### Table of Contents

- Introduction
- Writing and annotating digits
- Addition
- Subtraction
- Mixed addition and subtraction
- Multiplication by a single digit
- Multiplication by multiple digits
- Division by multiple digits

### Introduction

We all know the basic techniques taught to us in school for doing arithmetic on paper. While they work, I've always thought they were a bit more mentally demanding, error-prone, and wasteful of space than should really be necessary, and at some point I came up with a modified set of techniques that aim to reduce space requirements and minimize the chance for error. A good technique should try to achieve the following properties:**Working memory requirements are minimized**– One of the main causes of error in arithmetic is trying to hold too many numbers in your head at once. Smarter people tend to have a stronger working memory, but everyone has a limit, and the limit tends to be rather small – only a handful of digits. A good technique will minimize demands on working memory.**Mental math requirements are minimized**– While some people can multiply and divide fairly large numbers in their heads, most people cannot. A good technique will require a minimum of mental calculation and rely almost entirely on the addition, subtraction, and multiplication tables we memorized as children. At the same time, a good technique should allow those with stronger mental math skills to put them to use.**Pen movements are short and local**– When adding or subtracting a long column of numbers, for example, moving the pen to the top of the column to add a carry or borrow mark above the top row makes it easy to lose your place in the column. A good technique keeps all pen movements within a short, predictable, and fixed distance of each other, so you never lose track of your position.**Passes over the numbers are minimized**- Often, processing the numbers in a particular order will allow you to make only a single pass over them, while processing them in a different order may require multiple passes. A single-pass technique saves time.**Space on paper is minimized**- Excessively verbose techniques consume too much paper space, sometimes causing you to run out and have to cram things where they don't belong. A good technique minimizes the vertical and horizontal space required.

### Writing and annotating digits

Everyone already knows how to write digits, right? Yes, but I propose a method of annotating digits to allow them to be incremented, decremented, or added to without rewriting them. I'll explain why in a moment.^{·}

^{:}

^{/}

^{4}

_{2}

^{4:}

Try to see these as just another way of writing the digit, such that `5` and
`4 ^{.}` are exactly the same thing, and both are read as "five". Note
that sometimes the numbers can exceed nine or even be less than zero, and so technically
are not digits:

`9`is ten, and

^{.}`0`is -1.

^{/}The point of these annotations is to allow modifications to be made to numbers
without erasing or crossing out numbers on paper or having to finalize a number in your
head before committing it to paper. For example, carrying and borrowing, or revising a
quotient up or down in long division, can be a simple matter of adding a dot or slash.
The answer to `123` plus `456` could be written as
`1 ^{4}2^{5}3^{6}`. This is the same as writing 579.
Although the example seems absurd, it is useful when combining multiple answers together,
and can allow a large multiplication like 2348723436 x 7 to be carried out mentally and
the answer written down with almost no effort. (The answer is

`14`.)

^{2}1^{2}8^{5}6^{4}9^{1}4^{2}1^{2}8^{2}1^{4}2
Within this document, numbers will often be shown spaced out, as in
`1 2 3 ^{2}`. This is simply the number 125. This to not to suggest that you
put spaces between digits when writing on paper, but rather reflects a limitation of this
document's format. On paper, numbers should align into columns. When adding a superscript
to a digit, it should fit within the same column, and not push over subsequent digits,
but that's hard to do in HTML without reserving extra space for potential superscripts.

#### Normalizing numbers

At some point, a number like`14`may need to be converted back into a normal number.

^{2}17^{3}5^{4}21 4You should generally work from right to left, to ensure you only need to make a single pass over the numbers. The rightmost digit is^{2}1 7^{3}5^{4}2^{ }^{ }9^{ }2

`2`, which is already in normal form. The next digit is

`5`, which is nine (5 + 4 = 9).

