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${\displaystyle {\frac {4}{10}}={\frac {1}{3}}+{\frac {1}{15}}\;\;\;;\;\;\;{\frac {5}{10}}={\frac {1}{2}}\;\;\;;\;\;\;{\frac {6}{10}}={\frac {1}{2}}+{\frac {1}{10}}}$

${\displaystyle {\frac {7}{10}}={\frac {2}{3}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {8}{10}}={\frac {2}{3}}+{\frac {1}{10}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {9}{10}}={\frac {2}{3}}+{\frac {1}{5}}+{\frac {1}{30}}}$

${\displaystyle {\frac {6}{10}}={\frac {1}{2}}+{\frac {1}{10}}\;\;\;;\;\;\;{\frac {7}{10}}={\frac {2}{3}}+{\frac {1}{30}}}$

${\displaystyle {\frac {8}{10}}={\frac {2}{3}}+{\frac {1}{10}}+{\frac {1}{30}}\;\;\;;\;\;\;{\frac {9}{10}}={\frac {2}{3}}+{\frac {1}{5}}+{\frac {1}{30}}}$

${\displaystyle S=1+1/2+1/4={\frac {7}{4}}}$ and

${\displaystyle T=1+2/3+1/3=2}$.

Then for the following multiplications, write the product as an Egyptian fraction.

${\displaystyle 9:{\bigg (}{\frac {1}{2}}+{\frac {1}{14}}{\bigg )}S=1\;\;\;;\;\;\;10:{\bigg (}{\frac {1}{4}}+{\frac {1}{28}}{\bigg )}S={\frac {1}{2}}\;\;\;;\;\;\;11:{\frac {1}{7}}S={\frac {1}{4}}}$

${\displaystyle 12:{\frac {1}{14}}S={\frac {1}{8}}\;\;\;;\;\;\;13:{\bigg (}{\frac {1}{16}}+{\frac {1}{112}}{\bigg )}S={\frac {1}{8}}\;\;\;;\;\;\;14:{\frac {1}{28}}S={\frac {1}{16}}}$

${\displaystyle 15:{\bigg (}{\frac {1}{32}}+{\frac {1}{224}}{\bigg )}S={\frac {1}{16}}\;\;\;;\;\;\;16:{\frac {1}{2}}T=1\;\;\;;\;\;\;17:{\frac {1}{3}}T={\frac {2}{3}}}$

${\displaystyle 18:{\frac {1}{6}}T={\frac {1}{3}}\;\;\;;\;\;\;19:{\frac {1}{12}}T={\frac {1}{6}}\;\;\;;\;\;\;20:{\frac {1}{24}}T={\frac {1}{12}}}$

${\displaystyle 22:{\bigg (}{\frac {2}{3}}+{\frac {1}{30}}{\bigg )}+x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{5}}+{\frac {1}{10}}}$

${\displaystyle 23:{\bigg (}{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{10}}+{\frac {1}{30}}+{\frac {1}{45}}{\bigg )}+x={\frac {2}{3}}\;\;\;\rightarrow \;\;\;x={\frac {1}{9}}+{\frac {1}{40}}}$

${\displaystyle 24:x+{\frac {1}{7}}x=19\;\;\;\rightarrow \;\;\;x=16+{\frac {1}{2}}+{\frac {1}{8}}}$

${\displaystyle 25:x+{\frac {1}{2}}x=16\;\;\;\rightarrow \;\;\;x=10+{\frac {2}{3}}}$

${\displaystyle 26:x+{\frac {1}{4}}x=15\;\;\;\rightarrow \;\;\;x=12}$

${\displaystyle 27:x+{\frac {1}{5}}x=21\;\;\;\rightarrow \;\;\;x=17+{\frac {1}{2}}}$

${\displaystyle 28:{\bigg (}x+{\frac {2}{3}}x{\bigg )}-{\frac {1}{3}}{\bigg (}x+{\frac {2}{3}}x{\bigg )}=10\;\;\;\rightarrow \;\;\;x=9}$

