Difference between revisions of "ספר המספר / אברהם אבן עזרא"

Prologue

Numeration

והנה הַסְּפָר תשעה מספרים‫^[4]

• The resemblance of the ranks to the rank of units

• units – as foundations of all numbers

For nine is the end of every number and these are called units that are the first rank.

• tens and their resemblance to the units

Because ten resembles to one and twenty resembles to two.

ועשרים דומים לשנים שהם שני עשרות‫^[6]

The exceptional pronunciation of the number twenty

It would be appropriate to call it "‛esrayim", as "matayim" [= two hundred] is called from "meah" [= one hundred] and "alpayim" [= two thousand] from "eleph" [one thousand], but for the following numbers that are šelošim [thirty] to tiš‛im [ninty] it has been treated in their usage.

רק בעבור חבריו הבאים אחריו שהם שלשים עד תשעים נהגו כמנהגם‫^[7]

So šelošim is derived from šaloš [= three] and so all of them.

• hundreds and their resemblance to the units

One hundred resembles to one as well as to ten; two hundred resembles to two as well as to twenty.

ומאתים דומה לשנים גם לעשרים‫^[9]

• thousand and upwards

likewise the thousand and the ten thousand, because they are the beginning of decades of the numbers that follow them, which are 1, 10, 100; 2, 20, 200.

• The sign that the nine units are building all numbers – the presentation of the multiplication of units by nine through the arrangement of the nine units in a circle

The sign for this: when you draw a circle and write nine digits around it.

You multiply nine by itself; the reason is its being a square whose length is as its width; see it and so it is:

• The square is eighty-one, so the one is on the left of the nine, which is the beginning of the units; and the 8, which corresponds to eighty in the decade, is on the right of the nine.

• If you multiply 9 by 8, the product is 72; 2 is on its left and 7, which corresponds to 70, is on its right.

• If you multiply 9 by 7, the product is 63; 3 is on its left and 6, which corresponds to 60, is on its right.

• If you multiply 9 by 6, the product is 54, 4 is on its left and 5, which corresponds to 50, is on its right.

Since the number five is mean between the 9 digits, therefore it is called round number, for it circles around itself as its square contains five.

• When you multiply 9 by 5, the matter rotates in a circle, for the units are to the right of 9 and the decades to its left; because the product is 45; 5 is on the right side of 9 and the decades are on its left, which is 4 for the 40.

• When you multiply 9 by 4, the product is 36; 3 corresponds to thirty.

• When you multiply 9 by 3, the product is 27; 2 corresponds to twenty.

• When you multiply 9 by 2, the product is 18; 1 corresponds to ten.

The Positional Decimal System

• The written numerals

Therefore, the wise men of India have denoted all their numbers by nine and they make forms for the 9 digits. I wrote instead of them: א ב ג ד ה ו ז ח ט [the first nine Hebrew letters]

ואני כתבתי במקומם א' ב' ג' ד' ה' ו' ז' ח' ט‫'‫^[23]

• The written ranks and their writing order

• tens

• units and tens

If you have a number of units and the beginning of the decades, which are tens, always write first the number of the units, then the number of the decades.

• tens

If there is no number in the units, but it has a number in the second rank, which is the tens, one puts a shape of a wheel first to indicate that there is no number in the first rank and then write the number that he has in the tens afterwards.

• hundreds

• tens and hundreds

If his decade consists of the hundreds and the tens, one writes a wheel in the first [rank], then the number of the tens in the second [rank] and the number of hundreds in third [rank].

• thousands and upwards

If one has a number of thousands [he writes it] in the fourth [rank], the number of tens of thousands in the fifth and the number of the hundreds of thousands in the sixth.

• Threefold cycle of the decimal ranks – units, tens, hundreds

For 1, 10, 100 repeat as thousands in the fourth, as thousands of thousands in the seventh and as thousands of thousands of thousands in the tenth and so on endlessly.

• units and hundreds

If there is a number of units and hundreds, but no tens, one writes the number of the units in the first [rank], a wheel in the second [rank] and the number of hundreds in third [rank].

• zero – a placeholder digit

In this way one puts at first or in the middle two wheels or as much as needed endlessly.

Table of Contents

• Chapter One: multiplication of a number by itself or by other; multiplication of one number by two numbers or more; multiplication of multiple numbers by multiple [numbers]

• Chapter Two: division of decades by units; two decades by units; large decades by small decades; decades and units by units; including a discussion on the checking of multiplication and division

• Chapter Three: addition of a number to a number; units to decades or decades to decades

• Chapter Four: subtraction of a number from a number; units from decades or decades from decades; including a discussion on the checking of addition and subtraction

Chapter One – [Multiplication]

Shortcuts

I have already mentioned what the ranks of the number are. When you have two numbers to multiply a decade by a decade, be it by itself, or by another, look for their analogous in the first rank, see how much is the product of the multiplication of one by the other and keep it. Then, see how many ranks the two numbers are, whether they are in the same rank or two [different] ranks, and know what the sum of the numbers of the two ranks is. Always subtract one for the foundation and put the reserved number in the remaining number of the ranks.

בין שיהיה על עצמו
או על אחר
בקש דמיונו במעלה הראשונה וראה כמה המחובר מכפל זה על זה ושמור אותו
ואחר כך ראה כמה מעלות שני החשבונות בין שיהיה במעלה אחת או בשתים מעלות ודע כמה המחובר ממספר שתים מעלות וגרע לעולם אחד למוסד
ובקש במספר המעלה הנשאר במספר השמור‫^{[I 1]}

• Example: we wish to multiply thirty by two hundred.

The number that is analogous to thirty is three and the number that is analogous to two hundred is two. We multiply 2 by 3, the result is six and this is the reserved number. We seek the ranks: thirty is of the second rank, which are the tens; we take two for it. Since the two hundred is of the third rank, for they are the hundreds, we take three for it and add the two to them; they are five. We subtract one to the foundation; four remain. As we already know that the fourth rank is for the thousand and the reserved number is six, the result is six thousand.

כפלנו ב' על ג' והנה עלו ששה וזהו החשבון השמור
ונשוב לבקש המעלות והנה שלשים מהמעלה השנית שהם עשרות והנה נקח לו שנים
ובעבור כי המאתים מהמעלה השלישית כי הם מאות נקח לו שלשה ונחבר עליו השנים ויהיה חמשה
נחסר אחד למוסד ישארו ארבעה וכבר ידענו כי המעלה הרביעית היא לאלף
והמספר השמור היה ששה והנה העולה ששת אלפים‫^{[I 4]}

• Another example: we wish to multiply two hundred by seven hundred.

We multiply two by seven; it is 14 and this is the reserved. two hundred is of the third rank and 7 hundred is also of to the third rank; we take for them 6 and subtract one, it is 5. The beginning of the fifth rank is ten thousand and the reserved number is 14, so we take ten thousand of this number, the result is one hundred thousand and forty thousand.