^{4}1 4The next number is^{2}1^{.}7^{3}5^{4}2^{ }^{ }0^{ }9^{ }2

`7`, which equals ten. This is larger than a single digit, so write "0" and carry the 1 to the next digit by adding a dot. (Adding a dot alters the original number, so if that's undesirable you can make a copy first, simply remember that the next digit needs to be one greater, or use multiple passes as shown below.)

^{3}1 4The next digit is^{2}1^{.}7^{3}5^{4}2 1 6 2^{ }0^{ }9^{ }2

`1`, which equals two, and the subsequent one is

^{.}`4`, which equals six. Finally,

^{2}`1`is just one. So write "2", "6", and "1". The result is 162,092.

#### Normalizing numbers without altering them

If you wanted to preserve the original number (`14`) rather than altering it as part of the normalization, normally I'd recommend simply remembering when you need to adjust the next digit, but to reduce memory demands to the minimum you can make multiple passes instead. (Or, just make a copy and alter the copy.) Using the same example as above, we get to this point:

^{2}17^{3}5^{4}21 4After normalizing the^{2}1 7^{3}5^{4}2^{ }1 0^{ }9^{ }2

`2`and

`5`, the next digit is

^{4}`7`, which equals ten. Simply write both digits below.

^{3}1 4The next digit is^{2}1^{ }7^{3}5^{4}2^{ }1^{.}0^{ }9^{ }2

`1`. Since there's already a

`1`digit below, add them together, incrementing the digit using a dot.

1 4The subsequent digit is^{2}1^{ }7^{3}5^{4}2 1 6^{ }1^{.}0^{ }9^{ }2 1 6^{ }2^{ }0^{ }9^{ }2

`4`, which equals six. Finally,

^{2}`1`is just one. So write "6" and "1". The result is

`161`, but this needs to be further normalized in a second pass by converting the

^{.}092`1`to

^{.}`2`.

### Addition

This addition technique is simple, and has these properties.- Work is from right to left and top to bottom. Although some people extol the virtues
of working left to right – and there
*are*benefits when you only want an estimate instead of the exact answer – working from right to left requires only a single pass over the columns with no fix-up work at the end. - Carrying only requires making a single mark next to the pen location, even with a cascading carry (e.g. with a string of 9's).
- The technique requires only the ability to add a single digit to a number from 0 to 10 (producing values from 0 to 19), and requires only a single digit to be kept in memory.

#### Example

1 2 3 4 5 6 9 8 1 2 3 4 5 6 0 +Align the numbers and work down the columns from right to left, adding each digit to the number in your head and moving the pen to each digit as you add it. For the rightmost column, think something like "6 + 2 is 8", and write "8" at the bottom.4 3 2 1 0 08

1 2 3 4 5 6 9 8 1 2 3 4 5In the next column, we encounter our first carry. When the sum exceeds 9, subtract 10 (by dropping the leading digit), and place a dot in the column to the left of the digit that caused the overflow. Think something like "5 + 1 is 6, plus 6 is carry and 2." You can abbreviate this to "5, 6, carry and 2", but it's important to say "carry and 2", not something like "2 with a carry". The reason is that this leaves the last word in your mind as "2", the number you're trying to remember. If "carry" is the last word in your mind, you may forget the number. Write "2" at the bottom.^{.}6 0 +4 3 2 1 0 02 8

1 2 3 4 5 6 9Repeat for the next column, thinking "4 + 8 is carry and 2, plus 6 is 8, plus 1 is 9". Note that^{.}8 1 2 3 4 5^{.}6 0 +4 3 2 1 0 09 2 8

`5`is read as "six", because that's what it is. You can abbreviate the words to "4, carry and 2, 8, 9". Write "9" at the bottom.

^{.}1 2 3 4 5 6Do the next column. Note that this time the carry dot is placed into an empty column. Empty columns are implicitly zero, so an empty space with a carry dot has a value of 1. (You could also just write a "1" there.)^{.}9^{.}8 1 2 3 4 5^{.}6 0 +4 3 2 1 0 09 9 2 8

`9`is read as "ten".