${\displaystyle 29:{\frac {1}{3}}{\Bigg (}{\bigg (}x+{\frac {2}{3}}x{\bigg )}+{\frac {1}{3}}{\bigg (}x+{\frac {2}{3}}x{\bigg )}{\Bigg )}=10\;\;\;\rightarrow \;\;\;x=13+{\frac {1}{2}}}$

${\displaystyle 30:{\bigg (}{\frac {2}{3}}+{\frac {1}{10}}{\bigg )}x=10\;\;\;\rightarrow \;\;\;x=13+{\frac {1}{23}}}$

${\displaystyle 31:x+{\frac {2}{3}}x+{\frac {1}{2}}x+{\frac {1}{7}}x=33\;\;\;\rightarrow }$

${\displaystyle x=14+{\frac {1}{4}}+{\frac {1}{56}}+{\frac {1}{97}}+{\frac {1}{194}}+{\frac {1}{388}}+{\frac {1}{679}}+{\frac {1}{776}}}$

${\displaystyle 32:x+{\frac {1}{3}}x+{\frac {1}{4}}x=2\;\;\;\rightarrow \;\;\;x=1+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{114}}+{\frac {1}{228}}}$

${\displaystyle 33:x+{\frac {2}{3}}x+{\frac {1}{2}}x+{\frac {1}{7}}x=37\;\;\;\rightarrow \;\;\;x=16+{\frac {1}{56}}+{\frac {1}{679}}+{\frac {1}{776}}}$

${\displaystyle 34:x+{\frac {1}{2}}x+{\frac {1}{4}}x=10\;\;\;\rightarrow \;\;\;x=5+{\frac {1}{2}}+{\frac {1}{7}}+{\frac {1}{14}}}$

${\displaystyle 35:{\bigg (}3+{\frac {1}{3}}{\bigg )}x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{5}}+{\frac {1}{10}}}$

${\displaystyle 36:{\bigg (}3+{\frac {1}{3}}+{\frac {1}{5}}{\bigg )}x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{4}}+{\frac {1}{53}}+{\frac {1}{106}}+{\frac {1}{212}}}$

${\displaystyle 37:{\bigg (}3+{\frac {1}{3}}+{\frac {1}{3}}\cdot {\frac {1}{3}}+{\frac {1}{9}}{\bigg )}x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{4}}+{\frac {1}{32}}}$

${\displaystyle 38:{\bigg (}3+{\frac {1}{7}}{\bigg )}x=1\;\;\;\rightarrow \;\;\;x={\frac {1}{6}}+{\frac {1}{11}}+{\frac {1}{22}}+{\frac {1}{66}}}$

${\displaystyle V={\bigg (}d-{\frac {1}{9}}d{\bigg )}^{2}h}$

${\displaystyle ={\frac {64}{81}}d^{2}h}$

to calculate the volume of a cylindrical grain silo with a diameter of 9 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits. Furthermore, given the following equalities among other units of volume, 1 cubic cubit = 3/2 khar = 30 heqats = 15/2 quadruple heqats, also express the answer in terms of khar and quadruple heqats.

${\displaystyle =960\;\;\;khar}$

${\displaystyle =4800\;\;\;quadruple\;\;\;heqat}$

${\displaystyle ={\bigg (}1185+{\frac {1}{6}}+{\frac {1}{54}}{\bigg )}\;\;\;khar}$

${\displaystyle ={\bigg (}59+{\frac {1}{4}}+{\frac {1}{108}}{\bigg )}\;\;\;hundred\;\;\;quadruple\;\;\;heqat}$

${\displaystyle V={\frac {2}{3}}{\Bigg (}{\bigg (}d-{\frac {1}{9}}d{\bigg )}+{\frac {1}{3}}{\bigg (}d-{\frac {1}{9}}d{\bigg )}{\Bigg )}^{2}h}$

${\displaystyle ={\frac {2048}{2187}}d^{2}h}$

to calculate the volume of a cylindrical grain silo with a diameter of 9 cubits and a height of 6 cubits, directly finding the answer in Egyptian fractional terms of khar, and later in Egyptian fractional terms of quadruple heqats and quadruple ro, where 1 quadruple heqat = 4 heqat = 1280 ro = 320 quadruple ro.

