Babylonian Mathematics Explained: Base 60, YBC 7289 and Plimpton 322

Babylonian mathematics is the arithmetic of Old Babylonian scribes in what is now Iraq, about 1900 to 1600 BCE. It used base 60 place value written with two wedge signs, and it divided by multiplying with a reciprocal from a table. YBC 7289 gives √2 to about six decimal places; Plimpton 322 lists Pythagorean triples.

This post works it by hand. Every problem in our fixed problem set is converted to base 60 and solved in the scribes' own layout, column by column, with each carry written out. Every number was checked with Python. Then we read YBC 7289 and Plimpton 322 line by line, rebuild the square root method that later carried Heron's name, and follow base 60 into the clock on your wall.

Key takeaways

• Babylonian numbers use two signs, a vertical wedge for 1 and a corner wedge for 10. Every base 60 digit from 1 to 59 is a cluster of up to five tens and nine ones.
• Old Babylonian notation had no zero and no "sexagesimal point". The same written number could mean 30, 30 x 60 or 1/2, and the reader decided from context. A medial placeholder sign appears only in later texts.
• A fraction 1/n ends after a finite number of base 60 digits when n has no prime factors other than 2, 3 and 5. Scribes called these numbers regular and kept tables of their reciprocals, so dividing by 36 became multiplying by 0;1,40.
• Dividing by 7, 13, 19 or 37 has no exact base 60 answer. Scribes approximated, and one tablet says outright that 7 does not divide.
• YBC 7289 gives the square root of 2 as 1;24,51,10, which is 1.41421296... and differs from the true value by about 6 parts in 10 million.
• Plimpton 322 has 15 rows built on Pythagorean triples. Whether it is a list of triples, a teacher's aid built from reciprocal pairs, or a ratio based trigonometric table is still argued in print.
• Base 60 survives in hours, minutes and seconds, and in degrees, arcminutes and arcseconds, which the International Bureau of Weights and Measures still lists as accepted non-SI units.

The problem set for this series

This post belongs to the fast calculation methods series, which follows our mathematics traditions series. Both work the same fixed problems so you can compare methods across posts. The answers below were checked in Python. The right hand column gives each number in base 60, using the modern convention explained in the next section.

If you want to see the same problems done other ways, the Indian and Vedic mathematics post works them with the Vedic sutras, the Chinese mathematics post works them on counting rods, the Japanese mathematics post works them on the soroban, and the three way comparison puts them side by side.

Two of the seven problems do not suit the Babylonian method at all. Dividing by 37 and finding 1/19 both need the reciprocal of a number that has a prime factor other than 2, 3 or 5, and base 60 has no finite answer for those. We work them anyway and show exactly where and why the method strains. That is part of the comparison.

A short timeline

Dates for the second millennium BCE follow the conventional middle chronology, and the scholars we cite give ranges, not single years. Where sources differ, the table says so.

How does base 60 work?

Two signs make 59 digits

A Babylonian scribe pressed a reed stylus into wet clay. One stroke made a vertical wedge worth 1. A sideways stroke made a corner wedge worth 10. Every digit from 1 to 59 was a cluster of corner wedges followed by vertical wedges. As MacTutor puts it, the Babylonians only had to learn two symbols to produce their base 60 positional system.

In the diagrams below, Y stands for a vertical wedge and < for a corner wedge. Real scribes stacked the wedges in conventional patterns so a reader could count them at a glance, which Bill Casselman's photographs of YBC 7289 show clearly.

digit   wedges              digit   wedges
  1     Y                     10    <
  2     YY                    20    <<
  3     YYY                   30    <<<
  7     YYYYYYY               38    <<< YYYYYYYY
  9     YYYYYYYYY             49    <<<< YYYYYYYYY
 25     << YYYYY              59    <<<<< YYYYYYYYY

Inside a digit the counting is decimal: tens and ones. Between digits the counting is base 60. That is why historians describe the system as base 60 with a base 10 substructure.

Place value, and the notation used in this post

Each digit is worth 60 times the digit to its right. Modern writers separate base 60 digits with commas and mark the boundary between whole number and fraction with a semicolon. This post uses that convention throughout.

Here is the number 7,429 as a scribe would have pressed it, three digit groups with gaps between them:

  YY        YYY       <<<< YYYYYYYYY
  2    ,    3    ,    49
  (x 3,600)  (x 60)   (x 1)

No zero and no point: floating values

Two features of Old Babylonian writing trip up every modern reader.

First, there was no sign for an empty place. MacTutor points out that the numbers 1 and 1,0 (that is, 1 and 60) were written identically, and that an empty place in the middle of a number was simply a gap or nothing at all. Later Babylonian texts did introduce a placeholder sign, but only for empty places in the middle of a number, never at the end.

Second, there was no sexagesimal point. The tablet writes digits; the reader supplies the scale. Casselman calls it "a true floating point system": the written 30 on YBC 7289 could be 30 or 1/2. Friberg, quoted on the Stony Brook catalogue of tables, says every Babylonian number is written as if it were an integer, and its intended value can be its nominal value times any power of 60.

This sounds like a defect, and for record keeping it was. For calculation it was a feature. If you only care about the digits, multiplication by a reciprocal never needs you to track where the point goes. The scribe worked with digit strings and set the scale at the end, from the meaning of the problem. Engineers using a slide rule worked the same way three and a half thousand years later.

We will see the gap problem in action when we square 85, because the answer 2,0,25 has an empty middle place.

Why 60?

Nobody knows. MacTutor lists the main theories and is sceptical of all of them. Theon of Alexandria, in the 4th century CE, said 60 was chosen because it has many divisors. Otto Neugebauer proposed that a decimal count was adapted to weights and measures that used thirds. Others proposed astronomical reasons, finger counting on the joints of one hand, or the merging of two peoples with different bases. The honest summary is that base 60 came from the Sumerians, and the reason is not settled.

What is not in doubt is the arithmetic consequence. 60 = 2 x 2 x 3 x 5, so it has twelve divisors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60. That property is what made Babylonian division fast, and it is the subject of the next section.

Regular numbers: why base 60 makes division fast

Which fractions end

In decimal, 1/n has a finite expansion only when n has no prime factors other than 2 and 5, because 10 = 2 x 5. So 1/8 = 0.125 ends, but 1/3 = 0.333... never does.

In base 60 the rule is the same with one extra prime: 1/n ends exactly when n has no prime factors other than 2, 3 and 5. Historians call such numbers regular. So 1/3 = 0;20 and 1/9 = 0;6,40 both end in base 60, while decimal has no finite form for either.

