Column Division – An Alternative to Long Division

Many people have bad memories of long division.  It’s one of the first algorithms a young student learns in math class that contains multiple steps that all build on each other, so one miscalculation at the beginning can throw the whole answer off.  It also uses a combination of different operations, including division, multiplication, and subtraction, which makes it difficult to complete most problems mentally.

		1	5	4	R:2
8	1	2	3	4
		8
		4	3
		4	0
			3	4
			3	2
				2

1,234 ÷ 8 = 154 R:2

Column division is an alternative to long division, and is inspired by a specific case of synthetic division in which x is a power of 10.  The algorithm for column division is to subtract once in the beginning, and then alternate between adding and multiplying. 

Example of Column Division

For example, to divide 8 into 1,234, start by placing the 8 and 1,234 as follows:

8		1
		2
		3
		4

Subtract 10 – 8 = 2 and place it beside the 8.  This 2 will be the multiplier:

8	2	1
		2
		3
		4

Now alternate between adding and multiplying.  First, “add” 1 + 0 = 1 and place it in the second section.  Add two dashes behind the 1 (the first row will always have 2 less dashes than the number of digits in the dividend, and 4 – 2 = 2.)

8	2	1		1	_	_
		2
		3
		4

Multiply 2 · 1 = 2 and place it in the second row of the first section (beside the 2):

8	2	1		1	_	_
		2	2
		3
		4

Add 2 + 2 = 4, and place it in the second row of the second section, and add one less dash than the row above it:

8	2	1		1	_	_
		2	2		4	_
		3
		4

Multiply 2 · 4 = 8 and place it in the third row of the first section (beside the 3):

8	2	1		1	_	_
		2	2		4	_
		3	8
		4

Add 3 + 8 = 11, and place it in the third row of the second section, and add one less dash than the row above it (so no dashes):

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1
		4

Multiply 2 · 11 = 22 and place it in the fourth row of the first section (beside the 4):

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1
		4	22

Add 4 + 22 = 26, and since we ran out of room with the dashes, this represents a remainder, so place it in the third row of the third section:

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1	26
		4	22

Since the divisor 8 divides into the remainder 26 at least 3 times, place the 3 in the fourth row of the second section, and place 8 · 3 = 24 in the fourth row of the third section:

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1	26
		4	22			3	24

Subtract the numbers in the last column:

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1	26
		4	22			3	24
							2

Add the numbers in the second to last column:

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1	26
		4	22			3	24
				1	5	4	2

The answer can be read from the bottom row:

8	2	1		1	_	_
		2	2		4	_
		3	8		1	1	26
		4	22			3	24
				1	5	4	2

Therefore, 1,234 ÷ 8 = 154 R:2.

Example of Column Division with a Large Remainder

Sometimes the remainder is too large to divide easily, in which case you can repeat the process of column division underneath the problem with the remainder as the new divisor.  For example, to divide 7 into 12,345, start by placing the 7 and 12,345 as follows:

7		1
		2
		3
		4
		5

Subtract 10 – 7 = 3 and place it beside the 8.  This 3 will be the multiplier:

7	3	1
		2
		3
		4
		5

Now alternate between adding and multiplying.  First, “add” 1 + 0 = 1 and place it in the second section.  Add three dashes behind the 1 (the first row will always have 2 less dashes than the number of digits in the dividend, and 5 – 2 = 3.)

7	3	1		1	_	_	_
		2
		3
		4
		5

Multiply 3 · 1 = 3 and place it in the second row of the first section (beside the 2):

7	3	1		1	_	_	_
		2	3
		3
		4
		5

Add 2 + 3 = 5, and place it in the second row of the second section, and add one less dash than the row above it:

7	3	1		1	_	_	_
		2	3		5	_	_
		3
		4
		5

Multiply 3 · 5 = 15 and place it in the third row of the first section (beside the 3):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15
		4
		5

Add 3 + 15 = 18, and place it in the third row of the second section, and add one less dash than the row above it:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4
		5

Multiply 3 · 18 = 54 and place it in the fourth row of the first section (beside the 4):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54
		5

Add 4 + 54 = 58, and place it in the fourth row of the second section, and add one less dash than the row above it (so no dashes):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8
		5

Multiply 3 · 58 = 174 and place it in the fifth row of the first section (beside the 5):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8
		5	174

Add 5 + 174 = 179, and since we ran out of room with the dashes, this represents a remainder, so place it in the fourth row of the third section:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174

Since the remainder 179 is too large to divide easily into 7, repeat the process of column division by placing it as the new divisor underneath the problem:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1
		7
		9

Once again, alternate between adding and multiplying.  First, “add” 1 + 0 = 1 and place it in the second section.  This time add one dash behind the 1 (the first row will always have 2 less dashes than the number of digits in the dividend, which now has 3 digits, and 3 – 2 = 1.)