${\displaystyle ={\bigg (}2275+{\frac {1}{2}}+{\frac {1}{32}}+{\frac {1}{64}}{\bigg )}\;\;\;quadruple\;\;\;heqat}$

${\displaystyle +{\bigg (}2+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{36}}{\bigg )}\;\;\;quadruple\;\;\;ro}$

${\displaystyle l=10\;\;\;cubit}$

${\displaystyle l_{3}=3+{\frac {1}{3}}\;\;\;cubit}$

${\displaystyle {\begin{bmatrix}{\frac {100}{10}}&q.\;heqat&=&10&q.\;heqat\\{\frac {100}{20}}&q.\;heqat&=&5&q.\;heqat\\{\frac {100}{30}}&q.\;heqat&=&(3+{\frac {1}{4}}+{\frac {1}{16}}+{\frac {1}{64}})&q.\;heqat\\&&+&(1+{\frac {2}{3}})&q.\;ro\\{\frac {100}{40}}&q.\;heqat&=&(2+{\frac {1}{2}})&q.\;heqat\\{\frac {100}{50}}&q.\;heqat&=&2&q.\;heqat\\{\frac {100}{60}}&q.\;heqat&=&(1+{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}})&q.\;heqat\\&&+&(3+{\frac {1}{3}})&q.\;ro\\{\frac {100}{70}}&q.\;heqat&=&(1+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{32}}+{\frac {1}{64}})&q.\;heqat\\&&+&(2+{\frac {1}{14}}+{\frac {1}{21}}+{\frac {1}{42}})&q.\;ro\\{\frac {100}{80}}&q.\;heqat&=&(1+{\frac {1}{4}})&q.\;heqat\\{\frac {100}{90}}&q.\;heqat&=&(1+{\frac {1}{16}}+{\frac {1}{32}}+{\frac {1}{64}})&q.\;heqat\\&&+&({\frac {1}{2}}+{\frac {1}{18}})&q.\;ro\\{\frac {100}{100}}&q.\;heqat&=&1&q.\;heqat\\\end{bmatrix}}}$

${\displaystyle l_{2}={\bigg (}3+{\frac {1}{4}}+{\frac {1}{8}}{\bigg )}\;\;\;khet}$

${\displaystyle A_{1}={\bigg (}13+{\frac {1}{2}}+{\frac {1}{4}}{\bigg )}\;\;\;setat+{\bigg (}3+{\frac {1}{8}}{\bigg )}\;\;\;cubit\;\;\;strip}$

${\displaystyle A_{2}={\bigg (}9+{\frac {1}{2}}+{\frac {1}{4}}{\bigg )}\;\;\;setat+{\bigg (}9+{\frac {1}{4}}+{\frac {1}{8}}{\bigg )}\;\;\;cubit\;\;\;strip}$

${\displaystyle A_{3}={\bigg (}7+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}{\bigg )}\;\;\;setat}$

${\displaystyle ={\bigg (}{\frac {1}{2}}+{\frac {1}{8}}{\bigg )}\;\;\;setat+{\bigg (}7+{\frac {1}{2}}{\bigg )}\;\;\;cubit\;\;\;strip}$

