Division

When we perform division, we group a number (known as the dividend or numerator ) , which is a group of 1(s) , into smaller groups of another smaller number ( known as the divisor or denominator) and determining the number of such groups of the divisor ( known as the quotient ) which can be formed from the dividend or numerator.
Division is written in 1 of 2 ways :

In this form the the numerator is 8 , the denominator is 2 and the quotient is 4
In the second form 8 ÷ 2 = 4 the dividend is 8 , the divisor is 2 and the quotient is 4
Just as Multiplication is repeated addition so is Division repeated subtraction.

During division we repeatedly subtract an amount equals to the value of the divisor from the dividend until the dividend equals 0 or is less than the value of the divisor. A count is taken of the number of such subtractions and the final count is the quotient. If the final value of the dividend is not 0 but a number which is smaller than the divisor then this number is known as the remainder.

9 ÷ 2 = 4 r 1 ( r is usually used as a symbolic representation of Remainder )

As can be observed from the table, repeated subtraction is the basis for division. However, this method is very time and effort intensive - imagine a number in the range of thousands.
Before we expand more on the methods to do division it is appropriate to review the laws/principles of division.

Commutative Law

The Commutative law does not apply to subtraction and division.
7 - 5 is not equal to 5 - 7 therefore the order of subtraction is important.
8 ÷ 2 = 4 but 2 ÷ 8 = ¼ , therefore the order of division is important.

Distributive law

Where multiplication is concerned distribution can be applied to the dividend or numerator but not to the divisor or denominator.

E.g. let’s say you have 24 oranges and you want to determine the number of groups of 4 oranges.

If the divisor is split up say into 3 + 1 then we are performing 24 ÷ ( 3 + 1 ) or 24 ÷ 3 + 1
24 ÷ ( 3 + 1 ) => 24 ÷ 3 + 24 ÷ 1
24 ÷ 1 = 24 and so without even examining the other term of 24 ÷ 3 we know the answer is wrong.
24 ÷ 3 + 1

This is not the same as 6 groups of 4 which is supposed to be the correct answer.

Even, if we distribute 2 groups of 3 oranges among the remaining 6 groups, we would still get this

However, the dividend/numerator may be split up without affecting the answer.

24 ÷ 4 = ( 4 + 4 + 4 + 4 + 4 + 4 ) ÷ 4
= 4 ÷ 4 + 4 ÷ 4 + 4 ÷ 4 + 4 ÷ 4 + 4 ÷ 4 + 4 ÷ 4
= 1 + 1 + 1 + 1 + 1 + 1
= 6
24 ÷ 4 = ( 20 + 4 ) ÷ 4
= 20 ÷ 4 + 4 ÷ 4
= 5 + 1
= 6

The laws have been dispensed so let’s jump into the algorithms of division.
As stated earlier on, the basis for division is repeated subtraction but that in its basic form is very
inefficient. However, in whatever form it takes, or algorithm is used there is no mistake that
we cannot escape from repeated subtraction. Hence, we just have to make it more efficient. What
if we instead of subtracting 1 group of the divisor at a time subtract in big block or group of the
divisor say in 1000, 100, 50 , 10 etc. This way we can reduce the dividend very quickly and find the quotient efficiently.
Let’s try with a number as big as 987 and something within the range of lower elementary kids
like 64 ÷ 4

Note: I could have mentally broken
40 into 2 x 20 and end up with
3 x 2 x 20 and get quotient of 20
instead of having to carry 120
to the next stage - however with
small kids I got cautious.
Numbers in bracket eg. (120) are
remainders that need to
be carried forward to the next stage.