והנה מאתים מהמעלה השלישית וז' מאות גם כן מהמעלה השלישית נקח להם ששה ונחסר אחד הנה חמשה
וראש המעלה החמישית עשרת אלפים והשמור היה י"ד
הנה נקח במספר הזה עשרת אלפים ויהיה העולה מאה אלף וארבעים אלף‫^{[I 6]}

You can proceed endlessly according to this procedure.

If there are two numbers whose distance from a decade is as much as the distance of two other numbers, only that the one is deficient and the other exceeding, know how much the square of the decade, always subtract from it the square of the excess and the deficit and the remainder is the sought number.

דע כמה מרובע מספר הכלל וגרע ממנו לעולם מרובע החשבון היתר והחסר והנשאר הוא החשבון המבוקש‫^{[I 8]}

• Example: we wish to multiply 29 by 31.

^{[I 9]}

The decade is thirty, its square is 9 hundred, because three by three are nine; the excess and the deficit is one, its square one; we subtract it from the square of the decade, the remainder is the sought after, which is 899.

• Another example: we wish to multiply 66 by 54.

The decade is 60, the deficit and the excess is 6, the square of the decade is 3 thousand and 600. We subtract from it 36, which is the square of the deficit and the excess, the remainder is the sought after.

• Another example: one number is 250 and the other number 350.

The decade is 300, its square is 90 thousand. We subtract from it the square of 50, which is the deficit and the excess, and it is 2500. The remainder is the sought after.

According to this procedure we can proceed with other numbers that are similar to these, such that the deficit is the same as the excess.

Another important method that I have found is with thirds: We take one-third of the number and we know how much its square is. We take its closest higher decade and subtract from it the square of the third; the remainder is the sought after.

^{[I 17]}

• Example: we wish to know how much is the square number of 3.

We take its third, which is one; its square is one; ten is its closest decade. We subtract from it the square of one, which is the third; 9 remain and this is the sought after.

• Another example: we wish to know the square of 15.

Its one-third is 5, the square of which is 25, the analogous number in the closest decade is 250. Subtract from it the square of 5, which is the third; 225 remain.

• Another example: we wish to know how much is the square of 24.

Its one-third is 8, the square of which is 64 and its analogous number in the closest higher rank is 640. We subtract from it the square of the third, which is 64; 576 remain and this is the sought after.

If the number does not have a whole third and there is an excess of one, subtract the one from the number and calculate the sought number in the procedure that I have shown you. Add to the result the number that has a [whole] third and the [original] number itself, so the sum is the sought after.

• Example: we wish to know the square of 7.

It has no third, so we subtract from it one, which is the excess, one-third of the remainder is two, the square of which is four and the closest analogous decade is 40. We subtract from it 4, which is the square of the third; 36 remain; which is the square of 6. We add to it the 6, which has a third, and the 7 that was our number originally, both together are 13, the sum is 49, and this is the square of 7.

• Another example: we wish to know how much is the square of 22.

We subtract one; 21 remain, its third is 7, the square of which is 49 and its closest decade is 490. We subtract from it 49, which is the square of the third; 441 remain, which is the square of 21. We add 21 as well as 22 together, which are 43; the sum is 484 and this is the square of 22.

If there are 2 between our number and the number that has a third, we do the opposite by adding one to our number and know how much is the square of the number that has a third. We subtract from it the number that has a third and the number that we had, and the remainder is the sought after.

נעשה להפך שנוסיף על המספר שלנו אחד ונדע כמה מרובע מספר שיש לו שלישית ונחסר ממנו כמספר שיש לו שלישית וכמספר שהיה לנו והנשאר הוא המבוקש‫^{[I 28]}

• Example: we wish to know how much is the square of 23.

Since it does not have a whole third, we add one; it is 24, its third is 8, the square of which is 64 and the analogous number is 640. We subtract from it 64, which is the square of the third; 576 remain, which is the square of 24. We subtract from this number 47, which is the sum of 24 with 23; 529 remain and this is the sought after.

• units by units - one step

Know that if there are two digits to multiply by one another, once will be enough for you.

^{[I 31]}

• units by two-digit number - two steps

If one digit by two digits, you have to do this twice.

^{[I 32]}

• units by three-digit number - three steps

If by three [digits], three times.

• two-digit number by two-digit number - four steps

If they are two numbers by two numbers, you have to do this 4 times.

• Example: we wish to multiply 13 by 28.

We multiply 10 by 20, which is a decade, also 10 by 8; the result is 280. Then we multiply 3 by 20 and by 8; the result is 84. So the total is 364.

• two-digit number by two-digit number with the same digit of tens - three steps

If there is one decade shared by the two numbers, three times are enough for you.

• Example: we wish to multiply 13 by 16.

10 is shared by the two numbers. We add 3 with 6 and with 10, so our number is 19. We multiply it by ten; the result is 190. We multiply the two small numbers, which are 3 by 6; the result is 18. So the total is 208.

Sometimes twice alone are enough for you.

• Example: we wish to multiply 24 by 26.

20 is shared by the two numbers. We add 4 with 26, which is the greater number; the sum is 30. We multiply 20 by 30; the result is 600. We multiply the smaller [numbers] by each other; the result is 24. So the sought for is 624.

• three-digit number by three-digit number - nine steps

If you multiply 3 digits by 3 [digits], you have to do this nine times.

• If it is an even number, the product is also an even number.

The digit of the units indicates if the number is even:

The essence of the even number is in the units.

Because every decade is an even number.

• If one number is even and the other is odd, meaning that it is not even, whichever one of them is, the product is also an even number.

• If the one number is odd and the other is also, the product is also odd.

Written calculations

• The multiplicand of which the highest rank is smaller is written on the top line:

The paved way is that you put the smaller number in the top row; the meaning of the smaller number is the one whose decade is smaller; do not consider the units.

• the multiplicand of which the highest rank is larger is written on the lower line:

Put the number whose decade is greater in another lower row below.

• the number that consists of a greater number of ranks is written on the top line:

If, however, the number, whose decade is smaller, consists of more digits than the number whose decade is greater, then put this in the top row without consideration.

• After you put the digits, as many as they are, in the top row and the others in the bottom row, multiply the first from the top row by the first from the bottom row, and write the result corresponding to the first [that in the] top row.

Then, multiply the first top digit by the second bottom digit and write [the result] in the third row corresponding to the second top digit.

So on for all bottom digits with the first upper digit.

• The product of two digits is equal to units and tens:

If, when you multiply the first upper digit by its corresponding one in the bottom [row] there is a decade and units in the product, write the units in the place to which it belongs and write the number of the decade as the following digit.

• After you have finished multiplying the first digit of the top row by all the digits of the bottom row, start to multiply the second digit of the top row by the first digit of the bottom row and write the result in the third row corresponding to the second upper [digit].

Then, multiply the second [of the] top by the second that is in the bottom row and write it in the third row as the third digit, which is second to the digit from with which you have now started.

• Then, start with the third upper digit, multiply it by the first of the bottom row, and write the result corresponding to the third [digit] with which you have started.

• Then, start adding up the interim products of the top row with the bottom:

• If there are no tens there, write down the sum.

