After 2050 BC, a form of Egyptian multiplication doubled answers to problems. The doubled answers reported the arithmetic correctness of answers. Scholars noted that many initial and intermediate calculations were missing, and searched for the ab initio information. Finding few ciphered hieratic letters were transliterated into modern numbers leaving many hieratic arithmetic statements untranslated into modern arithmetic. Ahmes, for example, doubled the RMP 38 answer, substituting 22/7 for pi, to correct inventory losses of grain. The traditional pi value of 256/81 over-estimated grain in hekats, a context that scholars reported little. Scribes showed off skills with RMP 2/n table unit fraction series within finite arithmetic statements, another topic that scholars reported little before 2000 AD. Ahmes solved 87 problems, several complex ones, in the Rhind Mathematical Papyrus (RMP) by using an Egyptian fraction notation that early scholars understood on the additive level, but not on the deeper scribal level until the 21st century. The RMP included selections of least common multiples that scaled 2/n table answers, algebra answers, geometry answers, arithmetic progression answers, hekat, hin, dja, ro, and pesu inventory controls of grain, bread, and beer calculated in the notation, topics that drew many scholarly discussions.
False position was suggested by 20th century scholars to decode the division aspect of the mathematical texts. Scribes commonly used Old Kingdom doubling as proofs during the Middle Kingdom. In the 21st century scholars have decoded several ab initio fragments, i.e. newly parsing RMP 38, that pointed out hieratic multiplication and division operations were inverse operations in a manner that followed modern-like conventions, thereby refuting the false position suggestion.
Traditional 20th century scholars reported that Egyptian Middle Kingdom scribes were confined to applications of practical arithmetic with problems additively addressing how a number of loaves were equally shared, and little more. One exception was Richard Gillings. He reported rhetorical algebra's alternando [(y/x) =(q/p)], and dividendo [(y -x)/x = (q -p)/p] as proportions that mixed beer raw materials in RMP 73 and 75, and a harmonic mean that mixed ingredients in sacrificial bread in RMP 76 pesu formulas. Problems in the Moscow Mathematical Papyrus and Rhind Mathematical Papyrus (RMP) also expressed instructional views of bread and beer production beyond scribal practical limitations. Four ancient texts cover abstract definitions of number, and higher forms of arithmetic. Abstract definitions appear in the Akhmim Wooden Tablet, the Egyptian Mathematical Leather Roll, the Kahun Papyrus, and the Rhind Mathematical Papyrus. The abstract arithmetic scaled hekat and other weights and measures units. The hekat used Eye of Horus quotients and Egyptian fraction remainders scaled to ro, 1/320 of a hekat, and other sub-units. Five hekat two-part statements were used in the Akhmim Wooden Tablet and applied 30 times in the Rhind Mathematical Papyrus, and additional times in the Ebers Papyrus clarifying the arithmetic operations of Egyptian fraction arithmetic.
Circa 2700 BC Egyptians introduced the earliest fully developed base 10 numeration system. Though it was not a positional system, it allowed the use of large numbers and also fractions in the form of unit fractions and Eye of Horus fractions, or binary fractions. 
By 2700 BC, Egyptian construction techniques included precision surveying, marking north by the sun's location at noon. Clear records began to appear by 2000 BC citing approximations for π and square roots. Exact statements of number, written arithmetic tables, algebra problems, and practical applications with weights and measures also appeared after 2050 BC in hieratic script, with several problems solved by abstract arithmetic methods.
The Akhmim Wooden Tablet (AWT) listed five divisions of a unit of volume called a hekat beginning with a hekat unity valued at (64/64). The hekat unity was divided by 3, 7, 10, 11 and 13, and recorded by exact unit fraction answers. The first half of the answers cited binary quotients, (64/64/n). For example (64/64) was divided by 3, with a missing intermediate steps reporting a quotient 21/64 and a remainder 1/192. The scribe wrote the quotient 21/64 as (16 + 4 + 1)/64 obtaining (16 + 4 + 1)/64, or (1/4 + 1/16 + 1/64) hekat. The remainder scaled 1/192 to ro units, 1/320 of a hekat, scaling (1/192)*(5/5) by writing (5/3)*(1/320) and finally (1 + 2/3)ro.