Note: The end part of both examples could be done neatly with
multiplication if feasible with individual kid, after all it is
merely knowing that 27 = 6 x 4 r 3 and
24 = 6 x 4. This variant is the basis for “Long Division “

dividend = quotient x divisor + remainder
or
dividend - quotient x divisor = remainder
Hence, at any time we would want to find the highest value of quotient such that
quotient x divisor is closest to the dividend without going over it.
e.g. 87 ÷ 6
14 x 6 = 84, hence the quotient is 14
13 x 6 = 78 not near enough to 87
15 x 6 = 90 overshot
Note: at times I do overshoot but will correct after the fact by reducing the quotient accordingly
15 x 6 = 90 → exceed 87 by 3
subtract 1 from 15 = 14 leaving a 6 to subtract 3 giving the remainder of 3
quotient = 14 , remainder = 3
e.g. 67 ÷ 7
10 x 7 = 70 overshooting 67 by 3
By subtracting 1 from 10 = 9 and thus leaving a 7 to subtract 3 I get a remainder of 4
Answer : quotient = 9 and remainder 4

Alternate Method 2 variant 2

What multiplier times divisor gives a product that is closest to dividend but yet not surpassing it ?
E.g. 987 ÷ 6

Long Division as taught in school is very efficient when the divisor is small and falls in the range of the multiplication tables ie 2 - 12. The higher or bigger the divisor gets the more difficult the division becomes. In fact, the difficulty increase at a very fast rate - even for adults. Try 3 digits divisor and you will understand what I mean. You will be taking quite a few side-tracks to perform multiplication. However, such problems exist for higher grades kids and do not apply to lower primary or elementary schoolers. And even for those there is always the calculator.

At lower grade school the only division kids need to do should be covered by long division. However, if for whatever reason long division becomes a struggle then do a hybrid by building rectangles and performing repeated subtraction at as low level as it helps the kid to understand division.

The following few examples will cover dividend up till 300.
Let’s start with simple ones.

Enter the Abacus

987 ÷ 7

Each bead on upper deck = 5 x base value. Hence, each column is worth 15 x base value. A bead on the Unit column carries a value of 1 whereas a bead on the Hundredth column carries a value of 100.

Subtract 700 and add 100 groups of 7

Subtract 140 and add 20 groups of 7 ( -200 + 50 +10)

Replace the 2 upper beads on the T column with a bead on the H column

Subtract 70 and add 10 groups of 7 Note: I actually carried out
-100 + 50 - 20

Subtract 70 and add 10 groups of 7

Finally subtract 7 and add 1 group of 7

As you can see from the 7 pictures finding quotient with the Abacus is a breeze. It’s all a matter of moving beads up and down. And, the order of the columns to start with is irrelevant to the machine because it takes care of carries and borrows.
For more on the Abacus please refer to my page on the topic.

My page on Abacus

Repeated Subtraction - improved

Divisor	Quotient per 100	Left-overs		Divisor	Quotient per 50	Left-overs
2	50	0		2	25	0
3	33	1		3	16	2
4	25	0		4	12	2
5	20	0		5	10	0
6	16	4		6	8	2
7	14	2		7	7	1
8	12	4		8	6	2
9	11	1		9	5	5
10	10	0		10	5	0
11	9	1		11	4	6
12	8	4		12	4	2
13	7	9		13	3	11
14	7	2		14	3	8
15	6	10		15	3	5

The table on the left shows you what happens when you div 100 by divisors from 2-15. This table is key to what we intend to do next. The one on the right is for 50 dividing by the divisors shown. The second table is to help those who have not yet mastered their time-tables.

Lets do an example : 987 ÷ 7
987 = 900 + 80 = 7
Step1 : start with 900 we get quotient 14x9 r 14 x 2

Step 2 : add the 4 to 87 = 91 ; mental division 91 ÷ 7 = 13

For younger kids

Step 2a : 10 x 7 = 70 : 91 - 70 = 21
21 = 3 x 7
87 ÷ 7 = 13 r 0

Step 3 : quotient = 128 + 13 = 141
remainder = 0

This method is fast and can be even more efficient than long division

Note: When you divide by 10 you just remove a zero from the dividend.
example: 100 ÷ 10 = 10
120 ÷ 10 = 12
1200 ÷ 10 = 120

When you divide by 5 the trick is to multiply the dividend by 2 and then remove the zero.

example: 65 ÷ 5
65 x 2 = 130 so 65 ÷ 5 = 13
70 ÷ 5
70 x 2 = 140 so 70 ÷ 5 = 14

For bigger kids that have started or learnt decimals - instead of removing a zero
you move the decimal 1 place to the left.