^{.}1 2 3 4 5 6Do the next two columns and complete the addition. The answer is 599,928.^{.}9^{.}8 1 2 3 4 5^{.}6 0 +4 3 2 1 0 05 9 9 9 2 8

### Subtraction

Subtraction is similar to addition and has the same virtues.- Work is from right to left and top to bottom. Working from right to left requires only a single pass over the columns with no fix-up work at the end.
- Borrowing only requires making a single mark next to the pen location, even with a cascading borrow (e.g. with a string of 0's).
- The technique requires only the ability to subtract two numbers from 0 to 10, and requires only a single digit to be kept in memory.

#### Example

1 2 3 4 9 8 4 5 3 1 4 2 –Align the numbers and work down the columns from right to left. At the top of each column, place the digit in your memory and then subtract the subsequent digits from it, moving the pen to each digit as you subtract it. For the rightmost column, think something like "9 - 5 is 4, minus 2 is 2, minus 1 is 1", and write "1" at the bottom.2 7 5 11

1 2 3 4 9 8 4 5 3 1In the next column, we encounter our first borrow. When the remainder becomes negative, add 10 to it and place a dot in the column to the left of the digit that caused the underflow. Think something like "4 - 4 is 0, minus 4 is borrow and 6, minus 5 is 1". You can abbreviate this to "4, 0, borrow and 6, 1", but it's important to say "borrow and 6", not something like "6 with a borrow". The reason is that this leaves the last word in your mind as "6", the number you're trying to remember. If "borrow" is the last word in your mind, you may forget the number. Write "1" at the bottom.^{.}4 2 –2 7 5 11 1

1 2 3 4 9Repeat for the next column, thinking "3 - 8 is borrow and 5, minus 2 is 3, minus 7 is borrow and 6", which you can abbreviate to "3, borrow and 5, 3, borrow and 6".^{.}8 4 5 3 1^{.}4 2 –26 1 1^{.}7 5 1

`1`is read as "two", because that's what it is. Instead of borrowing by decrementing a digit at the top, we borrow by incrementing the digit to the left. Note that one of the borrow dots is placed into an empty column. Empty columns are implicitly zero, so an empty space with a dot has a value of 1. (You could also just write a "1" there.) Write "6" at the bottom.

^{.}1 2 3 4 9Do the next two columns and complete the subtraction. The answer is 5611.^{.}8 4 5^{.}3 1^{.}4 2 –25 6 1 1^{.}7 5 1

#### Ten's complement

Ten's complement is used to deal with negative results in subtraction. It's also useful for making change.100000 –Start with a power of ten and a smaller number to subtract. Work from right to left. First, the rightmost "0" digits of the lower number become zero in the result. Then, subtract the first non-zero digit from 10 (10 - 6 = 4) and the rest from 9 (9-7 = 2, 9-4 = 5, etc). You can easily do this in your head without needing to write the power of ten.2476075240

$200.00 –A customer is purchasing $136.53 worth of goods and paid $200 in cash. How much change should he receive? The subtraction is very similar to ten's complement. Since 200 is not a power of 10, the rule must be generalized slightly: perform ten's complement for the digits under the string of trailing zeros. Under the first non-zero digit in the top number (as viewed from the right), subtract normally but make the result one less than it would normally be. Beyond the first non-zero digit, subtract normally.$136.53$63.47

$1 5 2.0 0 –A customer is purchasing $136.53 worth of goods and paid $152 in cash. How much change should he receive? Perform ten's complement under the trailing zeros, do the next column normally but subtract one ("2 - 6 is borrow and 6, minus 1 is 5"), and subtract the final two columns normally.$1 3$ 1 5.4 7^{.}6.5 3

#### Subtraction with a negative result

Consider 123 - 251. Normally you would just arrange it with the bigger number on top, knowing the result is negative (*a*-

*b*= -(

*b*-

*a*)).