${\displaystyle ={\frac {1}{2}}\;\;\;setat+10\;\;\;cubit\;\;\;strip}$

2) Consider a right regular square pyramid whose base, the square face is coplanar with a plane (or the ground, say), so that any of the planes containing its triangular faces has the dihedral angle of ${\displaystyle \theta }$ with respect to the ground-plane (that is, on the interior of the pyramid). In other words, ${\displaystyle \theta }$ is the angle of the triangular faces of the pyramid with respect to the ground. The seked of such a pyramid, then, having altitude ${\displaystyle a}$ and base edge length ${\displaystyle b}$, is defined as that physical length ${\displaystyle S}$ such that ${\displaystyle {\frac {S}{1\;\;\;royal\;\;\;cubit}}=}$ ${\displaystyle \cot {\theta }}$. Put another way, the seked of a pyramid can be interpreted as the ratio of its triangular faces' run per one unit (cubit) rise. Or, for the appropriate right triangle on a pyramid's interior having legs ${\displaystyle a,{\frac {b}{2}}}$ and the perpendicular bisector of a triangular face as the hypotenuse, then the pyramid's seked ${\displaystyle S}$ satisfies ${\displaystyle \cot {\theta }={\frac {b}{2a}}={\frac {S}{1\;\;\;royal\;\;\;cubit}}}$. Similar triangles are therefore described, and one can be scaled to the other.

3) A pyramid has an altitude of 250 (royal) cubits, and the side of its base has a length of 360 (royal) cubits. Find its seked ${\displaystyle S}$ in Egyptian fractional terms of (royal) cubits, and also in terms of palms.

${\displaystyle ={\bigg (}5+{\frac {1}{25}}{\bigg )}\;\;\;palm}$

${\displaystyle S=5\;\;\;palm+1\;\;\;finger}$

${\displaystyle a=8\;\;\;cubit}$

${\displaystyle {\begin{bmatrix}{\frac {2}{3}}\cdot {\frac {2}{3}}={\frac {1}{3}}+{\frac {1}{9}}&;&{\frac {1}{3}}\cdot {\frac {2}{3}}={\frac {1}{6}}+{\frac {1}{18}}\\{\frac {2}{3}}\cdot {\frac {1}{3}}={\frac {1}{6}}+{\frac {1}{18}}&;&{\frac {2}{3}}\cdot {\frac {1}{6}}={\frac {1}{12}}+{\frac {1}{36}}\\{\frac {2}{3}}\cdot {\frac {1}{2}}={\frac {1}{3}}&;&{\frac {1}{3}}\cdot {\frac {1}{2}}={\frac {1}{6}}\\{\frac {1}{6}}\cdot {\frac {1}{2}}={\frac {1}{12}}&;&{\frac {1}{12}}\cdot {\frac {1}{2}}={\frac {1}{24}}\\{\frac {1}{9}}\cdot {\frac {2}{3}}={\frac {1}{18}}+{\frac {1}{54}}&;&{\frac {2}{3}}\cdot {\frac {1}{9}}={\frac {1}{18}}+{\frac {1}{54}}\\{\frac {1}{4}}\cdot {\frac {1}{5}}={\frac {1}{20}}&;&{\frac {2}{3}}\cdot {\frac {1}{7}}={\frac {1}{14}}+{\frac {1}{42}}\\{\frac {1}{2}}\cdot {\frac {1}{7}}={\frac {1}{14}}&;&{\frac {2}{3}}\cdot {\frac {1}{11}}={\frac {1}{22}}+{\frac {1}{66}}\\{\frac {1}{3}}\cdot {\frac {1}{11}}={\frac {1}{33}}&;&{\frac {1}{2}}\cdot {\frac {1}{11}}={\frac {1}{22}}\\{\frac {1}{4}}\cdot {\frac {1}{11}}={\frac {1}{44}}&&\\\end{bmatrix}}}$

${\displaystyle q=6(2n+1)}$

${\displaystyle 200}$

${\displaystyle 133+{\frac {1}{3}}}$

${\displaystyle 100}$

${\displaystyle {\bigg (}1+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}1+{\frac {1}{4}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}1+{\frac {1}{8}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}1+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}{\frac {1}{2}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle {\bigg (}{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}{\bigg )}\;heqat}$

${\displaystyle 15+{\frac {1}{3}}+{\frac {1}{26}}+{\frac {1}{78}}}$

${\displaystyle O_{8}=26+{\frac {2}{3}}\;\;\;quadruple\;\;\;heqat}$

${\displaystyle O_{6}=20\;\;\;quadruple\;\;\;heqat}$

${\displaystyle O_{4}=13+{\frac {1}{3}}\;\;\;quadruple\;\;\;heqat}$

the quantity of finished food product units ${\displaystyle p}$ yielded from exactly one heqat of raw food material.