Notice the last row. Decimal handles 1/37 far better than base 60 does, because 37 divides 999 = 10³ - 1, so the decimal repeats every 3 digits, while in base 60 the repeat takes 12 digits. Neither base is better for every divisor. Base 60 wins for divisors built from 2, 3 and 5, which are the divisors that came up most in Babylonian weights, measures, and rations.

The standard table of reciprocals

Because 1/n is finite for regular n, a scribe could write down a table of reciprocals and divide by any number in it with a single multiplication. Old Babylonian schools taught a standard table. The Stony Brook catalogue of Old Babylonian tables transcribes one student copy, written in the formula "igi n gál.bi m", which means "the reciprocal of n is m". Here is the full standard table, with each value recomputed in Python:

igi  2   gal.bi  30          igi 27   gal.bi  2,13,20
igi  3   gal.bi  20          igi 30   gal.bi  2
igi  4   gal.bi  15          igi 32   gal.bi  1,52,30
igi  5   gal.bi  12          igi 36   gal.bi  1,40
igi  6   gal.bi  10          igi 40   gal.bi  1,30
igi  8   gal.bi  7,30        igi 45   gal.bi  1,20
igi  9   gal.bi  6,40        igi 48   gal.bi  1,15
igi 10   gal.bi  6           igi 50   gal.bi  1,12
igi 12   gal.bi  5           igi 54   gal.bi  1,6,40
igi 15   gal.bi  4           igi 1,4  gal.bi  56,15
igi 16   gal.bi  3,45        igi 1,21 gal.bi  44,26,40
igi 18   gal.bi  3,20
igi 20   gal.bi  3
igi 24   gal.bi  2,30
igi 25   gal.bi  2,24

Read the table as floating values. "igi 4 gál.bi 15" means 1/4 = 0;15. "igi 1,4 gál.bi 56,15" means 1/64 = 0;0,56,15. The scribe does not need to know which; the digits are the same.

Look at which numbers are missing: 7, 11, 13, 14, 17, 19 and every other number with a prime factor above 5. MacTutor quotes a scribe giving a number close to 1/7 and adding that "an approximation is given since 7 does not divide." Tables were not limited to the school list either. A Seleucid era tablet by the scribe Inaqibit-Anu lists the reciprocals of 136 of the 231 six place regular numbers whose first digit is 1 or 2, and Donald Knuth called its author, in 1972, "the first man in history to solve a computational problem that takes longer than one second of time on a modern electronic computer". Two tables giving approximate reciprocals for some non-regular numbers are also known, one covering 56 to 80.

Multiplication tables

A reciprocal table is useless without fast multiplication, and the schools taught that too. Each Old Babylonian multiplication table took a head number and listed its multiples for 1 to 20, then 30, 40 and 50. The Stony Brook catalogue shows four such tables, and notes that the numbers 1 to 20 plus 30, 40 and 50 appear in the central column of each one. With those 23 entries, any multiplier from 1 to 59 is at most two lookups: 37 is 30 + 7, and 49 is 40 + 9.

The two kinds of table work as a pair. Dividing by 36 means multiplying by 1,40, so the speed of every division depends on how quickly the scribe can multiply by the digits of a reciprocal. A school that drilled both tables had, in effect, a division machine for every regular divisor.

Tables of squares existed too. MacTutor reports two tablets found at Senkereh in 1854 that give the squares of the numbers up to 59 and the cubes up to 32.

Division in one move: 7,429 ÷ 36

Our fixed problem divides by 37, which is not regular. Before we take that on, look at what the method does with its neighbour 36, which is. This is the case the system was built for.

Step 1. Look up the reciprocal. The table says igi 36 gál.bi 1,40, so 1/36 = 0;1,40.

Step 2. Write the dividend in base 60. 7,429 = 2,3,49.

Step 3. Multiply digit by digit. The multiplier 0;1,40 has two digits, worth 1/60 and 40/3,600. Each digit of 2,3,49 is multiplied by each digit of 0;1,40, and each product goes into the column given by the sum of the two place values.

Step 4. Carry from the right. Every column must end up under 60.

1/3,600s:  1,960       = 32,40   write 40, carry 32
1/60s:     169 + 32    = 201  = 3,21   write 21, carry 3
1s:        83 + 3      = 86   = 1,26   write 26, carry 1
60s:       2 + 1       = 3            write 3
result:    3,26;21,40

Step 5. Read the answer. 3,26;21,40 = 206 + 21/60 + 40/3,600 = 206.36111..., and 7,429 ÷ 36 = 206.36111... exactly. The quotient is 206 and the fraction 0;21,40 is 13/36, so the remainder is 13.

There was no trial quotient and no guessing. Decimal long division of 7,429 by 36 needs three rounds of "how many times does 36 go into this?", each followed by a multiplication and a subtraction. The Babylonian route needed one table lookup, six digit products, three carries, and nothing else. That is the real meaning of "base 60 makes division fast": for regular divisors, division has no search step.

Working the problem set in base 60

How to read the layouts

Every multiplication below uses the same layout. Write both numbers in base 60. Multiply every digit of one by every digit of the other. Drop each product into a column by place value. Add each column. Then carry from the right, writing each column's value modulo 60 and carrying the rest. This is the long multiplication you learned at school, run in base 60 instead of base 10.

The difference is the size of the digits. A base 60 digit product can be as large as 59 x 59 = 3,481, which is 58,1. So a single digit product may itself need table work, and carries can be large. Fewer columns, heavier columns: that trade runs through every example.

98 x 97

Convert. 98 = 1,38 and 97 = 1,37.

Partial products. The scribe needs four digit products.

The product 38 x 37 is not a single table entry. Using the tables, split 37 = 30 + 7 and look up 38 in the 30 times and 7 times rows, or split both digits: 30 x 30 = 900, 30 x 7 = 210, 8 x 30 = 240 and 8 x 7 = 56, which add to 1,406. In base 60 that is 23,26.

Carry from the right.

1s:       1,406         = 23,26   write 26, carry 23
60s:      75 + 23       = 98 = 1,38   write 38, carry 1
3,600s:   1 + 1         = 2           write 2
result:   2,38,26

Check. 2 x 3,600 + 38 x 60 + 26 = 7,200 + 2,280 + 26 = 9,506. Correct.