• If there is ten, write one after it.

• If there is more, write the excess aside from the sum you have and instead of the ten write one next to it.

Proceed so with all the interim products of the top and bottom rows and take out the excesses aside from the tens.

• After you know how much the sum in the third row is, count its ranks and see: if they are as much as the number of the ranks of the two upper rows minus one, know that your calculation is correct.

If the product of last digit in the top row multiplied by the last digit in the bottom row is a decade, the number of ranks in the third row is the same as to the number of [ranks of] the two upper rows without the subtraction of one.

• Example: we wish to multiply 127 by 355.

We write 127 in the top row and the number 355 beneath it, each characters in its place, like this.

1 2 7     3 5 5

2 3 3 5   1 1 5     6 1     3 5 5

the product

4 5 0 8 5

[Illustration of the procedure:]

127		127		127		127
355		355		355		355
		35		335		2335
				5		15

• We multiply 7 by 5; the result is 35. We write 5 in the first rank and 3, which is 30, in the second rank.

• We also multiply 7 by the second bottom 5; the result is 35. We write 5 in the second rank beneath the 3 and 3 in the third [rank].

• We also multiply the first 7 by the bottom 3; the result is 21. We write 1 in the third [rank] beneath 3 and 2 in the fourth [rank].

[Illustration of the procedure:]

	127		127		127
	355		355		355
	2335		2335		2335
	15		115		115
	1		1		61

• We also multiply the middle upper 2 by the first 5 from the bottom [row]; the result is 10. We write 1 in the third ]rank] beneath the 1.

• We also multiply the upper 2 by the second lower 5; the result is also 10. We write 1 beneath 2 in the fourth [rank].

• We also multiply the top 2 by the bottom 3; the result is 6. We write it under 1 in the fourth [rank].

[Illustration of the procedure:]

	127		127		127
	355		355		355
	2335		2335		2335
	15		115		115
	61		61		61
	5		55		355

• We also multiply the last upper 1 by the first lower 5; the result is 5. We write it in the third [rank].

• We also multiply 1 by the second lower 5; the result is 5. We write it in the fourth [rank].

• We also multiply the upper 1 by the bottom 3; the result is 3. We write it in the fifth [rank] after 2.

Thus, the multiplication is complete.

We add up all these numbers: all that is of the same rank together and what exceeds over ten or equal to ten, write it after that rank.

The total product is 45 thousand and 85.

וכפול ז' על ה' עלו ל"ה וג' עשרו' וג' עשרו' היו בידך בין הכל עלו ל"ח כתו' ח' במדרגה השנית ושמור ג' מאות
עוד כפול ז' על ג' שהם עלו כ"א וג' שהיו בידך היו כ"ד מאו' כתו' ד' במדרגת המאות והכ' שהם ב' אלפי' כתו' במדרגת האלפי' כי עתה אין עוד אות בטור השפל לכפול האות הראשון עמו

וכפול ב' על ה' השני עלו י' וא' שהיה שמור הנה י"א כתו' א' במדרגת המאות ובידך שמור א' שהוא אלף
עו' כפול ב' על ג' הנה הם ו' וא' הנה ז' כתו' ו' במדרגת האלפי‫'

וא' על ה' השני וכתו' במדרגת האלפי' ה‫'
וכפול א' על ג' וכתו' במדרגת העשרת אלפי' ג‫'
וראה ועשה

Chapter Two – Division

Introduction - Preliminary definitions

^{[II 2]}

• Every number is half of the sum of the numbers on either side - one is half the sum of its one side

It does with one side what every [other] number does with its two sides.

For two is the preceding side for the number three and four is the other following side, and the sum of the two sides is 6, which is double the number three, and so for every number.

But the one has no side that precedes it and following it is one side, which is two, and it is double the one.

They did this because the solar year has 12 lunar months and because there is no number, smaller than 12, that has as many parts as it, for it has whole integers in its half, its third, its quarter, its sixth and its half of one-sixth.

Because this number has more whole units than 12; for it has one-half, one-third, one-fifth, one-sixth and one-tenth.

So, the number of degrees of the celestial sphere is 360.

This number has one-half, one-third, a quarter, one-fifth, one-sixth, one-eighth, one-ninth and one-tenth only the seventh is missing.

When you multiply this number by 7, the result is two thousand and 520, and this number includes all fractions up to tenths.

• When the astrologers multiply degrees by degrees, the product is degrees, which are whole units.

• Similarly, when they divide degrees by degrees, the quotient is degrees, which are whole units.

Written calculations

• Example: we wish to divide 9 thousands by 70.

According to the following diagram.

0 0     2 6 4   9 0 0 0   1 2 8     7 0

[Illustration of the procedure:]

				0		00
		2		26		264
9000		9000		9000		9000
		1		12		128
70		70		70		70

• We put 70 in the bottom row according to its ranks and consider all as units. We give it 1.

We write it in the second rank back from the last digit of the first row, because the 70 is in the second rank in the lower row and we write it in the middle. We are left with two.

• We return it back, so it is twenty. We divide [them] by 7, we give it 2 and write it back before what was written first; we are left with 6.

• We return it back, so it is sixty. We divide them by 7, we give it 8. Write it back.

We are left with 4, which is in the second rank, so it is 40 and the divisor is greater than it.

• Example: we wish to divide 20 thousands by 90.

According to the following diagram.

0 0     0 2 2 2   2 0 0 0 0     2 2 2       9 0

[Illustration of the procedure:]

				0		00
		02		022		0222
20000		20000		20000		20000
		2		22		222
90		90		90		90

• Since the number 9 is greater than 2, we return it back, so it is 20. We divide it by 9; the result is 2. We write it back in the third rank, which is the second to the digit that we divided; 2 remain.

• We return them back to the third rank, so they are 20. We divide them by 9; the result is 2, and 2 remain.

• We return them back to the second rank, they are twenty. We divide them by 9; the result is and 2 remain, which are twenty, because they are in the second rank in the top row. This number is smaller than our number, so we write back a zero, because no units resulted, and it cannot be taken out completely.

• Example: we wish to divide 4 thousands and 32 by 30.

0 0     1 1 1   4 0 3 2   1 3 4     3 0

[Illustration of the procedure:]

				0		00
		1		11		111
4032		4032		4032		4032
		1		13		134
30		30		30		30

• We divide 4 by 3; we get 1. We write it back in the rank of the hundreds, because it is second to it; we are still left with 1.

• We return it back to the rank of the hundreds; so, they are 10. We divide them by 3; the quotient is 3, 1 is left to us.

• We return it back to the rank of the tens; so, they are 10. We add them to what is written in the second rank; they are 13. We divide them by 3 and give it 4; so, we are left with 1.

• We return it back to the first rank; they are 12 that cannot be divided, because the remainder is less than the divisor, and everything is already taken out.

• Example: we wish to divide 8213 by 353.

0     0 1 9   1 2 0   2 1 5 4 8 2 1 3     2 3   3 5 3

[Illustration of the procedure:]

				1		1
		2		2		215
8213		8213		8213		8213
		2		2		2
353		353		353		353

• The 94 that we wrote is the remainder of the number that cannot be divided.