1 2 3 – 2 5 1 –Here is 123 - 251 rearranged to put the number with the larger magnitude on top. Although there's a plus sign, the operation is performed via subtraction (because the negative sign in the result requires effectively reversing the operations).2 5 1-> +1 2– 1 2 8^{.}3

1 2 3 –Do the subtraction as normal. In the 100s column, you need to borrow 1000 but have no number to borrow from. So the answer is 872 - 1000, due to your "debt" of 1000.8 7 2^{.}2^{.}5 1

1 2 3 | 1 2 3 –872 - 1000 equals -(1000 - 872). Use ten's complement to quickly transform each digit in your head, or do the subtraction explicitly.| –^{.}2^{.}5 1^{.}2^{.}5 1~~8 7 2~~| 8 7 2 – 1 2 8 | | –1 0 0 0 | +| – 1 2 8^{.}8^{.}7^{.}2

### Mixed addition and subtraction

Because addition and subtraction work the same way, dotting the digit to the left on an overflow or underflow, it's very simple to mix them together. Just don't forget which lines are adding and which lines are subtracting! (It may help to group them together.)#### Example

6 3 4 1 + 1There's a lot of carrying and borrowing in this example, but it works just like the addition and subtraction examples above. Go down each column, from right to left, adding to or subtracting from the digit in your head, and carrying or borrowing by adding a dot to the left on overflow or underflow.^{.}9^{.}7 1 – 3^{.}8^{.}4^{.}4 +14 6 6 1^{.}9^{.}3

### Multiplication by a single digit

Multiplication by a single digit is a common operation that can be performed quickly and easily on paper. This technique only requires the ability to multiply two digits together. Multiplication by a single digit can be done in your head at the same time as you write a number down.#### Example

In this example, we'll multiply 23,156 by 7.2 3 1 5 6x7Write the problem like this.

2 3 1 5 6x7 1 4Multiply the first digit (2) by 7, producing 14. Write the result below.

2 3 1 5 6x7 1 4Multiply the second digit (3) by 7, producing 21. Each multiplication produces either one or two digits. The ten's digit, if any, is added to the previous digit (usually as a superscript). The one's digit is written in its own column. So, write the 2 as a superscript of the 4 and write the 1 in its own column.^{2}1

2 3 1 5 6x7 1 4Multiply the third digit (1) by 7, producing 7. Because it has no tens column, there is nothing to add to the previous digit. Just write 7 in a new column.^{2}1 7

2 3 1 5 6x7 1 4Continue until you get to the end. You can use the result directly, or normalize it into 162,092.^{2}1 7^{3}5^{4}2

1 4Alternatively, you can perform the multiplication as you write, with no need to write the original number or its factor. Simply multiply each digit pair, adding any tens digit to the previous digit (e.g. as a superscript). This saves time and space when the operation is used as part of a larger multiplication or division.^{2}1 7^{3}5^{4}2

### Multiplication by multiple digits

Multiplication by multiple digits is a straightforward combination of multiplication by single digits and addition.#### Example

In this example, we'll multiply 268 by 4613.2 6 8x4 6 1 3Write the two numbers on the same line without a gap between them, starting with the shorter number (or better, the number with fewer non-zero digits) and separating them with a small 'x'.

2 6 8x4 6 1 3 0 8Start by multiplying the first digit (2) by 4613. To make alignment easy and automatic, ensure that the first digit multiplication always produces a two-digit product, e.g. 2 x 4 = 08, as shown here.^{1}2 2 6

2 6 8x4 6 1 3 0 8Now multiply the second digit (6) by 4613. Because the first digit product is always two digits, the number will always shift one place further to the right on each line.^{1}2 2 6 2 4^{3}6 6^{1}8

2 6 8x4 6 1 3 0 8Now multiply the third digit (8) by 4613.^{1}2 2 6 2 4^{3}6 6^{1}8 3 2^{4}8 8^{2}4