What it cost. Four digit products, one of which needed up to four table lookups, then three carries. Compare the Vedic method in the Indian mathematics post, which uses the closeness of 98 and 97 to 100 to finish in two small steps. Base 60 has no such shortcut here because 98 and 97 are not close to a power of 60. The nearest round base 60 number is 2,0 = 120, which is 22 and 23 away.

47 x 53

Convert. Both numbers are below 60, so each is a single base 60 digit: 47 and 53.

Route 1: one digit product. The scribe would split 53 = 50 + 3 and use the 47 times table, if one existed, or split 47 = 40 + 7 and use the 53 row in the tables of 40 and 7. Either way, 47 x 50 = 2,350 and 47 x 3 = 141, sum 2,491, which is 41,31 in base 60 (41 x 60 = 2,460, plus 31).

Route 2: a table of squares. MacTutor's overview gives the identity that makes a table of squares into a multiplication machine:

ab = ((a + b)² - (a - b)²) / 4

With a = 53 and b = 47, a + b = 100 = 1,40 and a - b = 6. Here is the layout:

Even simpler, because 47 and 53 sit symmetrically around 50: 47 x 53 = 50² - 3² = 2,500 - 9 = 2,491. In base 60, 50 squared is 41,40, and 41,40 - 9 = 41,31.

A caution on history. MacTutor presents the quarter squares identity as a way a table of squares could do all multiplication, and the tables of squares are real. Whether scribes routinely multiplied this way, rather than with their many multiplication tables, is not settled by the sources we fetched. Treat Route 2 as an illustration of what the tables made possible, not a documented school drill.

386 x 524

Convert. 386 = 6,26 (6 x 60 = 360, plus 26). 524 = 8,44 (8 x 60 = 480, plus 44).

Partial products.

In base 60 the middle products are 264 = 4,24 and 208 = 3,28, and the last is 1,144 = 19,4. A scribe would get 26 x 44 from tables as 26 x 40 + 26 x 4 = 1,040 + 104.

Carry from the right.

1s:       1,144          = 19,4    write 4, carry 19
60s:      472 + 19       = 491 = 8,11   write 11, carry 8
3,600s:   48 + 8         = 56           write 56
result:   56,11,4

Check. 56 x 3,600 + 11 x 60 + 4 = 201,600 + 660 + 4 = 202,264. Correct.

What it cost. A three digit by three digit decimal product needs nine single digit products. The base 60 version needs four digit products, because each number is only two base 60 digits long. That is the genuine advantage of a large base: fewer columns. The price is that each digit product is bigger and may need two to four table lookups, so the total work is similar. The base 60 saving is in the bookkeeping, not the arithmetic.

85²

Convert. 85 = 1,25.

Square it like a binomial. (1,0 + 25)² = 1,0 squared + 2 x 25 x 1,0 + 25².

Carry.

1s:       625           = 10,25   write 25, carry 10
60s:      50 + 10       = 60 = 1,0   write 0, carry 1
3,600s:   1 + 1         = 2          write 2
result:   2,0,25

Check. 2 x 3,600 + 0 x 60 + 25 = 7,225. Correct.

The empty middle place. Here is the zero problem in a real calculation. The answer has nothing in the 60s place. An Old Babylonian scribe would write 2, then a gap, then 25. On clay that could look like 2,25 (which is 145) or 2,0,25 (which is 7,225), and only context or spacing tells them apart. MacTutor discusses a real case, the Louvre tablet AO 17264, where the scribe squares 147 = 2,27 and gets 21,609 = 6,0,9, another answer with an empty middle place. Late Babylonian scribes eventually added a placeholder sign for exactly this situation.

The Vedic comparison. The Vedic rule for squares ending in 5 gives 8 x 9 = 72, then 25, so 7,225 in two steps. In base 60, 85 does not end in 5 in any useful sense: its last digit is 25. The trick does not transfer, because it depends on the base being 10.

√7,744

Convert. 7,744 = 2,9,4.

A scribe had three routes, and the post shows all three because they teach different things.

Route 1: read a table backwards. Tables of squares and square roots existed. If the scribe's table ran far enough, the entry "1,28 squared is 2,9,4" gives the answer directly: 1,28 = 88. Here is the square, laid out, so you can confirm the table entry:

1s:       784           = 13,4    write 4, carry 13
60s:      56 + 13       = 69 = 1,9   write 9, carry 1
3,600s:   1 + 1         = 2          write 2
result:   2,9,4

So 1,28 squared is 2,9,4, which is 7,744.

Route 2: the Babylonian (Heron) iteration. Start from a guess a and improve it with a + (N - a²) / (2a). The trick in base 60 is to pick a guess whose double is regular, so the division is a table lookup. A good first guess is 1,30 = 90, because 2a = 3,0 = 180 is regular and its reciprocal is 0;0,20.

One step took the guess from 90 to 88.0222, an error of 0.0222, or 1 part in about 4,000. The next step would land within 0.000003 of 88, but the new divisor 2 x 1,28;1,20 is not regular, so the scribe would have to approximate its reciprocal. The practical move is to round 1,28;1,20 to the whole number 1,28 and confirm it by squaring, which Route 1 already did. Squaring 1,28 gives exactly 2,9,4, so the root is exact.

The same step can be written as an average: divide 7,744 by 90 by multiplying by igi 1,30 = 0;0,40, giving 1,26;2,40, then average with 1,30 to get 1,28;1,20. That is the form Heron gives, and the form in the section on YBC 7289 below.

Route 3: factor out regular squares. 7,744 = 64 x 121 = 2⁶ x 11². The factor 64 is a regular square (8 squared), so the scribe can take it out: √7,744 = 8 x √121 = 8 x 11 = 88. The leftover 121 is irregular, so this route only works if the scribe already knows 11 squared. Friberg argues that taking out square regular factors was part of how Old Babylonian scribes computed square roots, as the Plimpton 322 literature notes.

7,429 ÷ 37

The problem. 37 is prime and greater than 5, so 1/37 has no finite base 60 expansion. It repeats with period 12:

1/37 = 0;1,37,17,50,16,12,58,22,42,9,43,47, then 1,37,17,... again.

There is no line for 37 in the standard table. A Babylonian scribe could not divide by 37 in one move. Here is what the method can do instead.

Step 1. Find a regular number next to a multiple of 37. MacTutor shows the scribal trick for 1/13: write 1/13 = 7/91, and since 91 is close to the regular number 90, take 1/13 as roughly 7 x 1/90. For 37, Python finds that 37 x 81 = 2,997, which is next to the regular number 3,000 = 50,0. So

1/37 = 81/2,997, approximately 81/3,000 = 81 x 0;0,1,12 = 0;1,37,12.