The scribe combined quotients and remainders into one statement. The 1/3rd of a hekat was recorded as: (1/4 1/16 1/64)hekat (1 2/3)ro). Scribal addition and multiplication signs were not recorded. The 1/3 of a hekat series was written from right to left. The scribe proved answers by multiplying by initial divisors, finding the initial hekat unity value of (64/64) five times. The AWT scribe used an exact partitioning method, a method that was shortened by Ahmes and other Middle Kingdom scribes by omitting the proof step. However, Ahmes' partitioning steps did follow the AWT's two-part structure, using it 29 times in Rhind Mathematical Papyrus #81, and additional times in other problems.
Hana Vymazalova published in 2002 a fresh copy of the AWT that showed that all five AWT divisions had been exact, by parsing the proof steps, and returning the five division answers to (64/64). Vymazalova thereby updated Daressy's 1906 incomplete discussion of the subject that had only found 1/3, 1/7 and 1/10 answers as exact.
Beyond the (64/64)/n = (Q/64) hekat + (5R/n)ro (with Q a quotient, and R a remainder) formula two additional formula reveal early scribal thinking. The first formula allowed divisors to exceed n the limit 1/64 < n < 64 = 64. Sub-units of the hekat were developed,i.e. hin, dja and ro. In RMP 81 data 29 examples contrast two-part statements to equivalent (10/n)hin and (320/n) statements. The medical texts and 2,000 prescriptions extended the one-part method to (64/n)dja ingredients.
Ahmes went beyond the divisor 1/64 < n < 64 limit by writing quotient and remainder arithmetic in one other way. A second method increased the size of the numerator. A two-part hekat method used in RMP 35 divided 100 hekat by 70. Ahmes wrote (6400/64)/70 as (91/64 + 30/4480)hekat. The quotient was written (64 + 16 + 8 + 2 + 1)/64 or (1 1/4 1/8 1/32 1/64)hekat. Ahmes scaled the remainder (30/4480) by (5/5) obtaining (150/70)*(1/320) by writing (2 1/7)ro. The combined (1 1/4 1/8 1/32 1/64)hekat (2 1/7) ro answer was written down from right to left and used no arithmetic addition or multiplication signs, thereby following the Akhmim Wooden Tablet methodology.
Our understanding of ancient Egyptian mathematics has been impeded by the reported paucity of available sources. The most famous such source is the Rhind Mathematical Papyrus, a text that can be read by comparing many of its elements against other texts, i.e., the Egyptian Mathematical Leather Roll and the Akhmim Wooden Tablet. The Rhind papyrus dates from the Second Intermediate Period (circa 1650 BC), but its author, Ahmes, identifies it as a copy of a now lost Middle Kingdom papyrus. The Rhind papyrus contains a table of 101 Egyptian fraction expansions for numbers of the form 2/n, and 84 word problems, the answers to which were expressed in Egyptian fraction notation.
The RMP also includes formulas and methods for addition, subtraction, multiplication and division of sums of unit fractions. The RMP contains evidence of other mathematical knowledge,  including composite and prime numbers; arithmetic, geometric and harmonic means; and understanding of both the Sieve of Eratosthenes and perfect number theory. It also shows how to solve first order linear equations  as well as summing arithmetic and geometric series. 
Henry Rhind's estate donated the Rhind papyrus to the British Museum in 1863. Also included in the donation was the Egyptian Mathematical Leather Roll, dating from the Middle Kingdom era. Like the Rhind papyrus, the Egyptian Mathematical Leather Roll contains a table of Egyptian fraction expansions.
The Berlin papyrus, written around 1300 BC, shows that ancient Egyptians had solved two second-order, one unknown, equations that some have called Diophantine equations. The Berlin method for solving x2 + y2 = 100 has not been confirmed in a second hieratic text, though it has been confirmed by a second Berlin Papyrus problem. 
Sources other than the ones mentioned above include the Moscow Mathematical Papyrus, the Reisner Papyrus, and several other texts including medical prescriptions found in the Ebers Papyrus. math answers for dicussing this is what i need =]
Two number systems were used in ancient Egypt. One, written in hieroglyphs, was a decimal based tally system with separate symbols for 10, 100, 1000, etc, as Roman numerals were later written, and hieratic unit fractions. The second, written in a new ciphered one-number-to-one-symbol system was a digital system that was not similar to hieroglyphic system. The hieroglyphic number system existed from at least the Early Dynastic Period. The hieratic system differed from the hieroglyphic system beyond a use of simplifying ligatures for rapid writing and began around 2150 BC. Hieratic numerals used one symbol for each number replacing the tallies that had been used to denote multiples of a unit. For example, two symbols had been used to write three, thirty, three hundred, and so on, in a system that was superseded by the hieratic method. Later hieroglyphic numeration was modified and adopted by the Romans for official uses, and Egyptian fractions in everyday situations.