2) 3 + 1/2 heqats of meal produce 80 loaves of bread. Find the meal per loaf ${\displaystyle m}$ in heqats and ro, and find the pefsu ${\displaystyle P}$ of these loaves with respect to the meal. Express them as Egyptian fractions.

${\displaystyle P={\bigg (}22+{\frac {2}{3}}+{\frac {1}{7}}+{\frac {1}{21}}{\bigg )}{\frac {loaf}{heqat_{meal}}}}$

${\displaystyle P={\bigg (}12+{\frac {2}{3}}+{\frac {1}{42}}+{\frac {1}{126}}{\bigg )}{\frac {loaf}{heqat_{meal}}}}$

${\displaystyle y=2000}$

${\displaystyle {\begin{bmatrix}houses&7\\cats&49\\mice&343\\spelt&2401\\heqat&16807\\Total&19607\\\end{bmatrix}}}$

Moreover, one of Ahmes' methods of solution for the sum suggests an understanding of finite geometric series. Ahmes performs a direct sum, but he also presents a simple multiplication to get the same answer: "2801 x 7 = 19607". Chace explains that since the first term, the number of houses (7) is equal to the common ratio of multiplication (7), then the following holds (and can be generalized to any similar situation):

${\displaystyle \sum \limits _{k=1}^{n}7^{k}=7{\bigg (}1+\sum \limits _{k=1}^{n-1}7^{k}{\bigg )}}$

That is, when the first term of a geometric sequence is equal to the common ratio, partial sums of geometric sequences, or finite geometric series, can be reduced to multiplications involving the finite series having one less term, which does prove convenient in this case. In this instance then, Ahmes simply adds the first four terms of the sequence (7 + 49 + 343 + 2401 = 2800) to produce a partial sum, adds one (2801), and then simply multiplies by 7 to produce the correct answer.

${\displaystyle {\begin{bmatrix}1&heqat&=&10&hinu\\{\frac {1}{2}}&heqat&=&5&hinu\\{\frac {1}{4}}&heqat&=&(2+{\frac {1}{2}})&hinu\\{\frac {1}{8}}&heqat&=&(1+{\frac {1}{4}})&hinu\\{\frac {1}{16}}&heqat&=&({\frac {1}{2}}+{\frac {1}{8}})&hinu\\{\frac {1}{32}}&heqat&=&({\frac {1}{4}}+{\frac {1}{16}})&hinu\\{\frac {1}{64}}&heqat&=&({\frac {1}{8}}+{\frac {1}{32}})&hinu\\\end{bmatrix}}}$

Begin with the statement that "10 fattening geese eat 2 + 1/2 heqats in one day". In other words, the daily rate of consumption (and initial condition) ${\displaystyle i}$ is equal to 2 + 1/2. Determine the number of heqats which 10 fattening geese eat in 10 days, and in 40 days. Call these quantities ${\displaystyle t}$ and ${\displaystyle f}$, respectively.

Multiply the above latter quantity ${\displaystyle f}$ by 5/3 to express the amount of "spelt", or ${\displaystyle s}$, required to be ground up.

Multiply ${\displaystyle f}$ by 2/3 to express the amount of "wheat", or ${\displaystyle w}$, required.

Divide ${\displaystyle w}$ by 10 to express a "portion of wheat", or ${\displaystyle p}$, which is to be subtracted from ${\displaystyle f}$.