• When we divide 8, which is the last [digit] of the upper row, by 3, which is the last [digit] of the lower row, we give it two, and because the 3 of the lower row is third, we return it back to the third [digit] from there, and it reaches the rank of tens. We are left with two on the digit 8.

• We leave there one, because one is enough for us, and return the one back to the two; they are 12. We subtract from them 10, which is double five, which is the middle in the bottom row, so 2 are left on the 2.

• We leave there one and return one back on the one, which is third; they are 11. We subtract from them 6 for 3, which is first in the bottom row; so, 5 remain.

Thus, all 3 lower [digits] have what they are entitled to.

We divide again, because are left with 1 on the 8, 1 on the 2 and 5 on the 1.

[Illustration of the procedure:]

					0
	0		01		019
	12		120		120
	215		215		2154
	8213		8213		8213
	23		23		23
	353		353		353

• We return the 1, which is on the 8 back on the 1 above the 2; they are 11. We divide them by the 3, which is in the bottom row; they are 3. We write it corresponding to the first digit, which precedes the two; so, we are left with 2.

• We return them back on the 5; they are 25. We give 15 to the 5 that is middle in the lower row; 10 remain.

• We return one back on the 3; 9 remain. The 1 with the 3 are 13; we take from them 9 and 4 remain.

• The 9 on the 5, because we cannot take the 3 out of the 10 first, although it was enough for it, for it is not the same rank now; although it was taken first, it was first third to it, since the last of the bottom row took from the 8, but now it took from the 2, and since it is third, it must take from the same rank. So, the remainder is 94.

• Another example.

2       3 0   0 5 1   3 6 0 5 9 3 8 1     3 1   2 9 6

[Illustration of the procedure:]

				0		05
		3		36		360
9381		9381		9381		9381
		3		3		3
296		296		296		296

• We want to divide 9 by two, but we cannot give it 4, because then only 1 remains, and when you return it back, with the 3 they are 13, but 4 times 9 is 36. So, we give it 3; 3 remain.

• We return all back to the second to it, which is 3; they are 33 there. We give 3 to the 9; they are 27; 6 remain on the 3.

• We take one from them and leave 5. We return it back on the 8, which is the third from the last of the top row; with the 8 they are 18. We give all to the 6, which is the third of the bottom row. We write a zero on the 8, because there is nothing left on the 8.

[Illustration of the procedure:]

			2		2
	3		3		30
	05		051		051
	360		360		3605
	9381		9381		9381
	31		31		31
	296		296		296

• We divide again, because not everything has been taken out yet. We take two, from the 5 that we left on the 3, which is the second in the top row; [the multiple] is one. We write this one on the 6 after the 3, which we put on the 9 in the first division, which is the third in the bottom row, then everything has already been removed.

• We take one from the 3 that is above the 3; two remain. We give the one to the zero; they are 10. We give 9 to the 9 that is in the bottom row; 1 remains on the zero.

• We return it back on the 1, which is the fourth foremost in the top row; they are 11. We give it 6; 5 remain on the 1.

There remainders on the top row are: 5, zero and 2, which are 205. These cannot be divided further, because the 3 [digits] in the bottom row are 296. So, the taken for each [= the quotient] is 31.

• Another example.

1 0     0 8 1 8   2 4 4 2   3 5 6 4   5 4 0 9 3       1 8   2 9 4 5

[Illustration of the procedure:]

				2		24
		3		35		356
54093		54093		54093		54093
		1		1		1
2945		2945		2945		2945

	24
	3564
	54093
	1
	2945

• We want to divide the top row by the bottom row. We cannot give it 2, because then only 1 remains and with 4 they are 14, but we have to divide it by 9 twice. So, we only give it 1 and write it corresponding to the 9 that is in the top row, which is fourth to the 5 back, as the two of the lower row, which is fourth to the first 5 in the bottom row. 3 remain on the 5.

• We take one from them; 2 remain on the 5. We return it back on the 4; they are 14. 9 are taken; 5 remain.

• The 4 in the bottom row must take from the third in the top row, because it is third, but we cannot, because the third in the top row is a zero. So, we return one from the 5 that we left on the 4 back on the zero; they are 10. 4 are taken; 6 remain on the zero.

• The 5 that is the beginning of the bottom row, which is fourth, takes from the fourth in the upper row, which is 9; 4 remain on the 9.

We divide again, because not everything has been taken out yet. The remainders from the top row are the 3, which is the first and fifth, 4 on the 9, 6 on the zero, 4 on the 4, and 2 on the 5.

[Illustration of the procedure:]

			1
	08		08
	24		244
	3516		3516
	54093		54093
	18		18
	2945		2945

	1		10
	081		0818
	2442		2442
	3516		3516
	54093		54093
	18		18
	2945		2945

• We return the 2 back on the 4; they are 24. We give 8 to the 2, which is the fourth in the lower row; 8 remain on the 4.

• We return 7 of them back on the 6, and 1 remains on the 4; they are 76. We take 72 for the 9; 4 remain instead of the 6.

• If we return 3 of them, the 4 in the lower row indeed can take from the 3 together with the 4 that follows it, which are 34, but only two will remain, and when we return it to the 3, they are only 23, whereas the 5 have to take 40.

Therefore, we return the whole 4 back to the 4 and write a zero instead of the 4 corresponding to the zero in the upper row; they are 44. The 4 take 32; 12 remain.

• We take 4 from them, because less is not enough for us; 8 remain on the 4 that are on the 9. With the 3 they are 43. The 5 take from 40.

3 remain on the 3, 8 on the 9, which are 80, a zero, which takes the number out of the hundreds and into the thousands, and 1 on the fourth that is on the 4, which is one thousand. These cannot be divided, because the divisor is greater than it; this is 2 thousand 9 hundred and 45.

• Another example: we wish to divide 68 thousands, 9 hundred and 21 by 7 thousands and 53.

This is the diagram:

4   0 5 4 7 4 6 8 9 2 1         9   7 0 5 3

[Illustration of the procedure:]

						4
		05		0547		05474
68921		68921		68921		68921
		9		9		9
7053		7053		7053		7053

• The 7, which is fourth in the bottom row, cannot take from the 6, which is the fifth in the top row, since we cannot give the 7 what it needs.

• We return all of it back on the 8; they are 68. So, from the 8, which is the fourth in the top row, we give 9 to the 7; we take it out on the 3, which is the fourth in the bottom row; so, 5 remain on the 8.

• The second to the 7 does not take anything because it is a zero.

• The 5 must take from the 2, the fourth in the top row, which is the third in the division, but cannot take it.

• So, we take 5 from the 9 that follows it. We return it back on the 2; they are 52, and 4 remain on the 9. The 5 take 45 out of 52; 7 remain on the 2.

• We take 3 from the 7 and return them back on the 1; they are 31. The 3 take 27; 4 remain on the 1.

• Another example: we wish to divide 680402 by 2009.