The Rhind Mathematical Papyrus was written in hieratic. It contains examples of how the Egyptians did their mathematical calculations. Fractions were denoted by placing a line over the letter n associated with the number being written, as 1/n. This method of writing numbers came to dominate the Ancient Near East, with Greeks 1,500 years later using two of their alphabets, Ionian and Doric, to cipher all of their numerals, alpha = 1, beta = 2 and so forth. Concerning fractions, Greeks wrote 1/n as n', so Greek numeration and problem-solving adopted or modified Egyptian numeration, arithmetic and other aspects of Egyptian math.
Example from the Rhind Papyrus
5 + 1⁄2 + 1⁄7 + 1⁄14 (= 5 5⁄7)
Egyptian multiplication was done by repeated doubling of the number to be multiplied (the multiplicand), and choosing which of the doublings to add together (essentially a form of binary arithmetic), a method that links to the Old Kingdom. The multiplicand was written next to the figure 1; the multiplicand was then added to itself, and the result written next to the number 2. The process was continued until the doublings gave a number greater than half of the multiplier. Then the doubled numbers (1, 2, etc.) would be repeatedly subtracted from the multiplier to select which of the results of the existing calculations should be added together to create the answer.
As a short cut for larger numbers, the multiplicand can also be immediately multiplied by 10, 100, etc.
For example, Problem 69 on the Rhind Papyrus (RMP) provides the following illustration, as if Hieroglyphic symbols were used (rather than the RMP's actual hieratic script).
|To multiply 80 × 14|
|Egyptian calculation||Modern calculation|
The / denotes the intermediate results that are added together to produce the final answer.
Hieratic and Middle Kingdom math followed this form of hieroglyphic multiplication.
Subtraction defined in the Egyptian Mathematical Leather Roll (EMLR), an 1800 BC document, included four additive or identity methods, followed by one non-additive, abstract, method that was used five to fifteen times for the 26 EMLR series listed, that looked like this:
1/pq = (1/A)* (A/pq)
with A = 3, 4, 5, 7, 25, citing A = (p + 1) 10 times.
1/8 was written using A = (2 + 1)= 3, the A = (p + 1) case, as used in the RMP 24 times, seeing p = 2, q = 4 and A = 25, following
A = 3: 1/8 = (1/3)*(3/8) = 1/3*(1/4 + 1/8) = 1/12 + 1/24
A = 25: 1/8 = 1/25*(25/8) = 1/5*(25/40)= 1/5 *(24/40 + 1/40)
= 1/5*(3/5 + 1/40) = 1/5*(1/5 + 2/5 + 1/40) = 1/5 *(1/5 + 1/3 + 1/15 + 1/40) = 1/25 + 1/15 + 1/75 + 1/200
with the out-of-order 1/25 + 1/15 sequence marking the scribal method of partition.
Confirmation of the EMLR (1/A)* (A/pq), with A = (p + 1) rule is found 24 times in the RMP 2/nth table, using the form
2/pq = (2/A)* (A/pq), with A = (p + 1)
example, 2/27, a = 3, q = 9
2/27 = 2/(3 + 1)*(3 + 1)/9 = 1/4*(1/3 + 1/9) = 1/12 + 1/36
Another subtraction method is seen in the RMP 2/nth table as first suggested by F. Hultsch in 1895, and confirmed by E.M. Bruins in 1944, or
2/p - 1/A = (2A - p)/Ap
2/p = 1/A + (2A -p)/Ap
where the divisors of A, from the first partition, were used to additively find (2A - p), thereby exactly solving (2A -p)/Ap.