2 6 8x4 6 1 3 0 8Finally, add the three numbers together, working from right to left as usual. The result is 1,236,284.^{1}2 2 6^{.}2 4^{3}6^{.}6^{1}8+1 2^{.}3^{.}2^{̥4}8^{.}8^{2}4^{ }3^{ }6^{ }2^{ }8^{ }4

#### Zeros on the left

2 0 3x1 2 3 4 0 2 4 6 8 0 0 3 6 9Note that if one of the digits on the left side is zero, then the product is obviously zero. You can either write a zero line to help maintain alignment, or omit the line if you can remember to shift the next line two places. Both approaches are shown here.^{1}2 2 0 3x1 2 3 4 0 2 4 6 8 3 6 9^{1}2

### Division by multiple digits (long division)

This method for dividing by multiple digits is similar to the standard method, but has the following features:- It doesn't require trial division by or finding multiples of the full divisor. You only ever need to divide by a single digit, though if you can do more then progress may be faster. The choice of multiple does not need to be exact.
- It doesn't require additional space for multiplying the divisor by the chosen multiple or holding more than one digit in your head or multiplying more than two digits at a time.

#### Basic example

In this example, we'll divide 894,387 by 741.7 4 1x 8 9 4 3 8 7Begin by writing the divisor followed by a small 'x', and on the next line writing the dividend, offset to the right by one digit. Mentally round the divisor up to a value that you can easily divide by (ignoring trailing zeros). In this example, we'll round 741 up to 800 and do mental division by 8. This is not required, but it simplifies the work. Don't round down, to avoid dealing with negative numbers.

7 4 1x1Take some number of digits from the dividend that you can mentally divide by 8 (ignoring any remainder). In this case, we'll take one digit (underlined). Obviously 8/8 = 1, so multiply 1 by the true divisor (741) and write it below the digits you chose. Be careful to align it correctly. Add the multiplier (1) to the quotient above the last digit of the subtrahend. Then subtract.89 4 3 8 7 - 7 4 1 = 1 5 3 3 8 7

7 4 1x1 1 8 9 4 3 8 7 - 7 4 1 =Next, we'll take two digits (15). We can't take one digit because 1/8 = 0 so we wouldn't make any progress. 15/8 = 1, so again write 741 below, being careful with alignment. Again, add the multiplier (1) above the last digit of the subtrahend. (If you're able to see that 1533 ≥ 741x2, feel free to choose 2 for the multiplier. The important thing is to make progress without excessive mental strain and without overestimating the quotient.) Subtract.1 53 3 8 7 -^{.}7^{.}4 1 = 7 9 2 8 7

7 4 1x1 1In this case we'll take three digits (792) because we can clearly see that 792/741 is just over 1, so write 741 below. The multiplier (1) is added to the quotient. Since there's already a 1 there, dot it, making it 2. Finally, do the subtraction.^{.}8 9 4 3 8 7 - 7 4 1 = 1 5 3 3 8 7 -^{.}7^{.}4 1 =7 9 28 7 - 7 4 1 = 5 1 8 7

There's no need to always do the trial division by 8(00), or by 741. If it's convenient to divide by 750, go for it. The important things are to divide by some number greater than or equal to the true divisor, and to use the true divisor when you subtract from the dividend. The trial division is only to find an approximate multiplier for the true divisor. There's also no need to ensure the multiplier is a single digit.

7 4 1x1 1We'll take two digits (51). 51/8 = 6. Use the 'multiplication by a single digit' algorithm to compute the product of 6 and 741 as you write it. (It'll look like^{.}0 6 8 9 4 3 8 7 - 7 4 1 = 1 5 3 3 8 7 -^{.}7^{.}4 1 = 7 9 2 8 7 - 7 4 1 =5 18 7 - 4^{.}2^{2}4 6 = 7^{ }4 1

`42`, i.e. 4446.) Then add 6 to the quotient, inserting the zero where needed, and subtract.