97 ÷ 5
97 x 2 = 194 so 97 ÷ 5 = 19.4

Another example
859 ÷ 6
800 → 8 x 16 + 8 x 4 →128 r 32 → 128 + 5 r 2= 133 r 2
59+2 = 61 → 10 r 1
Answer : 143 r 1 

The kiddie Abacus 987 ÷ 7

Kiddie Abacus Page

Charge to the Target

This strategy is used in conjunction with multiplication to form the basis of the next
technique I will describe.

Dividend ÷ divisor = quotient + remainder

Now , if multiplier x divisor + remainder = dividend then, the multiplier is the quotient that we are trying to find. Hence, we will use multiplication to charge to the quotient and then slow down if necessary until we find a multiplier, that produces exactly the dividend or is close enough, such that the (multiplier + 1) x divisor > dividend. For example, when we divide a number by 8
we will get remainders that range from 0 to 7 - 0 meaning the number is exactly divisible by 8. Therefore, we are trying to find a multiplier that will produce a product that is so close to the dividend that when the result of the multiplication is subtracted from the dividend the remainder will be less than 8. Lets use 57 for illustration.
7 x 8 = 56 , 57 - 56 = 1
Hence, 7 is the number we want. If we had tried with 6 : 6 x 8 = 48 and 57 - 48 = 11 which is greater than 8. When this happens, we know we are near but not near enough.

If the Dividend is within 1000 we will start with a multiplier of 100 and if that fails we will change
to 50. If ,of course, a divisor that is higher than 50 is known then we should use it.

Example 856 ÷ 7
Since 700 ( 100 x 7 ) < 856 we start with that. step 1 : 856 - 700 = 156 I am tempted to use mental Arithmetic at this juncture but in the interest of education I will continue to perform the charge with easily derived multiplier. 2 x 7 = 14 so we will use 20 step 2 : 20 x 7 = 140 : 156 - 140 = 16 step 3: 2 x 7 = 14 ; 16 -14 = 2 ; r = 2 quotient = 100 + 20 + 2 = 122 If we ignore all the comments, we can appreciate that this method is as efficient as Long Division and if not even more.

Example 2

657 ÷ 8

50 x 8 : 400 ; 657 - 400 = 257
8 x 30 = 240 : 257 - 240 = 17
2 x 8 = 16 : remainder = 1
Quotient = 50+30+2 = 82 r = 1

Example 3

794 ÷ 9

50 x 9 = 450 : 794 - 450 = 344
9 x 30 = 270 : 344 - 270 = 74
9 x 8 = 72 : remainder = 2
Quotient = 50 + 30 + 8 = 88

For younger kids

50 ÷ 7
5 x 7 = 35 : 50 -35 = 15
2 x 7 = 14: 15 - 14 = 1
Quotient = 5 + 2 = 7 r = 1
( of course 7 x 7 = 49 )
100 ÷ 7
10 x 7 = 70
100 - 70 = 30
7 x 4 = 28 : 30 - 28 = 2
Quotient = 10 + 4 = 14 : r = 2

78 ÷ 6
10 x 6 = 60
78 - 60 = 28
6 x 4 = 24
28 - 24 = 4
Quotient = 10 + 4 = 14 : r = 4

95 ÷ 3
3 x 3 = 9
30 x 3 = 90
95 - 90 = 5
5 - 3 = 2 ( 1 x 3 = 3 )
Quotient = 30 + 1 = 31 : r = 2

Note: In short time, with enough practices all the steps could be executed mentally.