0   3       1 1 4 6   0 7 7 7 3 0 6 8 0 4 0 2       3 3 8     2 0 0 9

Digits by digits so that there are 2 zeros in the top row and in the bottom row.

• We give the 2 that is the fourth in the bottom row, 3 from the 6 that is the sixth in the top row.

The zero that is the second in the bottom row has to take from the 8, but it takes nothing.

The zero that is the third in the bottom row has to take from the zero that is the third in the top row, but cannot take.

• The 9 that is the first in the bottom row must take from the 4 in the top row but cannot. Therefore, we have to return from the 8 that is the second in the top row, what is enough for it. We return 1 back, because it is enough for us. We write 7 on the 8, and the 1 that we have returned back on zero; it is taken as tens for us. That is still not enough. We take from them 3, return them back; 7 remain on the zero, and with the 3 they are 30 on the 4.

So, 7 remain instead of 4, as well as zero and two that are the first in the top row, 7 on the zero, and 7 on the 8.

• We divide again, because not everything has been taken. We take 3 from the 7 that is on the 8 and put them beneath the first zero, which is fourth from the 8.

1 remains on the 8.

• The zero has to take the from the zero but cannot.

• The zero that follows is has to take from its rank, which is the 7 on the 4 but cannot.

• The 9 must take from the zero that is in its rank, because it is the fourth in the division, but cannot, since there is nothing on the zero and it is impossible to return it back on the two, because the two is not in its rank. So, we return 3 from the 7 that is on the preceding 4 back to zero; they are 30.

3 remain on the zero, as well as 4 on the preceding 4, 7 on the zero that precedes the 8 and 1 on the 8.

• We divide again because it is not all taken yet. The last in the lower row cannot take from the 1, so we return it back on the 7 that is on the zero; they are 17. We give it 8, take it out after the 3 that is the fourth in the division, because we divided from 7 that is on the zero.

1 remains there.

• The zero has to take from the 4, but cannot take.

• Also the other zero has to take from the 3 that is on the zero that is in the top row but cannot take.

• The 9 has to take from the 2, but cannot, so we return back. If we say to the 3 that is on the zero that to give to the 2 that is first to the zero, it does not have enough, because it only has 3. So, we take one from the 4, which replaces the 4, it is on the zero, with the 3 they are 13. We take 7 from them and return them back on the 2; they are 72. All is taken out by the 9.

We write a zero on the 2, because there is nothing left on it; 6 are left on the zero that is second to the 2, 3 on the 4 and 1 on the zero that is second to the 8. These remain as they cannot be divided, because the divisor is greater from them. The remainder is one thousand 3 hundred and 60 and the divisor two thousand and 9.

• Another example, which is more valuable and difficult than all previous calculations without zero.

Like the following and this is the diagram of the valuable and difficult:

• Example: we wish to divide 9 sevens by 4 nines.

0               0 1               2 5 5           0 8 6 6           1 9 3 0           7 5 5 5 5       0 8 8 6 6 6       1 4 9 2 3 0       7 7 5 5 5 5 6   0 8 8 8 1 1 2 0   1 4 4 4 4 4 4 5 2 7 7 7 7 7 7 7 7 7         7 7 7 8 5           9 9 9 9

First, I shall teach you how to divide after you see that the 7 is less than the 9.

• You have to return it back. This is the diagram. First, when you start dividing, give 7 to the 9 and write it back in the rank of the 4, which you are dividing, that is second to the 9.

The remainder is 14. Proceed so, until 7 remain in the first rank of the division, likewise 7 in the second [rank] of the division, 8 in the third of the division and 4 in the fourth of the division.

• We divide again: we return the 7 that is in the second [rank] back to the third. We divide from there and write 7 beneath the sixth 7, which is fourth of the division; the remainder is 14. Proceed so, until 7 remain, after it we write 8, then 5 and then 4.

• We divide again: we return the 7 that on the 8 and put it in the fourth rank; the remainder is 15 and what you leave is: 8, then 5, then 5 and then 4. Write 7 beneath the third that is the fourth of the division.

• We divide again: we return the 8 back to the 5 and put 8 beneath the eighth, which is 7 that is the fourth of the division. 5 remain on 5, then 5, then 5 and then 5.

• We divide again: we return the 5 back to the 5 and put 5 beneath 7 that is the ninth, which is the fourth of the division; the remainder is 10 and what you leave is: 5, then 5, then 6 and then 2.

^{[II 132]}

It cannot be divided further, because everything has already been removed with the 5 and because the remainder is 5 thousand 5 hundred and 62 and the divisor is 9 thousand 9 hundred and 99. The received number is 77 thousand 7 hundred and 85.

In total of the division of the number occurred: three 7, three 8, three 4, nine 5, one 6 and one 2.

The rule is that in all the divisions there were 4 [digits in the dividends] except the last one, where it decreased to 3 [digits].

• casting out by 9

• Know the scale of the number by which you have divided [= the divisor], whether it consists of one or more digits.

• Know also the scale of the quotient that you wrote between the two rows, whether it consists of one or more digits.

Multiply them by each other, the middle one by the lower one and know how much remains [after subtracting] the nines; this is the reserved, if you have no remainder that remains, because it is smaller than the divisor.

For, if there is a remainder, take its scale and add it to the reserved that you have; the sum is the actual reserved.

Look at the scale of the number that you have divided [= the dividend] that was in the top row, if it is equal to the scale of the reserved, you know that your calculation is correct.

• inverse operation - multiplication:

If you multiply the quotient by the divisor, then add to them the remainder from the division, then the sum is equal to the digits of the top row and the division is correct.

• Example: we wish to divide 83521 by 903.

• The 9 in the bottom row is greater than 8 that is the end of the top [row], so you return the 8 back to the 3; they are 83. We give 9 of the 83 to the 9; they are 81 and 2 are left on the 3.

• The zero cannot take from the top 5.

• The bottom 3 cannot take from the top 5, because it is not of the same rank, so return three from 5 back to 2; they are 32. We subtract 27 from them, which are 9 times 3, which is the first of the lower [row]; 5 remain on the 2 and 2 on the 5.

Since the lower 9 is in the third rank and 9 has already resulted in the division, we write it in the second rank, which is the third to the top 3 which you divided.

• We divide again and return the 2 that is on the 3 back to the 2 that is on the 5; they are 22. We subtract 9 twice from them, because 9 is the lower [digit]; 4 remain on the 5. We write the 2 in the first rank, which is the third to the top 5, which you divided in this division.

• The zero cannot take anything from the 5 that is on the 2.

• The lower 3 cannot take from the upper 1, because there is only one in it, so we return one from the 5 that is on the 2 back to the 1; they are 11. We take from them 6 for the lower 3, which are 2 times 3.

4 remain on the 5, 4 on the 2 and 5 on the 1, which cannot be divided.

• Another example: we wish to divide 11350 by 110.

• We take 1 from the last top 1 and write it in the third rank beneath the top 3; because the last lower 1 is in the third rank, so we remove the result from the last digit from which you started dividing.

Take also the second bottom 1 from the fourth upper 1.