^{2}467 4 1x1 1Obviously we were off by one, so add a dot to the quotient above the last digit of the subtrahend. The quotient is^{.}0 6^{.}8 9 4 3 8 7 - 7 4 1 = 1 5 3 3 8 7 -^{.}7^{.}4 1 = 7 9 2 8 7 - 7 4 1 = 5 1 8 7 - 4^{.}2^{2}4 6 =7 4 1- 7 4 1 = 0

`11`, i.e. 1207.

^{.}06^{.}#### Example with a larger multiplier and a fractional portion

In this example, we'll divide 15,435 by 12.1 2x 1 5 4 3 5Write the problem as described above.

1 2x1 2We'll choose three digits (154) because we know that a dozen dozen (12x12) is a gross (144), so 154/12 is a little over 12. (Even if it was actually 13, choosing 12 would be fine.) Add 12 to the quotient, with the last digit (2) aligned with the last digit of the subtrahend (4).1 5 43 5 - 1 4 4 = 1 0 3 5

1 2x1 2 1 5 4 3 5 - 1 4 4 =1 and 10 are both too small, but what if 103 is too difficult to mentally divide by 12? There are many options. For example, round 12 up to 20 and divide 103 by 20 (just as we rounded 741 to 800 in the example above). Or, reduce 103 to 72 and use use 72/12 = 6. To make progress, we just need to subtract some multiples of the divisor from the dividend without going negative. Any overestimate of the divisor or underestimate of the multiplier (so long as it's still above zero) will suffice.1 0 35

1 2x1 2 5 1 5 4 3 5 - 1 4 4 =We'll round 12 up to 20. 103/20 = 5, so use 60 (5x12 = 60) as the subtrahend and add 5 to the quotient over the last digit of the subtrahend. Subtract.1 0 35 -^{.}6 0 = 4 3 5

1 2x1 2 5Take two digits (43) with 43/12 = 3, and use a subtrahend of 36 (3x12). Add 3 to the quotient. Since there's already a 5 there, augment it with a 3 in superscript (or use the old trick to turn a 5 into an 8).^{3}1 5 4 3 5 - 1 4 4 = 1 0 3 5 -^{.}6 0 =4 35 - 3^{.}6 = 7 5

1 2x1 2 575/12 = 6 and 6x12 = 72. We can either stop here (because 3<12), with the answer being 1286 remainder 3 (i.e. 1286 and 3/12), or continue to get the decimal fraction. We'll continue.^{3}6 1 5 4 3 5 - 1 4 4 = 1 0 3 5 -^{.}6 0 = 4 3 5 - 3^{.}6 =7 5- 7 2 = 3

1 2x1 2 5You could realize that 3/12 = 1/4 = 0.25, thus making the answer 1286.25, or continue with the long division method by extending the dividend with zeros. We'll continue. 30/12 = 2 and 2x12 = 24.^{3}6.2 1 5 4 3 5 - 1 4 4 = 1 0 3 5 -^{.}6 0 = 4 3 5 - 3^{.}6 = 7 5 - 7 2 =3.0- 2.4 = 0.6

1 2x1 2 560/12 = 5, 5x12 = 60, and we've reached the end. The quotient is 1286.25.^{3}6.2 5 1 5 4 3 5 - 1 4 4 = 1 0 3 5 -^{.}6 0 = 4 3 5 - 3^{.}6 = 7 5 - 7 2 = 3.0 - 2.4 = 0.6 0- 0.6 0 = 0

#### Example rounding down, with a dividend that goes negative

In this example, we'll divide 4,098,711 by 1187. Instead of rounding the divisor up, we'll round it down. As a result, we may have to deal with negative dividends. Normally we wouldn't do that, but I want to show that it's possible. It also demonstrates how to recover if you accidentally overestimate the multiplier and wind up with a negative dividend.1 1 8 7x4Normally we would round 1187 up to 1200 or so, but let's say we wanted to round down to 1000 to make the mental division very easy. The danger of rounding down is that you may overestimate the multiplier. We'll begin with 4098/1000 = 4, and subtract 1187x4. (You should be able to see that 1187x4 would be too big and choose 3 instead, but let's pretend we didn't notice.)4 0 9 87 1 1 -^{.}4^{.}4^{3}2^{2}8 = 9 3 5 0 7 1 1