Find ${\displaystyle f-p=g}$. This is the amount of "grain", (or wedyet flour, it would seem), which is required to make the feed for geese, presumably on the interval of 40 days (which would seem to contradict the original statement of the problem, somewhat). Finally, express ${\displaystyle g}$ again in terms of hundreds of double heqats, double heqats and double ro, where 1 hundred double heqat = 2 hundred heqat = 100 double heqat = 200 heqat = 32,000 double ro = 64,000 ro. Call this final quantity ${\displaystyle g_{2}}$.

${\displaystyle t=25\;\;\;heqat}$

${\displaystyle f=100\;\;\;heqat}$

${\displaystyle s={\bigg (}1+{\frac {1}{2}}{\bigg )}hundred\;\;\;heqat}$

${\displaystyle +{\bigg (}16+{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}{\bigg )}\;\;\;heqat+{\bigg (}3+{\frac {1}{3}}{\bigg )}\;\;\;ro}$

${\displaystyle w={\bigg (}{\frac {1}{3}}+{\frac {1}{4}}{\bigg )}hundred\;\;\;heqat}$

${\displaystyle +{\bigg (}8+{\frac {1}{4}}+{\frac {1}{16}}+{\frac {1}{64}}{\bigg )}\;\;\;heqat+{\bigg (}1+{\frac {2}{3}}{\bigg )}\;\;\;ro}$

${\displaystyle p={\bigg (}6+{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}{\bigg )}\;\;\;heqat+{\bigg (}3+{\frac {1}{3}}{\bigg )}\;\;\;ro}$

${\displaystyle g={\bigg (}{\frac {1}{2}}+{\frac {1}{4}}{\bigg )}hundred\;\;\;heqat}$

${\displaystyle +{\bigg (}18+{\frac {1}{4}}+{\frac {1}{16}}+{\frac {1}{64}}{\bigg )}\;\;\;heqat+{\bigg (}1+{\frac {2}{3}}{\bigg )}\;\;\;ro}$

${\displaystyle g_{2}={\bigg (}{\frac {1}{4}}{\bigg )}hundred\;\;\;double\;\;\;heqat}$

${\displaystyle +{\bigg (}21+{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}{\bigg )}\;\;\;double\;\;\;heqat}$

${\displaystyle +{\bigg (}3+{\frac {1}{3}}{\bigg )}\;\;\;double\;\;\;ro}$

${\displaystyle t={\bigg (}12+{\frac {1}{2}}{\bigg )}\;\;\;heqat}$

${\displaystyle f=50\;\;\;heqat}$

${\displaystyle g_{2}={\bigg (}23+{\frac {1}{4}}+{\frac {1}{16}}+{\frac {1}{64}}{\bigg )}\;\;\;double\;\;\;heqat}$

${\displaystyle +{\bigg (}1+{\frac {2}{3}}{\bigg )}\;\;\;double\;\;\;ro}$

Suppose that four geese are cooped up, and their collective daily allowance of feed is equal to one hinu. Express one goose's daily allowance of feed ${\displaystyle a_{1}}$ in terms of heqats and ro.

Suppose that the daily feed for a goose "that goes into the pond" is equal to 1/16 + 1/32 heqats + 2 ro. Express this same daily allowance ${\displaystyle a_{2}}$ in terms of hinu.

Suppose that the daily allowance of feed for 10 geese is one heqat. Find the 10-day allowance ${\displaystyle a_{10}}$ and the 30-day, or one-month allowance ${\displaystyle a_{30}}$ for the same group of animals, in heqats.

Finally a table will be presented, giving daily feed portions to fatten one animal of any of the indicated species.