The true value starts 0;1,37,17,50, so the approximation is good to about 1 part in 1,000. Other choices work too: 37 x 13 = 481 is next to 480 = 8,0, giving 0;1,37,30, which is less accurate.

Step 2. Multiply. 7,429 = 2,3,49. Multiply by 0;1,37,12.

1/216,000s:  588            = 9,48    write 48, carry 9
1/3,600s:    1,849 + 9      = 1,858 = 30,58   write 58, carry 30
1/60s:       184 + 30       = 214 = 3,34      write 34, carry 3
1s:          77 + 3         = 80 = 1,20       write 20, carry 1
60s:         2 + 1          = 3               write 3
result:      3,20;34,58,48

That is 200.583, an estimate. The true quotient is 200.78378...

Step 3. Take the whole part and check it. The estimate says the quotient is 3,20 = 200. Multiply back: 37 x 3,20 has partial products 37 x 3 = 111 in the 60s column and 37 x 20 = 740 in the 1s column. Carry: 740 = 12,20, write 20 carry 12; 111 + 12 = 123 = 2,3. So 37 x 200 = 2,3,20 = 7,400.

Step 4. Subtract. 2,3,49 - 2,3,20 = 29. The remainder is 29, which is less than 37, so the quotient is 200 remainder 29. Correct.

Step 5, if you want the fraction. 29/37 in base 60 is 0;47,1,37,17,50,16,12,58,22,42,9,43, repeating with period 12. So 7,429 ÷ 37 = 3,20;47,1,37,17,50,... with no end.

Where it strains. Look at what happened. The approximate reciprocal gave 200.58 instead of 200.78. Because both are above 200 and below 201, the whole number part still came out right, but the method did not know that by itself. A weaker approximation can push the estimate across an integer: using 0;1,37,30 gives 201.20, and the check 37 x 201 = 7,437, which is larger than 7,429, is the only thing that catches the error. So for an irregular divisor the Babylonian method turns into estimate, multiply back, correct, which is just long division with an extra lookup. The fast path exists only for regular divisors.

Decimal does better on this one. 1/37 = 0.027027..., period 3, and 7,429 ÷ 37 by ordinary long division takes three short rounds. This is the clearest case in the problem set where the choice of base, not the cleverness of the method, decides the speed.

1/19 as a decimal

The problem. 19 is prime, so 1/19 repeats in both bases. In base 60 its period is 18, because 60 leaves remainder 3 when divided by 19, and the smallest power of 3 that leaves remainder 1 is the 18th. In base 10 the period is also 18.

What a scribe would do. No table entry exists. The scribe can approximate as before. 19 x 19 = 361, which is next to the regular 360 = 6,0, so 1/19 is close to 19/360 = 19 x 0;0,10 = 0;3,10, which is 0.05278. Better, 19 x 21 = 399 is next to 400 = 6,40, so 1/19 is close to 21/400 = 0;3,9, which is 0.0525. Neither is exact, and nothing in the method says how wrong they are.

Exact base 60 long division. To get the true expansion, divide by 19 one base 60 digit at a time. Start with remainder 1. At each step multiply the remainder by 60, divide by 19, write the quotient as the next digit, and keep the new remainder.

The remainder is back to 1 after 18 steps, so the digits repeat: 1/19 = 0;3,9,28,25,15,47,22,6,18,56,50,31,34,44,12,37,53,41, and again.

Decimal long division. The question asks for a decimal, and the same procedure in base 10 gives it.

Back to remainder 1 after 18 digits: 1/19 = 0.052631578947368421, repeating.

The verdict. The Babylonian method cannot do this problem in its own terms. It has no table entry for 19 and no built in long division. It can give a quick rough answer (0;3,9 or 0;3,10), and a modern reader can run long division in base 60, but that is our algorithm, not theirs. The Vedic post shows a much faster route to 1/19 in decimal using the "one more than the one before" sutra, because 19 sits next to 20.

Summary: what each problem cost in base 60

The counts below use one rule: a digit product counts as one operation if it is a single table entry and as up to four if both digits must be split into tens and ones. Carries count as one operation each. These are counts of written steps for a trained scribe, not timings. No measured speed data exists for Babylonian methods; there is no competition category for them and no study we could find.

YBC 7289: the square root of 2 on a hand tablet

What the tablet shows

YBC 7289 is a small round tablet, about 8 cm across, in the Yale Babylonian Collection. Casselman, who photographed it, dates it to between 1800 and 1600 BCE with unknown provenance. Yale News calls it one of the most reproduced cultural objects Yale owns and says it came to Yale in 1909 in J. Pierpont Morgan's collection; Casselman's page says it was purchased around 1912 by Morgan's agent. The two dates disagree, and we have not resolved which is right.

The tablet shows a square with both diagonals. Three numbers are written on it:

side (top left):      30                 <<<
along the diagonal:   1,24,51,10         Y  << YYYY  <<<<< Y  <
below the diagonal:   42,25,35           <<<< YY  << YYYYY  <<< YYYYY

The size and shape suggest a student's hand tablet, held in the palm for rough work. Neugebauer and Sachs first recognised its significance in their 1945 Mathematical Cuneiform Texts.

Reading the numbers

Read the diagonal number as 1;24,51,10:

1 + 24/60 + 51/3,600 + 10/216,000 = 30,547/21,600 = 1.414212962...

The true value of √2 is 1.414213562... The difference is about 0.0000006, which is about 6 parts in 10 million, or less than 1 part in 2 million relative to √2. Casselman's analysis puts it simply: the value is 1.41421296 to nine significant figures, against a true 1.41421356.

Squaring the tablet value in Python gives 1;59,59,59,38,1,40, just under 2. That tells us the scribe's value is a truncation: the true base 60 expansion of √2 starts 1;24,51,10,7,46,6, so 1;24,51,10 is exactly the first four digits with the rest cut off.

Now the third number. If the side is 30, the diagonal is 30 x 1;24,51,10:

Sum: 1s = 30 + 12 = 42; 1/60s = 25; 1/3,600s = 30 + 5 = 35. The diagonal is 42;25,35, exactly the number on the tablet. In decimal, 42.42638...