1 1 8 7x4 4 0 9 8 7 1 1 -We'll quickly use ten's complement to obtain the correct, negative dividend (-649,289). The easiest thing to do is make the dividend positive again by adjusting the quotient downward. In preparation for that, we'll leave a blank line where we'll write a positive term to add.^{.}4^{.}4^{3}2^{2}8~~= 9 3 5 0 7 1 1~~- 6 4 9 2 8 9

1 1 8 7x4Slash the 4 in the quotient to adjust it to 3, and write 1187 (and some trailing zeros) in the blank line. This is the value we're adding back to the dividend. Complete the subtraction to get the revised dividend. You can use the "making change" rule described in the "Ten's complement" section to simplify the work slightly. The new dividend is 537,711.^{/}4 0 9 8 7 1 1 -^{.}4^{.}4^{3}2^{2}8~~= 9 3 5 0 7 1 1~~1 1 8^{ }7 0 0 0 -^{.}6 4^{.}9 2 8 9 5 3^{ }7 7 1 1

1 1 8 7x4At this point you should be wary of overestimating the multiplier and avoid choosing 5377/1000 = 5, since 1187x5 is clearly too big, but let's say you do it anyway. Use ten's complement again to get the correct, negative dividend (-55789). Again we'll leave a blank line for the correction.^{/}5 4 0 9 8 7 1 1 -^{.}4^{.}4^{3}2^{2}8~~= 9 3 5 0 7 1 1~~1 1 8^{ }7 0 0 0 -^{.}6 4^{.}9 2 8 95 31 1 -^{ }7 7^{.}5^{.}5^{4}0^{3}5~~= 9 4 4 2 1 1~~- 5 5 7 8 9

1 1 8 7x4This time, instead of trying to decrement the quotient from 35(00) to 34(00), we'll recover a different way. We'll simply continue doing the division, but since the dividend is negative, it's not so bad to overestimate the magnitude of the multiplier because it'll make us go positive. So we'll take -5578/1000 = -5 and subtract -5x1187, which is the same as adding 5x1187 (^{/}5 0_{5}4 0 9 8 7 1 1 -^{.}4^{.}4^{3}2^{2}8~~= 9 3 5 0 7 1 1~~1 1 8^{ }7 0 0 0 -^{.}6 4^{.}9 2 8 9 5 3^{ }7 7 1 1 -^{.}5^{.}5^{4}0^{3}5~~= 9~~5 5^{ }4^{ }4^{ }2 1 1^{4}0^{3}5 -5 5^{.}7^{.}8^{.}9 = 3 5^{ }6 1

`55`). This puts us back in positive territory, with a dividend of 3561. To represent the addition of -50 to the quotient, you can either keep track of the negative values elsewhere or represent the -5 digit as

^{4}0^{3}5`0`. We'll do the latter.

_{5}1 1 8 7x4Finally, we take 3561/1000 = 3, subtract 1187x3, and we're done. The quotient is^{/}5 0_{5}3 4 0 9 8 7 1 1 -^{.}4^{.}4^{3}2^{2}8~~= 9 3 5 0 7 1 1~~1 1 8^{ }7 0 0 0 -^{.}6 4^{.}9 2 8 9 5 3^{ }7 7 1 1 -^{.}5^{.}5^{4}0^{3}5~~= 9~~5 5^{ }4^{ }4^{ }2 1 1^{4}0^{3}5 - 5 5^{.}7^{.}8^{.}9 =3- 3^{ }5^{ }6^{ }1^{ }3^{2}4^{2}1 = 0

`4`, which normalizes to 3453.

^{/}50_{5}3
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