• Since nothing remains in the fourth rank, we have to divide again from the top 3. We take from it 3 for the last bottom 1.

We write 3 in the first rank, which is the third for the upper 3 from which we started now to divide. Hence, we put a zero in the second rank.

This is because in the first division we started from the last [digit] of the upper [row], therefore we write the quotient at a distance of 3 ranks backwards, as much as the number of ranks of the last bottom [digit].

However, in this second division we started dividing from the top 3, so we write the result 3 ranks backwards and it arrives to the first rank.

For the second 1 in the bottom row, we take from the top 5; 2 are left on the 5 and these are twenty that cannot be divided.

So, the quotient is 103.

Chapter Three – Addition

Sums

• Example: we wish to know how much are the numbers that are summed from 1 to 11?

We know that the half of 11 is 5 and a half, we add to it one half, it is 6, then we multiply 11 by 6, they are 66 and this is the sum.

• Another example for an even number [of terms]: how much is the sum up to 18?

We take its half, which is 9, multiply 18 by it and the result is 162, then we have to multiply 18 again by one half, the result is 9, add it to 162, the result is 171 and it is the sum.

• Example for an odd number [of terms]: how much is the sum up to 11?

We add 1 and they are 12, half 11 is 5 and a half, we multiply 12 by 5 and a half, the result is 66 and so is the sum.

• For an even number [of terms]: to 18.

Its half is 9, we add one to 18, they are 19, we multiply 19 by 9, the result is 171.

• Example: we know that the square of 11 is 121, then we add to it 11, which is the root and it is the last number, the result is 132 and its half is 66.

• Example: one asks: I sum all the numbers until they reach a known number and the sum is 465. How much is the last number?

Hence, we double 465 and the result is 930. It is known that the preceding square is 900 and its root is 30, so it is the last summed number, for between the square and the double there is none but 30, thus it is the last number.

• Example: we wish to know how much is the sum up to 12?

Half the number is 6, the square is 36, double it, they are 72, add to it 6, which is the root, the result is 78 and it is the required.

• Example: we wish to know how much are the squares up to seven?

We already know that the sum is 28, thus we turn to know how much are two thirds of a seven plus one third - they are five, for two thirds of a six are four and we have two thirds more plus one third, so we multiply the sum by 5 and the result is 140 which is the required.

• Another example: how much is the sum of the squares up to 12?

It is known that its sum is 78 and its two thirds are 8, we multiply it by 78, the result is 624, we add to it a third of the sum, which is 26, the result is 650 and it is the required.

• Example: we wish to know how much is the sum of the cubes up to 5?

The sum up to 5 is 15, take its square, it is 225 and it is the [sum] of the cubes up to 5.

• Example: the sum of 3, 6, 12

They are 24 minus 3, thus 21 is the sum.

• Another example: 1, 2, 4

They are 8 minus 1, thus 7 and this is the sum.

• Such as, if one asks: how much are the doubles up to 32?

Double 32; it is 64, subtract one, which is the first; 63 remain and this is the required.

• If it is said: how much is the sum of the squares up to 8?

• Question: how much is the sum from 4 to 9?

Written calculations

Addition of Sexagesimal Fractions

• One according to the solar years, in which the years of the decades are summed twenty by twenty.

• The second way according to the lunar years, in which the years of the decades are summed thirty by thirty.

For, if the degrees are 17, the minutes are those that have passed from the 18th degree. Suppose the whole minutes are 15 and say the seconds are 45, so they are three-quarters of one minute of sixteen minutes.

Chapter Four – Subtraction

Written calculations

• Example: The top row is 5432 and the bottom row is 2379.

Each [digit] of the bottom should be subtracted from the top [digits], each one from its rank: 2 from 5 and 3 from 4.

But we cannot subtract 7 from 3, nor 9 from 2.

• So, we subtract 2 from 5; we are left with 3; we write it beneath the fourth rank.

• We subtract 3 from 4; we are left with one, but we do not write it down, we only put a zero instead, because we have to return one back, for the preceding digit that is in the lower row is greater than the corresponding upper one. So, they are 13. We subtract 7 from it; 6 remain. However, because the first digit in the bottom row is greater than the [first digit] in the top row, we have to return one back and write 5 in the third row. So, there are 12 above; we subtract 9, 3 remain.

This is the diagram:

Chapter Five - Fractions

• The prime numbers are two, three, five and seven.

• The composite numbers are four, six, eight and nine.

• The multiplication of one-third by one-third is one-ninth.

• The multiplication of one-quarter by one-quarter is 16 and the product is one, which is one-half of one-eighth.

Such as one part of 11 multiplied by one part of 11; it is one part of 121, which is the square.

• Example: we wish to multiply 3 quarters by 3 quarters.

• common denominator: The denominator is 4.

• We take 3 for each of the 3 quarters, so the product is 9

• We divide it by 16, which is the square of the denominator; it is one-half and half the eighth.

If you want, divide 9 by 4, the result is the same, because one-quarter of the quarter is half the eighth.

• Example: we want to multiply 3 fifths by 4 fifths.

• common denominator: The denominator is 5.

• We take 3 for the 3 fifths and 4 for the 4 fifths, we multiply 4 by 3, the result is 12 and this is the product

This is 2 fifths of the square and two-fifths of one-fifth.

• One says: multiply 2 thirds by 3 quarters.

• common denominator: We are looking for a denominator for both: we multiply 3 by 4 and this is the denominator

• We take 8 for the 2 thirds and 9 for the 3 quarters, we multiply 8 by 9, the result is 72 and this is a half of 144, which is the square of the denominator.

• The result of multiplication is one half

If you multiply 2 by 3, it is also half the denominator, which is 12.

Example. We have taken the one denominator 3, because it was said thirds, and the other denominator 4, because it was said quarters. We multiply the one denominator, which is 3, by the other denominator, which is 4, the result is 12 and it is the required, because we take the ratio of the product to it. We take two for the 2 thirds, because we have taken it from the 3, and three for the 3 quarter, because we have taken it from the 4. We multiply 2 by 3, the result is 6, which is half the product of the denominators.

• Question: how much are 4 sevenths multiplied by 7 ninths?

• common denominator: We seek for a common denominator: it is 63 by the multiplication of 7 by 9

• Its 4 sevenths are 36, for the seventh is 9 and its 7 ninths are 49, since the ninth is 7, we multiply 36 by 49, the result is 1764.

• The square of the denominator is 3969

When we divide our first number by 63, the result is 28, which are 4 by7, and they are 4 ninths of 63, or if you want to say that they are 3 sevenths and one-ninth of a seventh.

• If we calculate with 2 denominators, the product is 63 and our result from the multiplication is 28, so it is the same thing.

• Such as 2 thirds and 5 sixths and 4 sevenths.

• common denominator: Take one denominator for them: it is by multiplying 3 by 6; the result is 18. We multiply 18 by 7; the result is 126 and this is the denominator.

• 2 thirds of 126 are 84.

• 5 sixths are 105.

• 4 sevenths are 72.