2/19 - 1/12 = (24 - 19)/(12*19)
with the divisors of 12 = 6, 4, 3, 2, 1 being inspected to find (24 - 19) = 5 taken only from the divisors of 12. Optimally (3 + 2) was selected, by Ahmes and other scribes, over (4 + 1) such that,
2/19 = 1/12 + (3 + 2)/(12*19) = 1/12 + 1/76 + 1/114
Rational numbers could also be expressed, but only as sums of unit fractions, i.e. sums of reciprocals of positive integers, 2/3, and 3/4. The hieroglyph indicating a fraction looked like a mouth, which meant "part", and fractions were written with this fractional solidus, i.e. the numerator 1, and the positive denominator below. Special symbols were used for 1/2 and for two non-unit fractions, 2/3 (used often) and 3/4 (used less often).
Problem 25 on the Rhind Papyrus may have used the method of false position to solve the problem "a quantity and its half added together become 16; what is the quantity?" (i.e., in modern algebraic notation, what is x if x+½x=16).
1 2 / ½ 1 / Total 1½ 3
As many times as 3 must be multiplied to give 16, so many times must 2 be multiplied to give the answer.
1 3 / 2 6 4 12 / 2/3 2 1/3 1 /
Total 5 1/3 16
1 5 1/3 (1 + 4 + 1/3) 2 10 2/3
The answer is 10 2/3.
1 10 2/3 ½ 5 1/3
Total 1½ 16
A more likely and direct approach to solve this class of problem is given by: x + (1/2)x = 16, using these steps
1. (3/2)x = 16, 2. x = 32/3, 3. x = 10 2/3.
Problem 31 sets the problem "q quantity, its 2/3, its 1/2 and its 1/7, added together, become 33; what is the quantity?" In modern algebraic notation, "what is x if x + 2/3 x + 1/2 x + 1/7 x =33?" The answer is 14 1/4 1/56 1/97 1/194 1/388 1/679 1/776, or 14 and 28/97. To solve the problem as Ahmes wrote his answer 28/97 had to be broken up into 2/97 and 26/97, and solved the two separate vulgar fraction conversion problems using Hultsch-Bruins (without using false position, as other algebra problem may have been solved).
The remainder arithmetic solution, the historical method that is most likely, for x + (2/3)x + (1/2)x + (1/7)x = 33 looks like this:
1. 97/42 x = 33, 2. x = 1386/97, and 3. x = 14 + 28/97.
with, 2/97 - 1/56 = (112 - 97)/(56*97) = (8 + 7)/(56*97) = 1/679 1/776,
and 26/97 - 1/4 = (104-97/(4*97) = (4 + 2 + 1)/(4*97)= 1/97 1/194 1/388,
2/97 = 1/56 1/670 1/776,
26/97 = 1/4 1/97 1/194 1/388
such that, writing out x = 14 + 28/97 in an ordered unit fraction series
4. x = 14 1/4 1/56 1/97 1/194 1/388 1/679 1/776, as written by Ahmes.
The ancient Egyptians knew that they could approximate the area of a circle as follows:
Problem 50 of the Ahmes papyrus uses these methods to calculate the area of a circle, according to a rule that the area is equal to the square of 8/9 of the circle's diameter. This assumes that π is 4×(8/9)² (or 3.160493...), with an error of slightly over 0.63 percent. This value was slightly less accurate than the calculations of the Babylonians (25/8 = 3.125, within 0.53 percent), but was not otherwise surpassed until Archimedes' approximation of 211875/67441 = 3.14163, which had an error of just over 1 in 10,000. Interestingly, Ahmes knew of the modern 22/7 as an approximation for pi, and used it to split a hekat, hekat x 22/x x 7/22 = hekat; however, Ahmes continued to use the traditional 256/81 value for pi for computing his hekat volume found in a cylinder.
Problem 48 involved using a square with side 9 units. This square was cut into a 3x3 grid. The diagonal of the corner squares were used to make an irregular octagon with an area of 63 units. This gave a second value for π of 3.111...
The two problems together indicate a range of values for Pi between 3.11 and 3.16.
RMP 53 calculated the areas of two triangles by the formula: 1/2 base times the altitude, and the area of a third shape by another method. Cubit, khet lengths were used to find areas in setat and mh units. The setat was 100 by 100 cubits and a mh was 1/100 of a setat, one cubit by 100 cubits. RMP 54 partitioned a setat into 7/10, 14/10, 28/10 and 56/10 into setat and mh segments. RMP 55 divided 5 setat by 3/5 to obtain 3 setat, taking three multiples of 1/8 setat and remainder mh units and summing 1/2 setat and 3 1/4 1/8 setat plus 1/8 setat written in mh units.