The floating point reading

Because the scribe wrote no point, Fowler and Robson pointed out a second reading. If the side is 0;30, which is 1/2, then the diagonal is 0;42,25,35 = 0.70710648..., and 1/√2 = 0.70710678... So the tablet may show √2 and its reciprocal 1/√2 together, a reciprocal pair with a geometric meaning. Fowler and Robson find the reading attractive given how central reciprocal pairs are in Babylonian mathematics, but they also give reasons for scepticism. Both readings produce the same digits, which is the point.

How did the scribe get 1;24,51,10?

The honest answer is that the tablet does not say. The literature summarised in the Wikipedia article on YBC 7289, which cites Fowler and Robson 1998, says the student would likely have copied the value from a table of constants (YBC 7243 lists the same approximation for "the diagonal of a square"), and that an iterative procedure for computing such values appears on another tablet, BM 96957 + VAT 6598. The same value turns up again in Ptolemy's Almagest in the 2nd century CE, without an explanation of where it came from.

What we can show is that the Babylonian square root iteration produces it. Start at 1;30 and apply a new = (a + 2/a) / 2:

The second iterate, cut to four digits, is 1;24,51,10, the tablet's value. The third iterate agrees with √2 to 11 decimal places. Notice how the error roughly squares at each step: 0.08, then 0.002, then 0.000002, then 0.0000000000016. That doubling of correct digits is why the method is still used inside computers today.

There is a catch for the scribe. Step 2 divides 2 by 1;25, and 1;25 = 85/60 is not regular (85 = 5 x 17). So the scribe could not do that division exactly from a table. Whatever route produced 1;24,51,10, it involved an approximation somewhere. Fowler and Robson's paper is the place to read the competing reconstructions; we could not fetch its full text, so we do not summarise its arguments in detail.

The square root method that Heron wrote down

The rule

Heron of Alexandria, whom MacTutor dates to about 10 to 75 CE, gives the rule in Book I of his Metrica, with a worked example. In MacTutor's translation: since 720 has no rational side, take the next square, 729, whose side is 27. Divide 720 by 27 to get 26 2/3. Add 27, making 53 2/3, and take half, 26 5/6. Then 26 5/6 squared is 720 1/36, so the error in the square is 1/36. To do better, start again from 26 5/6.

MacTutor says the method was known to the Babylonians 2,000 years before Heron. It is often called the Babylonian method, and in modern numerical analysis it is Newton's method applied to x² - N.

Heron's example in base 60

The example runs cleanly in base 60, which is a nice confirmation of why the method suited the Babylonian toolkit:

Step 3 divides by 27, which is regular: igi 27 gál.bi 2,13,20, so 720 x 0;2,13,20 = 26;40. Every step is a table operation. A second pass from 26;50 gives 26;49,58,8,11,..., which squares to 720.0000003.

Why it works, in pictures

The Old Babylonian way of thinking about a square root was geometric: the root of N is the side of a square of area N. If a guess a is too big, then N/a is too small, and a rectangle with sides a and N/a has exactly the right area. Averaging the two sides makes the rectangle more nearly square. Repeat until it is a square to the precision you need.

 guess a = 1,30 (90)                 average side
 +------------------------+          +---------------------+
 |                        |          |                     |
 |   area 2,9,4 (7,744)   | N/a      |   area about 7,744  |  1,28;1,20
 |                        | =1,26;   |                     |
 |                        |  2,40    |                     |
 +------------------------+          +---------------------+
          1,30                              1,28;1,20

The Old Babylonian geometry of "cut and paste", described by historians of the period, uses the same kind of picture to solve quadratic problems. MacTutor gives one: a rectangle has area 1,0 and its length exceeds its width by 7. The scribe halves 7 to get 3;30, squares it to get 12;15, adds 1,0 to get 1,12;15, takes the square side 8;30, and subtracts 3;30 to get the width 5. In modern terms that is the quadratic formula for x² + 7x = 60, done as a recipe.

Plimpton 322: what is on the tablet

The object

Plimpton 322 is a broken clay tablet, about 13 cm wide and 9 cm tall, in the Rare Book and Manuscript Library at Columbia University. The Cuneiform Digital Library Initiative catalogues it as P254790, a mathematical table excavated at Larsa (modern Tell as-Senkereh) and dated to the Old Babylonian period, about 1900 to 1600 BCE. George Arthur Plimpton bought it from the dealer Edgar J. Banks in about 1922 and bequeathed it to Columbia in the mid 1930s. Robson dates its handwriting and format to the period 1822 to 1784 BCE, according to the summaries of her work we read. MacTutor gives a wider range, 1800 to 1650 BCE.

The surviving part has four columns and 15 rows. The left edge is broken off, so there may have been more columns. Each column has a heading in Akkadian, with some Sumerian logograms.

The four columns

The fourth column is a row number, "its line", 1 to 15. The second and third columns are headed with the term ÍB.SI₈ "of the width" and "of the diagonal"; Robson reads ÍB.SI₈ here as "square side". The first column has a damaged heading that Robson reconstructs as "the holding square of the diagonal from which 1 is torn out, so that the width comes up".

Here is the whole table, rebuilt in Python from the standard reconstruction. Columns 2 and 3 are integers s (short side, "width") and d (diagonal). The long side l is not written on the tablet, but in every row it is a regular number. Column 1 is (d/l)²; whether the leading 1 was written is one of the things the broken edge hides.

Check row 1: 169² - 119² = 28,561 - 14,161 = 14,400 = 120². And (169/120)² = 1.98340..., which is 1;59,0,15. Row 4 is the one that convinces most readers the triples were not found by trial: 12,709, 13,500 and 18,541 are not numbers you stumble on.

The first column falls steadily from about 1.983 to about 1.387. If you think in angles, which the scribes apparently did not, that is a slope running from about 45 degrees down to about 32 degrees in uneven steps of about one degree. Row 11 is the familiar 3, 4, 5 triangle scaled to 45, 60, 75, or read as 0;45, 1, 1;15.

The errors on the tablet

Most scholars agree the tablet has six errors, and the errors are evidence about how it was made. Here are the corrected values from the table above set against what the scribe wrote:

The row 13 error is the most revealing. 2,41 is 161, and 161² = 25,921 = 7,12,1. The scribe wrote down the square instead of its side, which suggests the work involved squares and square sides, not just a list of finished triples. MacTutor records that 8,1 was copied as 9,1 in row 9.

What was Plimpton 322 for? Three readings

The purpose of the tablet is not known. Here are the three main proposals, with the arithmetic of each run on row 1.