We multiply 84 by 105 and the product by 72. The result of their division is a part of the square of 126; the quotient is 40.

• Example: We multiply 2 by 5, the product is 10. We multiply 10 by 4; the result is 40 and it is a part of 126, which are 2 sevenths and 2 ninths of the seventh.

• Or you proceed like this: Multiply 3 by 6; it is 18 and this is the denominator. Multiply 2 by 5; they are 10. We take 4 sevenths from 10; they are 2 sevenths and 2 ninths of the seventh.

• Or multiply 6 by 7; they are 42 and this is the denominator. Multiply 5 by 4; they are 20. We take 2 thirds from 20; they are 2 sevenths and 2 ninths of the seventh.

• Or multiply 3 by 7; they 21 and this is the denominator. Then we multiply 2 by 4; the result is 8. We take 5 sixths and the result is a part of 21.

• Example: we wish to multiply 4 integers by 3 fifths.

• common denominator: The denominator is 5.

Since we have 4 integers, we take 20 for them. We multiply 4 by 3 and divide by 5; the result is 2 integers and 2 fifths.

• Example: we wish to multiply 4 integers and 2 fifths by 5 integers and 3 fifths.

As follows:

• First, we multiply the integers by the integers; the result is 20.

• Then we multiply 4 by 3; they are 12 fifths

• Also 2 by 5; they are 10 fractions of the same kind.

• So, 22 fractions.

Then, we multiply the fractions by fractions: 2 by 3; the result is 6 and they are fractions of fractions, in the third rank. We divide the integers by 5, which is the denominator; the result is one fraction and one remains in the third stage, which is a fraction of a fraction. We add the fraction we received to the 22 we had; it is 23. We divide it by 5; the result is 4 integers, and 3 remain. So, the integers are 24 and the fractions are 3, which are 3 fifths that are 15 of 25, which is the square and the fraction of the fraction, which is one-fifth of a fifth and these are 16 of 25.

The other method is:

• We take 20 for the 4 integers, add to them 2, which are the fractions; the result is 22 fifths, and this is the one number.

• The second in this way also is 28.

We multiply one by the other and divide the product by the square, which is 25; the result is 24 and remain 16, which cannot be divided.

• Another example: we wish to multiply 3 integers and 4 fifths by 2 integers and 3 fifths.

as follows:

• First, we multiply integers by integers; they are 6.

• Then we multiply integers by the opposite fractions:

• 3 by 3 are 9

• 2 by 4 are 8

• The total is 17 fractions of the upper rank.

• We multiply fractions by fractions, 4 by 3; they are 12 fractions of fractions of the second rank.

We divide them by 5; we get 2 fractions, and 2 fractions of fractions remain that cannot be divided. We add the quotient to the fractions, which are 17; they are 19. We divide by 5; the result is 3 integers. We add them to the integers that are 6; they are 9 and 4 remain, which are 20 fractions of fractions, and with the two we had, they are 22.

So, the total is 9 integers and 22 fractions of 25.

• According to this way: the one is 3 integers and 4 fifths and the other is 6 integers and 7 eighths.

as follows:

• We multiply integers by integers, 3by 6; the result is 18.

• We also multiply these 3 by the fractions of the number that are 7; the result is 21. We divide it by 8, because there are eighths; the result is 2 integers. So, 20 integers, and 5 eighths remain. We know that the denominator is 40, because you multiply 5 by 8, and these 5 eighths are 25 parts, for you multiply 5 by fifths, which are 5.

• We multiply again the 6 by 4 fifths; the result is 24. We divide it by 5; the result is 4 integers. So, the integers are 24, and 4 fifths remain that are 32 parts. When you multiply 4 by 8, we add to them the 25 parts that we had, the result is 57 parts. We make one integer of forty; they are 25 integers and 17 remain. Then, we multiply 4 by 7; the result is 28 parts. We add 17 to them; they are 45. We give a integer from the 40; the integers are 26 and the remainder is 5 parts that are one-eighth.

• Since the fractions are fifths and eighths, the denominator is forty. We take 120 for the number 3 and 32 for the 4 fifths, so the number is 152. Here is the diagram of the digits. We take 240 for the 6 integers, add to them 35; which are 7 eighths, so the second number is 275. We multiply one by the other; the result is 41 thousand and 8 hundreds; this is the figure.

We divide it by one thousand and six hundred, which is the square of the denominator; the result is 26 integers and 200 remain that cannot be divided.

Seek how much is 200 of the square, which is one thousand and six hundred; it is its eighth.

So, we divide 200 by 40; the result is 5, which is its eighth.

We set the one number, which is 3, as 15; add 4 to it; so, the first number is 19. We take 48 for the six; add 7 to it; the result is 55 and this is the second number. We multiply one by the other; the result is one thousand and 45. We divide it by the product, which is 5 by 8 that is 40; the result is 26 integers and 5 remain, which are one-eighth.

• Example. How much is three-sevenths by 5 parts of 11? For if is the integer that man cannot express.

• We multiply the fractions by the fractions; the denominator is 77.

We now have to beware that the fractions are reversed, because each of the 7 become 11 and each of the 11 become 7.

We take 33 for the 3 sevenths, and we take 35 for the 5 parts of 11. We multiply these by the others; the result is one thousand and 155. We divide them by 77, so that we know how much are the fractions resulting from this product in ratio to 77, which are the integer that is 15; that is a little less than one-fifth out of 77.

In a very precise way: we divide these 15 by 11, which is the seventh; the result is one-seventh and 4 parts of 11.

If we divide by 7, the result is 2 parts of 11 and one-seventh of the part.

• Another example for 2 fractions that the man cannot express.

We take the one number 13 and the other 19

We want to multiply 9 parts of 13 by 17 parts of 19.

• common denominator: First, we look for the denominator by multiplying 13 by 19; the result is the number 247 and it is the denominator.

• Then, we multiply 9 by 19; the product is 171.

• We multiply 17 by 13; the product is 221.

• We multiply one by another; the product is 37 thousand and 791.

We divide it by 247; it is 153. The product of 9 by 17 is also 153; the ratio of it to 247 is as the ratio of the first number to the square of 247 and this is the required.

If you want to make it more precise, divide 153 by 19; they are 8 parts of 13 and one part of 19 of 13.

Or if you want, divide it by 13; they are 11 parts of 19 and 10 part of 13 of 19.

• Example. We multiply 2 thirds of one-quarter of one-fifth by [4] sevenths of one-eighth.

The fractions are numerous; but I shall teach you a short method how to proceed:

• Know that since you have sixths and an eighth, there is no need for a third and a quarter. So, we are looking for a number that has a fifth, a sixth, a seventh and an eighth.

We multiply 5 by 6; they are 30; also 30 by 7; they are 210; also 210 by 8; the result is one thousand 6 hundred and 80.

• We look for the first number. One-fifth of the denominator is 336; its quarter is 84; its 2 thirds are 56 and this is the one number.

• We already know that the eighth is 210; its seventh is 30; since they are 4 sevenths the second number is 120.