${\displaystyle a_{2}=1\;\;\;hinu}$

${\displaystyle a_{10}=10\;\;\;heqat}$

${\displaystyle a_{30}=30\;\;\;heqat}$

${\displaystyle {\begin{bmatrix}goose&({\frac {1}{8}}+{\frac {1}{32}})&heqat&+&(3+{\frac {1}{3}})&ro\\terp-goose&({\frac {1}{8}}+{\frac {1}{32}})&heqat&+&(3+{\frac {1}{3}})&ro\\crane&({\frac {1}{8}}+{\frac {1}{32}})&heqat&+&(3+{\frac {1}{3}})&ro\\set-duck&({\frac {1}{32}}+{\frac {1}{64}})&heqat&+&1&ro\\ser-goose&{\frac {1}{64}}&heqat&+&3&ro\\dove&&&&3&ro\\quail&&&&3&ro\\\end{bmatrix}}}$

${\displaystyle {\begin{bmatrix}&Loaves&Common\;food\\4\;fine\;oxen&24\;heqat&2\;heqat\\2\;fine\;oxen&22\;heqat&6\;heqat\\3\;cattle&20\;heqat&2\;heqat\\1\;ox&20\;heqat&\\Total&86\;heqat&10\;heqat\\in\;spelt&9\;heqat&(7+{\frac {1}{2}})\;heqat\\10\;days&({\frac {1}{2}}+{\frac {1}{4}})\;c.\;heqat&({\frac {1}{2}}+{\frac {1}{4}})\;c.\;heqat\\&+15\;heqat&\\one\;month&200\;heqat&({\frac {1}{2}}+{\frac {1}{4}})\;c.\;heqat\\&&+15\;heqat\\double\;heqat&{\frac {1}{2}}\;c.\;heqat&{\frac {1}{4}}\;c.\;heqat\\&+(11+{\frac {1}{2}}+{\frac {1}{8}})\;heqat&+5\;heqat\\&+3\;ro&\\\end{bmatrix}}}$

Problem 84 seems to suggest a procedure to estimate various food materials and allowances in similar terms as the previous three problems, but the extant information is deeply confused. Still, there are hints of consistency. The problem seems to start out like a conventional story problem, describing a stable with ten animals of four different types. It seems that the four types of animals consume feed, or "loaves" at different rates, and that there are corresponding amounts of "common" food. These two columns of information are correctly summed in the "total" row, however they are followed by two "spelt" items of dubious relationship to the above. These two spelt items are indeed each multiplied by ten to give the two entries in the "10 days" row, once unit conversions are accounted for. The "one month" row items do not seem to be consistent with the previous two, however. Finally, information in "double heqats" (read hundred double heqats, double heqats and double ro for these items) concludes the problem, in a manner reminiscent of 82 and 82B. The two items in the final row are in roughly, but not exactly, the same proportion to one another as the two items in the "one month" row.

"1... living forever. List of the food in Hebenti...

2... his brother the steward Ka-mose...

3... of his year, silver, 50 pieces twice in the year...

4... cattle 2, in silver 3 pieces in the year...

5... one twice; that is, 1/6 and 1/6. Now as for one...

6... 12 hinu; that is, silver, 1/4 piece; one...

7... (gold or silver) 5 pieces, their price therefor; fish, 120, twice...

8... year, barley, in quadruple heqat, 1/2 + 1/4 of 100 heqat 15 heqat; spelt, 100 heqat... heqat...

9... barley, in quadruple heqat, 1/2 + 1/4 of 100 heqat 15 heqat; spelt, 1 + 1/2 + 1/4 times 100 heqat 17 heqat...

10... 146 + 1/2; barley, 1 + 1/2 + 1/4 times 100 heqat 10 heqat; spelt, 300 heqat... heqat...

11... 1/2, there was brought wine, 1 ass(load?)...

12... silver 1/2 piece; ... 4; that is, in silver...

13... 1 + 1/4; fat, 36 hinu; that is, in silver...

14... 1 + 1/2 + 1/4 times 100 heqat 21 heqat; spelt, in quadruple heqat, 400 heqat 10 heqat...

15-18 (These lines are repetitions of line 14.)"

The first month of the inundation season, 23rd day, the commander (?) of the army (?) attacked (?) Zaru.

25th day, it was heard that Zaru was entered.

Year 11, first month of the inundation season, third day. Birth of Set; the majesty of this god caused his voice to be heard.

Birth of Isis, the heavens rained."

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