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7
		9

Multiply 3 · 1 = 3 and place it in the first section (beside the 7):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3
		9

Add 7 + 3 = 10, and place it in the second section, and add one less dash than the row above it (so no dashes):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0
		9

Multiply 3 · 10 = 30 and place it in the first section (beside the 7):

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0
		9	30

Add 9 + 30 = 39, and since we ran out of room with the dashes, this represents a remainder, so place it in the third section:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0	39
		9	30

Since the divisor 7 divides into the remainder 39 at least 5 times, place the 5 in the the second section, and place 7 · 5 = 35 in the third section:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0	39
		9	30				5	35

Subtract the numbers in the last column:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0	39
		9	30				5	35
								4

Add the numbers in the second to last column:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0	39
		9	30				5	35
				1	7	6	3	4

The answer can be read from the bottom row:

7	3	1		1	_	_	_
		2	3		5	_	_
		3	15		1	8	_
		4	54			5	8	179
		5	174
		1				1	_
		7	3			1	0	39
		9	30				5	35
				1	7	6	3	4

Therefore, 12,345 ÷ 7 = 1,763 R:4.

Example of Column Division with a Multiple Digit Divisor

Column division can also be adjusted to handle multiple digit divisors.  If the divisor has n digits, then we can group our numbers in sets of n digits, and we can subtract the divisor from 10ⁿ to obtain the multiplier.  For example, if we were to divide 99 into 12,345, the divisor 99 has 2 digits, so we would group our numbers in sets of 2 digits each:

99		1
		23
		45

Subtract 10² – 99 = 1 and place it beside the 99.  This 1 will be the multiplier:

99	1	1
		23
		45

Now alternate between adding and multiplying.  First, “add” 1 + 0 = 1 and place it in the second section.  Add one set of two dashes behind the 1 (the first row will always have 2 less groups of dashes than the number of groups of digits in the dividend, and 3 – 2 = 1.)

99	1	1		1	_ _
		23
		45

Multiply 1 · 1 = 1 and place it in the second row of the first section (beside the 23):

99	1	1		1	_ _
		23	1
		45

Add 23 + 1 = 24, and place it in the second row of the second section, and add one less groups of dashes than the row above it (so no dashes):

99	1	1		1	_ _
		23	1		24
		45

Multiply 1 · 24 = 24 and place it in the third row of the first section (beside the 45):

99	1	1		1	_ _
		23	1		24
		45	24

Add 45 + 24 = 69, and since we ran out of room with the dashes, this represents a remainder, so place it in the second row of the third section:

99	1	1		1	_ _
		23	1		24	69
		45	24

Since the divisor 99 divides into the remainder 0 times, place 0 in the third row of the second section, and place 99 · 0 = 0 in the third row of the third section:

99	1	1		1	_ _
		23	1		24	69
		45	24		0	0

Subtract the numbers in the last column:

99	1	1		1	_ _
		23	1		24	69
		45	24		0	0
						69

Add the numbers in the second to last column:

99	1	1		1	_ _
		23	1		24	69
		45	24		0	0
				1	24	69

The answer can be read from the bottom row:

99	1	1		1	_ _
		23	1		24	69
		45	24		0	0
				1	24	69

Therefore, 12,345 ÷ 99 = 124 R:69.

Proof

The proof involves running variables through the algorithm of column division and showing that the product of the divisor and the quotient simplifies to the dividend.  Let the divisor be p with m digits, and the dividend be d₁(10^m)^{n – 1} + d₂(10^m)^{n – 2} + d₃(10^m)^{n – 3} + … + d_{n – 1}(10^m)¹ + d_n(10^m)⁰.  (So if the divisor has one digit, then m = 1 and the d variables would represent each digit in the dividend).  Then x = 10^m – p  and column division with these variables would look as follows:

p	x	d1		(d1)(10m)n–2
		d2	d1x	(d1x + d2)(10m)n–3
		d3	d1x2 + d2x	(d1x2 + d2x + d3)(10m)n–4
		…	…	…
		dn–1	d1xn–2 + d2xn–3 + d3xn–4 … dn–2x	(d1xn–2 + d2xn–3 + d3xn–4 … dn–2x + dn–1)(10m)0	d1xn–1 + d2xn–2 + d3xn–3 … dn–2x2 + dn–1x + dn
		dn	d1xn–1 + d2xn–2 + d3xn–3 … dn–2x2 + dn–1x	0	0
				d1(10m)n–2 + (d1x + d2)(10m)n–3 + (d1x2 +d2x +d3)(10m)n–4 + … + (d1xn–2 + d2xn–3 + d3xn–4 … dn–2x + dn–1)(10m)0	d1xn–1 + d2xn–2 + d3xn–3 … dn–2x2 + dn–1x + dn

The product of the divisor and the quotient then simplifies to:

p · [d1(10m)n–2 + (d1x + d2)(10m)n–3 + (d1x2 + d2x + d3)(10m)n–4 + … + (d1xn–2 + d2xn–3 + d3xn–4 … dn–2x + dn–1)(10m)0 + (d1xn–1 + d2xn–2 + d3xn–3 … dn–2x2 + dn–1x + dn)/p]	(divisor times the quotient)
d1p(10m)n–2 + (d1x + d2)p(10m)n–3 + (d1x2 + d2x + d3)p(10m)n–4 + … + (d1xn–2 + d2xn–3 + d3xn–4 … dn–2x + dn–1)p(10m)0 + (d1xn–1 + d2xn–2 + d3xn–3 … dn–2x2 + dn–1x + dn)]	(distribute p)
d1{p[(10m)n–2 + x(10m)n–3 + x2(10m)n–4 + … + xn–2(10m)0] + xn–1} + d2{p[(10m)n–3 + x(10m)n–4 … + xn–3(10m)0] + xn–2} + d3{p[(10m)n–4 + … + xn–4(10m)0] + xn–3} + … + dn–2{p[ … x(10m)0] + x2} + … dn–1{p[ … (10m)0] + x} + … dn	(rearrange and factor according to d)
d1{[10m – x][(10m)n–2 + x(10m)n–3 + x2(10m)n–4 + … + xn–2(10m)0] + xn–1} + d2{[10m – x][(10m)n–3 + x(10m)n–4 … + xn–3(10m)0] + xn–2} + d3{[10m – x][(10m)n–4 + … + xn–4(10m)0] + xn–3} + … + dn–2{[10m – x][ … x(10m)0] + x2} + … dn–1{[10m – x][ … (10m)0] + x} + … dn	(substitute p = 10m – x)
d1{[(10m)n–1 – xn–1] + xn–1} + d2{[(10m)n–2 – xn–2] + xn–2} + d3{[(10m)n–3 – xn–3] + xn–3} + … + dn–2{[(10m)2 – x2] + x2} + … dn–1{[(10m)1 – x1] + x} + … dn(10m)0	(simplify using the difference of powers equations)
d1(10m)n–1 + d2(10m)n–2 + d3(10m)n–3 + … + dn–2(10m)2 + dn–1(10m)1 + dn(10m)0	(simplify)

which is the dividend.

Assessment

With the above examples, it may seem that column division is far superior to long division because it is quicker and easier to use.  Unfortunately, this is not always true.  For example, dividing 2 into 888, which is obviously 444 by long division, would require 14 iterations of column division before obtaining a remainder smaller than 2 · 10 = 20 to complete the problem:

2	8	8		8	_
		8	64	7	2	584
		8	576
		5		5	_
		8	40	4	8	388
		4	384
		3		3	_
		8	24	3	2	264
		8	256
		2		2	_
		6	16	2	2	180
		4	176
		1		1	_
		8	8	1	6	128
		0	128
		1		1	_
		2	8	1	0	88
		8	80
		8			8	72
		8	64
		7			7	58
		2	56
		5			5	48
		8	40
		4			4	40
		8	32
		4			4	32
		0	32
		3			3	26
		2	24
		2			2	22
		6	16
		2			2	18
		2	16		9	18
				44	4	0

As a general rule of thumb, the closer the divisor is to a power of ten, the easier it is to do column division, but the further away the divisor is from a power of ten, the harder it is to do column division.  So it is easier to divide by 8 (or 9, 98, or 99, etc.) by column division than it is by long division, but it is more difficult to divide by 2 (or 3, 4, 11, or 12, etc.) by column division than it is by long division.  However, column division still remains a fascinating alternative to long division, especially since the algorithms are so completely different.