111-10(2)	91
   3	111-10(3)	81
   $$\vdots$$	$$\vdots$$	$$\vdots$$
   11	111-10(11)	1

   After 11 subtractions, we can't take anymore. The final row of our table is our answer, it just needs to be rearranged.

   \[ \begin{align*} 111-10(11)&=1\\ 111&=10(11)+1\\ a&=b(q)+r\\ \end{align*} \]

   Our quotient is $11$ and our remainder is $1$. $10$ goes into $111$ elevent times with a remainder of 1.

2. A quick count of how many 3's are in 17 $$3,6,9,12,15,18$$ There are $5$, $6$ would be too many. $5$ threes gets us up to $15$ and we would need $2$ ones to get to $17$. We have

   $$17=3(5)+2$$

   Which means our quotient is $5$ and the our remainder is $2$

3. $23=4(5)+3$, so $q=5, r=3$.

The division algorithm tells us that if $123$ is divided by some number and give a remainder of $13$, then

$$ 123=bq+13 \Rightarrow 110=bq $$

$b$ and $q$ are being multiplied together to get 110.So $b$ must divide 110.

Divisors of 110 are $1,2,5,10,11,22,55,110$.

But not all of these values will work. For example,

$$123=1(2)+121$$

is a valid equation, but it doesn't line up with the definition of our divion algorithm. Namely, that remainder is way too big. That equation implies we divided by $1$, repeated our subtraction $2$ times, and had $121$ left over. We could have taken so many more $1$'s out of that dividend!

The condition we're missing is that the remainder must be smaller than the divisor. Or another way to think about it is that the divisor is larger than the remainder. We know the remainder is $13$ from the given problem. Of the divisors that are larger than 13, we know

$$ b\in{22,55,110} $$

These give us 3 valid solutions.

$$123=22(5)+13$$ $$123=55(2)+13$$ $$123=110(1)+13$$

Let's walk this one slowly using the same scaffold process.

Set up the division bracket:

Now we ask: how many groups of $7$ can we remove from the first usable chunk of the dividend? Since $2<7$, that answer is none. So we break that $2$ and instead we look at $23$. We can take $3$ groups of $7$ (because $3(7)=21$, but $4(7)=28$ is too large).

We are not done yet, so we bring down the $4$ to make $24$.

Now ask again: how many $7$'s are in $24$? That is $3$, since $3(7)=21$ and $4(7)=28$ is too big.

So the final result is

$$ 234\div7=33\text{ remainder }3. $$

Quick check with the division algorithm form:

$$ 234=7(33)+3 $$

and the remainder condition is satisfied since $0\le 3<7$.

We're start by setting up the saffold. I'll exclude the subscripts from the scaffolding for ease of reading, just remember everything here is in base 5.

Our first question is, how many $3_{\text{five}}$'s can be taken from $2_{\text{five}}$, but there are none. So we break and move to the next position and ask again. How many $3_{\text{five}}$'s can be taken from $21_{\text{five}}$? This is where our multiplication table comes in handy, go to the row that corresponds with $3$, and go through that row trying to get as close to $21$ but not going over. We find $3_{\text{five}}\times 3_{\text{five}} =14_{\text{five}}$. That means the closest we can get to $21_{\text{five}}$ is $14_{\text{five}}$ and that partial-quotient is $3_{\text{five}}$.

Now we break the $2_{\text{five}}$ we're left with by bringing down the $3_{\text{five}}$ and we go back to the table. In that $3_{\text{five}}$ row, how close can we get to $23_{\text{five}}$ without going over?

We do our final break and bring down that $4_{\text{five}}$ to finish up.

Back at the multiplication table we find $$3_{\text{five}}\times 3_{\text{five}}=14_{\text{five}}$$

And we're done!

$$2134_{\text{five}} \div 3_{\text{five}} = 343_{\text{five}}$$

We'll follow the exact same rhythm as the base-5 example: table first, then scaffold.

Here is the base-2 multiplication table:

For this problem, we mostly use the fact that multiplying by $1_2$ keeps the value the same, and multiplying by $0_2$ gives $0_2$.

Now set up the scaffold (again, I'll omit repeated base-two subscripts in the scaffold itself for readability):

The divisor is $101_2$, so we look for the first chunk big enough to divide by $101_2$. Since $11_2<101_2$, we start with $110_2$.

From the table, $101_2\times 1_2=101_2$, so the first quotient digit is $1_2$.

Bring down the next digit (a $1_2$) to get $11_2$. Since $11_2<101_2$, the next quotient digit is $0_2$.

Bring down the next digit (another $1_2$) to get $111_2$. Now it fits once: $101_2\times 1_2=101_2$.

Bring down the final digit ($0_2$) to get $100_2$. This is still less than $101_2$, so the last quotient digit is $0_2$.

So,

$$ 110110_{\text{two}}\div101_{\text{two}}=1010_{\text{two}}\text{ remainder }100_{\text{two}}. $$