Reading 1: a list of Pythagorean triples (Neugebauer and Sachs, 1945)

Pick two regular numbers p and q. Then s = p² - q², l = 2pq and d = p² + q² form a Pythagorean triple, and because p and q are regular, l is regular too. Row 1 comes from p = 12 and q = 5: s = 144 - 25 = 119, l = 120, d = 169.

Christopher Zeeman added an observation, reported by MacTutor: with this rule, exactly 16 triples meet a natural set of limits (q at most 60, the slope between 30 and 45 degrees, and a finite base 60 value in the first column), and 15 of them are on the tablet. Objections, as summarised by Buck and Robson: column 1 plays no part in this construction, the rule does not explain the order of the rows, and a number theory motive looks out of place in what we know of Old Babylonian mathematics.

Reading 2: reciprocal pairs and a teacher's list (Bruins, then Robson and Friberg)

Pick a regular number x and read its reciprocal 1/x from a table. Then form half the sum and half the difference. In modern terms, ((x + 1/x)/2)² - ((x - 1/x)/2)² = 1, so the two halves are the diagonal and width of a rectangle whose length is 1.

Here is row 1, step by step in base 60:

Everything in this reading is a table operation the scribes are known to have done: reciprocal lookup, halving, squaring, and removing common regular factors (the "trailing part" step). The column 1 heading, about a square "from which 1 is torn out so that the width comes up", matches step 5 directly: subtract 1 from (d/l)² and you get (s/l)². The row 15 error, 56 where 28 was expected, can be explained as removing a common factor of 2 from one column but not the other. Robson's 2001 paper in Historia Mathematica and her 2002 article in the American Mathematical Monthly argue that the tablet was probably a teacher's list of parameters for setting right triangle problems, made by a scribe at Larsa trained in exactly these procedures.

Reading 3: exact ratio based trigonometry (Mansfield and Wildberger, 2017)

Daniel Mansfield and Norman Wildberger, of UNSW Sydney, published "Plimpton 322 is Babylonian exact sexagesimal trigonometry" in Historia Mathematica 44(4), 2017. Their UNSW news release says the tablet describes right triangle shapes "using a novel kind of trigonometry based on ratios, not angles and circles", and calls it the world's oldest and most accurate trigonometric table, predating Hipparchus by more than 1,000 years. The idea is that a user could find the row whose ratio matched a known pair of sides and read off the others exactly, because all the ratios are finite base 60 numbers.

The objection from other historians is that the Babylonians appear not to have had a concept of measured angle, so calling the tablet trigonometry imports a modern idea. The Wikipedia summary of the scholarship says most current scholars consider a number theory motive anachronistic and rule out the trigonometric reading for similar reasons. Coverage of the 2017 paper mostly repeated the UNSW release, so treat the "oldest trigonometric table" phrase as the authors' claim, not a consensus.

Where that leaves us

All three readings agree on the arithmetic: every row is a right triangle whose long side is regular, so every ratio is exact in base 60. They disagree on intent. The reciprocal pair reading explains the most features of the actual tablet, including the column headings and several errors, and is the most widely held view in the sources we read. The tablet itself cannot settle it, because its left edge is gone.

For this series the useful lesson is narrower. Plimpton 322 exists because regular numbers made exact division possible. A table like this one is what you get when a calculating culture's fast path, the reciprocal table, is pushed as far as it will go.

Base 60 in your clock and your compass

Hours, minutes, seconds; degrees, arcminutes, arcseconds

Two everyday measurements still run in base 60. The International Bureau of Weights and Measures, in Table 8 of the SI Brochure, lists the minute (1 min = 60 s), hour (1 h = 60 min = 3,600 s) and day (1 d = 24 h) as long standing non-SI units of time, and the degree, minute and second of plane angle, with 1′ = (1/60)° and 1″ = (1/60)′. Astronomy still measures small angles in arcseconds.

The line of descent is indirect. Greek astronomers took base 60 fractions from Babylonian astronomy, and Ptolemy's Almagest in the 2nd century CE uses them throughout, including the value 1;24,51,10 for √2. Dividing the hour itself into sexagesimal minutes and seconds is later: al-Biruni did it around 1000 CE, and Sacrobosco around 1235. MacTutor's overview says the Babylonians divided the day into 24 hours of 60 minutes of 60 seconds; the medieval evidence suggests the minute and second of the hour arrived much later, so treat that line with care.

Our problem set as clock readings

A clock reading is a base 60 number with decimal digits, so several of our answers are readable as times. Every conversion below was checked in Python.

The 2,0,25 row is the zero problem again. We write "2:00:25" with a zero; an Old Babylonian scribe would have written 2 and 25 with a gap.

Try it on a clock

You already do Babylonian arithmetic when you add times. 1 hour 38 minutes plus 1 hour 37 minutes is 2 hours 75 minutes, which you carry to 3 hours 15 minutes. That carry of 60 is the same carry as in our 98 x 97 layout, where the 60s column summed to 75 + 23 = 98 and carried 1.

How fast is the Babylonian method, really?

The brief for this series says "fast" has to be shown, not asserted. Here is what the worked examples show.

Where it is fast. Division by a regular number. 7,429 ÷ 36 needed one table lookup and a plain multiplication with no trial quotients. For a scribe dividing rations, field areas, or bricks among workers, where divisors like 12, 30, 36, 45 or 1,12 were everywhere, this removed the slowest and most error prone step of arithmetic. Square roots by iteration are also fast in the sense that matters: the number of correct digits roughly doubles each step, as the YBC 7289 table shows.

Where it is neutral. Multiplication. Base 60 numbers have fewer digits (386 x 524 is a 2 by 2 digit product instead of 3 by 3), but each digit product is bigger and may need several table lookups. The total work is about the same as decimal. The win is fewer columns to line up.

Where it is slow. Division by any number with a prime factor above 5. Our 7,429 ÷ 37 and 1/19 needed approximations plus a check, or a long division the scribes did not use. Decimal handled 1/37 better, with a period of 3 against 12.

Memory load. A scribe working from tablets held very little in memory: the tables were written down. The working memory load per step is one partial product and one carry, but the carries can be two digit numbers (23, 30, 32 in our examples). Mental base 60 arithmetic without tables would be much harder than decimal mental arithmetic, because you would need a 59 by 59 multiplication table in your head instead of a 9 by 9 one. Nobody we know of trains to do this mentally, and there are no recorded speed results for it.

Error checking. The method has one strong built in check: multiply back. Every division in this post ended by multiplying the quotient by the divisor, which is how the 7,429 ÷ 37 approximation was caught and corrected. There is no Babylonian equivalent of casting out nines that we know of; in base 60, the analogous check would be casting out 59s, and we found no source showing scribes used it.

The soroban column refers to the Japanese mathematics post, and the Vedic column to the Indian mathematics post. The final post in this series puts every method from both series on one table.

A practice plan

If you want to work like a scribe, a few short sessions are enough to get the feel of it.

1. Day 1: write numbers. Convert ten numbers between 100 and 10,000 into base 60 by dividing by 3,600 and then 60. Convert back. Use clock times to check yourself: 7,429 seconds is 2:03:49.
2. Day 2: learn the reciprocal table. Copy the 26 line standard table above by hand twice. Then test yourself on 1/n for n = 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36, 40, 45, 48, 50.
3. Day 3: divide by multiplying. Do five divisions by regular numbers, for example 7,429 ÷ 36, 1,000 ÷ 48, 5,000 ÷ 45, 3,600 ÷ 16 and 720 ÷ 27. Use the column layout. Check each with a calculator.
4. Day 4: multiply in base 60. Redo 98 x 97, 386 x 524 and 85² from this post without looking, then try 123 x 456.
5. Day 5: square roots. Run the iteration for √2 from 1;30 for two steps, and for √7,744 from 1,30 for one step.
6. Day 6: irregular divisors. Find a regular number next to a multiple of 7, 13 or 37, build an approximate reciprocal, divide, and correct by multiplying back.

Expect day 4 to feel slow. It is supposed to. The base 60 multiplication table is large, and that is why the scribes wrote theirs on clay.

Common mistakes

• Treating base 60 digits as decimal. 2,3,49 is 7,429, not 2,349. Always expand by powers of 60.
• Forgetting the empty place. 85² is 2,0,25, not 2,25. When a column sums to exactly 60, write 0 and carry 1.
• Carrying 10 instead of 60. In base 60 you carry when a column reaches 60, and the carry can be large (1,960 carries 32).
• Fixing the point too early. A Babylonian number has no point. Work with digit strings and decide the scale at the end from the meaning of the problem.
• Looking up a reciprocal that does not exist. 7, 11, 13, 14, 17, 19, 37 and every other number with a prime factor above 5 have no finite reciprocal. If you find yourself using one, you are approximating.
• Trusting an approximate quotient. After dividing with an approximate reciprocal, always multiply back. The 0;1,37,30 approximation for 1/37 gives 201, and only the check shows it should be 200.
• Saying the Babylonians knew the Pythagorean theorem as a proof. The tablets show they used the relationship and generated large triples. They do not show a proof in the Greek sense.
• Calling Plimpton 322 a trigonometric table without qualification. That is one published reading; the reciprocal pair reading is more widely held.
• Calling YBC 7289's value "exact". 1;24,51,10 is a truncation, accurate to about 6 parts in 10 million.

FAQ

What is base 60, and how does it work?

Base 60, or the sexagesimal system, is place value counting where each place is worth 60 times the place to its right. Babylonian scribes wrote each digit from 1 to 59 with two wedge signs, one for 1 and one for 10, so 7,429 appears as 2,3,49: 2 x 3,600 + 3 x 60 + 49.

Why did the Babylonians use base 60?

No one knows for certain. The base came from the Sumerians. Theories include the many divisors of 60, weights and measures based on thirds, astronomy, finger counting, and the merging of two counting systems. MacTutor regards none as convincing. The useful consequence is that 1/n is finite in base 60 whenever n has no prime factors other than 2, 3 and 5.

Did the Babylonians have a zero?

Not in the Old Babylonian period. An empty place was a gap or nothing, so 1 and 60 looked identical. Later Babylonian texts used a placeholder sign for an empty place in the middle of a number, but not at the end, so it was not a full zero as a number in its own right.

How did Babylonians divide numbers?

They multiplied by a reciprocal from a table. To divide by 36 a scribe looked up igi 36, which is 1,40, and multiplied by 0;1,40. This only works exactly for regular numbers. For divisors like 7 or 37 they used approximations, and one scribe noted that "7 does not divide".

What is YBC 7289?

A small Old Babylonian school tablet in the Yale Babylonian Collection, dated to about 1800 to 1600 BCE. It shows a square with its diagonals, the number 30 on one side, and 1;24,51,10 and 42;25,35 on the diagonal. The first is √2 to about six decimal places; the second is 30 times that value.

How accurate is the Babylonian square root of 2?

1;24,51,10 equals 1.414212962..., against the true 1.414213562... The difference is about 0.0000006, less than 1 part in 2 million. It is the true base 60 expansion of √2 cut off after three fractional places.

What is the Babylonian method for square roots?

Start with a guess a, then replace it with the average of a and N/a. Each step roughly doubles the number of correct digits. Heron of Alexandria wrote it down with a worked example (√720) in the 1st century CE, and MacTutor notes it was known to the Babylonians long before. Today it is Newton's method for x² - N.

What is Plimpton 322?

An Old Babylonian clay tablet from Larsa, about 1800 BCE, held at Columbia University. Its 15 rows each relate to a right triangle with whole number sides whose long side is a regular number, such as 119, 120, 169 in row 1. It contains six scribal errors and its left edge is broken.

Is Plimpton 322 a trigonometric table?

Mansfield and Wildberger argued so in Historia Mathematica in 2017, describing a ratio based trigonometry. Many historians disagree, because the Babylonians appear to have had no concept of measured angle. Robson and others read it as a teacher's list built from reciprocal pairs. The question is open.

Why are there 60 minutes in an hour?

Because Greek astronomers inherited base 60 fractions from Babylonian astronomy, and medieval astronomers later applied them to the hour: al-Biruni around 1000 CE and Sacrobosco around 1235. The BIPM still lists the minute, hour, degree, arcminute and arcsecond as accepted non-SI units.

Where Safeguard fits

Safeguard does not do mathematics, and nothing in this post was computed by a Safeguard model; every number was checked with plain Python. The honest parallel is the scribes' habit of multiplying back. A Babylonian division by an approximate reciprocal was only trusted after the quotient was multiplied by the divisor and compared with the original, and a copied table was only as good as the tablet it was copied from, which is how six errors reached Plimpton 322. Software supply chains have the same two weaknesses at a much larger scale: components copied from upstream, and versions trusted without a check. The script behind this post's tables is small, but real projects pull in third party libraries that carry their own vulnerabilities, so knowing exactly which versions you ship and moving them to fixed versions quickly is the part of the job Safeguard is built for. Our post on using SBOMs in incident response shows what that check looks like when an advisory lands.

More in this series

Other posts in our fast calculation methods series:

• Chisanbop Explained: Korean Finger Math to 99, With Worked Examples
• Egyptian and Russian Peasant Multiplication: Doubling and Binary
• Fastest Mental Math Methods Compared: 14 Systems, Same 7 Problems
• Lattice Multiplication, Napier's Bones and the Slide Rule Explained
• Line Multiplication: The Stick Method, Japanese, Chinese or Neither?
• Maya Mathematics Explained: Base 20, Bar and Dot Numerals and Zero
• Mental Calculation Techniques of World Champions: Left to Right, Criss-Cross, Squares and Calendars
• Trachtenberg Speed System Explained: Rules for 11, 12 and Any Number

Fix it without waiting for upstream

Whatever the vulnerability, most of the risk sits in two gaps. The first is the time between a fix being published and that fix actually running in your production builds. The second, and worse, is the time before any fix exists at all, while a maintainer works out a patch and a release schedule. Waiting through both gaps is a habit, not a requirement. Safeguard gives teams three ways to close them.

1. Autonomous remediation: pull requests, not tickets

When a fixed version exists, Safeguard's Griffin models open a tested, policy gated pull request that moves the affected dependency to the safe version. You can run it on one repository or in bulk across every project that carries the vulnerable version. Remediation covers eight categories, including dependencies, licenses, security posture and findings. Reachability analysis ranks the vulnerable code paths your application actually calls first, so the first pull requests you merge are the ones that matter most. Autonomous remediation ships switched off, with every category set to Manual, so your team decides how much runs on its own and what still needs a human approval.

2. Gold Open Source: the safe version in seconds, free

gold.safeguard.sh is free and needs no login. Search any CVE or package to see affected and fixed versions, CVSS, EPSS, CISA KEV status and a recommended safe version, or use the "Am I affected?" checker. The same data is available through a free read API, README status badges and RSS threat feeds. The safeguard-sh/gold-scan-action GitHub Action can fail a build on a known exploited vulnerability, on any known vulnerability, or never, so a version you have just removed cannot quietly come back in next week's merge.

3. Gold Registry: start from components that are already fixed

registry.safeguard.sh provides hardened, drop in replacements for upstream packages and container images. Every Gold artifact is held continuously to zero exploitable critical or high CVEs, zero malware and a completed license check, with SLSA L3 or higher provenance, a Sigstore signature and a public evidence page you can hand to an auditor. An artifact loses its Gold badge if it regresses. Teams can point their own package directory at the registry so hardened equivalents resolve without changing how developers install anything.

When upstream has no fix yet, or will not ship one, Safeguard first works to get the fix accepted upstream. If the issue is severe or upstream will not move, Safeguard builds a fork, fixes the vulnerability there and publishes a hardened Gold build, so you are not stuck on a maintainer's release schedule. You can also request a hardened, compatibility tested build of a specific package. This fork and patch flow is in early access today, so ask which packages it covers before you plan around it.

One honest limit: when the vulnerable code lives in a vendor appliance, a hosted service or an operating system runtime you do not build yourself, the fix still has to come from that vendor or distribution. The steps above apply to the open source dependencies and container images you build and ship, which is where most teams carry the largest and least tracked share of their exposure.

See how Safeguard would help

These demos show dependency scanning and remediation workflows, not mathematics. You can download Safeguard and follow the CLI docs to try them on your own projects.

This demo is a real CLI session. After installing, safeguard opens the interactive terminal; asked to "fix the lodash CVE in this repo", it reads the finding and the advisory, asks before any command that changes something, edits package.json to lodash 4.17.21, runs npm install and npm test with approval, prints a summary and pushes nothing. The same flow applies to the JavaScript or Python tooling behind a teaching site, a base 60 converter or a table generator like the one used for this post.

Click through it at your own pace, or open it full screen.

This demo sets up the Safeguard runner on your own machine, approves it with its device code and scans from the terminal. Pointed at a project folder, the scan lists the packages and versions it depends on and their known vulnerabilities, so you know what to update before you share the project.

Click through it at your own pace, or open it full screen.

This demo connects Claude to the Safeguard MCP server with OAuth and lets you choose which tools Claude can use. If you use Claude to help write or review code for a project like a calculation notebook, the same connection gives Claude the Safeguard tools you choose, and no others.

Click through it at your own pace, or open it full screen.

Quick check

Sources

• MacTutor: Babylonian numerals (O'Connor and Robertson)
• MacTutor: An overview of Babylonian mathematics
• MacTutor: Pythagoras's theorem in Babylonian mathematics
• MacTutor: Heron of Alexandria biography
• Bill Casselman, UBC: YBC 7289 photographs and provenance
• Bill Casselman, UBC: Analysis of YBC 7289
• Bill Casselman, UBC: Mathematical commentary on YBC 7289
• Yale News: A 3,800-year journey from classroom to classroom (11 April 2016)
• CDLI: Plimpton 322, P254790
• Tony Phillips, Stony Brook: Old Babylonian multiplication and reciprocal tables
• UNSW Newsroom: Mathematical mystery of ancient clay tablet solved (August 2017)
• BIPM: The International System of Units (SI Brochure, 9th edition), Table 8
• D. Fowler and E. Robson, Square root approximations in Old Babylonian mathematics: YBC 7289 in context, Historia Mathematica 25(4), 1998, 366 to 378
• E. Robson, Neither Sherlock Holmes nor Babylon: a reassessment of Plimpton 322, Historia Mathematica 28(3), 2001, 167 to 206
• E. Robson, Words and pictures: new light on Plimpton 322, American Mathematical Monthly 109(2), 2002, 105 to 120
• D. F. Mansfield and N. J. Wildberger, Plimpton 322 is Babylonian exact sexagesimal trigonometry, Historia Mathematica 44(4), 2017, 395 to 419
• D. E. Knuth, Ancient Babylonian algorithms, Communications of the ACM 15(7), 1972, 671 to 677
• Wikipedia: Plimpton 322 (summary of the scholarship, used for the error list and column headings)
• Wikipedia: YBC 7289 (summary citing Fowler and Robson 1998)
• Wikipedia: Sexagesimal
• Wikipedia: Regular number
• O. Neugebauer and A. Sachs, Mathematical Cuneiform Texts (American Oriental Society, 1945), cited through MacTutor and Yale News

Last reviewed: 25 October 2026. The scholarly debate on Plimpton 322 is ongoing; check the journal articles above for the latest positions.