We multiply one by the other; the number is 10 thousand and 80. We divide this by one thousand 6 hundred and 80; the result is 6 and these 6 are one-fifth of one-eighth of one-seventh, for the seventh is 240, the eighth is 30 and the fifth is 6.

• Example: We wish to divide 3 integers and two-fifths by 2 integers and 4 sevenths.

• common denominator: The denominator is 35.

The 3 integers are 105 and the 2 fifths are 14; so, the number is 119. We turn the 2 integers into 70 and the 4 sevenths into 20; they are 90. We divide 119 by it; the quotient is one integer and 29 remain, which are two-ninths and one-tenth.

• Example of fractions alone: divide 7 ninths by 2 sevenths.

• common denominator: The denominator is 63.

• Its seven-ninths are 49.

• Its 2 sevenths are 18.

We divide one by the other; the result is 2 and 6 ninths and one-half of one-ninth.

• We add 2 fifths to 5 sevenths. How much is the sum?

• common denominator: The denominator is 35.

• Its 2 fifths are 14.

• Its 5 sevenths are 25.

• We sum them; the sum is 39.

We take one integer for the 35; 4 remain, which are 4 fifths of one-seventh that are one of 35.

• Question: we add to an amount its ninth and its tenth and they are 50.

• common denominator: We set the denominator as 90.

It is known that its ninth and its tenth are 19, we add them to 90; they are 109.

We also convert the 50 into ninths; they are 4 hundred and 50. We convert the 4 hundred and 50 into tenths; they are 4 thousand and 500.

We divide this by 109; the result is 41 integers and 31 parts of the divisor.

• Check: We check whether it is correct.

• We know that one- tithe of 40 are 4 integers.

• We take 109 for the remaining one, add to it 31 that exceed over the integers; the result is 140. Its tenth is 14 and these are the parts of the tenth that exceed over the integers.

• We take also the ninth of 36; they are 4 integers.

• 5 integers remain. We consider each of them as 109; they are 545. We add to them the 31 that exceed over the integers; they are 576. Its ninth is 64.

We add to them the 14 that resulted from the tenth; they are 78.

We also add to them the exceeding 31; they are 109, which is 1 integer.

The total is 50 integers.

• Another question: we take one-fifth of an amount, plus its seventh and its. How much is it from the value of the amount?

You can solve this problem in two ways.

• The first: the ninth is smaller than the other fractions; we consider them as 3 ninths. The difference between the fifth and ninth is 4 and the difference between the fifth and the seventh is 2. We now multiply 2 by 4; the product is 8. We convert the 7 into one-ninth; they are 4 ninths and one remains. The difference between the seventh and the ninth is 2. So, the amount is 4 ninths and 3 parts that are 3 fifths of one-seventh of one-ninth.

• The other way: we look for the denominator by multiplying 5 by 7; they are 35, and multiplying this by 9; the product is 315 and this is the denominator.

Then we add one-fifth of this denominator, its seventh and its ninth; they are 143. We divide them by 35; the result is 4 ninths and 3 parts of 35 remain, because 35 is the ninth and the 5 is one-seventh of one-ninth, therefore 3 parts are 3 fifths of one-seventh of one-ninth.

• Another question: an amount, we add to it its half, its third, its fifth and its sixth. The total sum is 40, how much is the amount?

We know that the half, the third and the sixth are one integer.

We think as if it was one, so it is two and we have the addition of the fifth.

We have to divide 40 by 2 and one-fifth, so the quotient is the amount. We take 5 for each of the integers, add to them 1, which is the fifth; they are 11. We also multiply 40 by 5, so that all is of one kind; they are 200. We divide them by 11; the result is 18 integers and 2 parts of 11.

• Check: We check this.

• We take its half, which is 9 integers and one part of 11.

We have 27 integers and 3 parts of 11.

• The third is 6 integers and 2 thirds of one.

So, we have 33 and 3 parts and 2 thirds.

• We take the fifth of 15, 3 integers.

So we have 36 integers.

• We still have to take one-fifth of 3 integers, so we take for each 11 and add to them the additional 2 parts; they are 35 and their fifth is 7.

We add to them the 3 parts and 2 thirds that we had.

• We take one-sixth of 18; they are 3 integers.

We add them to 36; they are 39 integers.

• Two-sixths are one-third

We add it to 10 parts and 2 thirds of a part that we had; they are 11 parts and it is one integer.

The total is 40.

We take one for their number and one for him; it is two; and a half of us; it is two and a half. We add a quarter of us; they are two and 3 quarters. Since we have quarters, we take 4 for each integer; they are 8. We add to them the 3 quarters, they are 11.

Since they said that they are one hundred with him, their number together with the addition is 99. We convert it into quarters; they are 396, divide them by 11; they are 36 and this is their number.

• Question: A man buys 100 liṭra for 100 zehuvim. He sells 50 [of them] at 1¼ liṭra for one zahuv, and the other 50 at (1‒¼) liṭra for one zahuv. We want to know: did he earn or lose?

We convert the first 50 into 200, because they are quarters, divide them by 5, because he sells one liṭra and one-quarter of a liṭra for one zahuv; they are 40 zehuvim. We also multiply the other 50 by 4; they are 200, divide them by 3, because he sells 3 quarters for one zahuv; they are 66 zehuvim and two thirds of one zahuv. We add the 40 to them; the profit is 6 zehuvim and two thirds.

• Another question: A man buys three fifths of a liṭra for one pašuṭ, then he sells four sevenths of a liṭra for one pašuṭ and he earns one pašuṭ. How much money did he have originally?

common denominator: Find the denominator, it is 35, which is the product of 5 by 7.

Its 3 fifths are 21.

Its 4 sevenths are 20.

The amount was 20 and the number of liṭra was 12.

• Question: a man buys four sevenths of a liṭra for one pašuṭ, then he sells five ninths of a liṭra for one pašuṭ and he earns one pašuṭ. How much money did he have originally?

It is known that 4 sevenths are more than 5 ninths.

common denominator: The denominator is

Its 5 ninths are 35

Its 4 sevenths are 36.

• Check: You can check it out. Since he bought 4 sevenths of a liṭra for one pašuṭ and his money is 35, he has 20 liṭra. Convert them to ninths; they are 180, divide this number by 5, because he sold 5 ninths for one pašuṭ; you get 36.

• If he said that he had earned 2 pešuṭim.

Multiply them by 35; they are 70.

• If he said 3 pešuṭim.

Multiply 35 by 3.

• Question: a man buys 9 parts of 17 of a liṭra for one pašuṭ, then he sold 10 parts of 19 parts of a liṭra for one pašuṭ and he earned one pašuṭ. How much money did he have originally?

common denominator: Search the denominator; it is 323.

It is known that 9 parts of 17 are 171.

10 parts of 19 are 170 and so was the amount.

The number of liṭra is 90.

• Example: we wish to subtract 4 ninths from 5 sevenths.

• common denominator: The denominator is 63

• Its 5 sevenths are 45:

• Its 4 ninths are 28:

• We subtract 28 from 45, 17 remain: