nyuuzyou/wikis huggingface dataset:
Greatest Common Measure of Commensurable Magnitudes

== Theorem ==
=== Porism ===
== Proof ==
Let $AB$ and $CD$ be two commensurable magnitudes such that $AB < CD$.
$AB$ either measures $CD$ or it does not.
Let $AB$ measure $CD$.
As $AB$ also measures itself, it follows that $AB$ is a common measure for $AB$ and $CD$.
As no magnitude can measure a smaller magnitude, it follows that $AB$ is the greatest common measure of $AB$ and $CD$.
:
Let $AB$ not measure $CD$.
Let the less be continually subtracted in turn from the greater.
We are given that $AB$ and $CD$ are commensurable.
From Incommensurable Magnitudes do not Terminate in Euclid's Algorithm, that which is left over will eventually measure the one before it.
Let $AB$ measure $ED$ and leave $EC$ less than $AB$.
Let $EC$ measure $FB$ and leave $AF$ less than $EC$.
Let $AF$ measure $CE$.
Since:
: $AF$ measure $CE$
and:
: $CE$ measure $FB$
it follows that:
: $AF$ measure $FB$.
But $AF$ also measures itself.
Therefore $AF$ measures the whole of $AB$.
But $AB$ also measures $DE$.
Therefore $AF$ also measures $DE$.
But $AB$ also measures $CE$.
Therefore $AF$ also measures the whole of $CD$.
Therefore $AF$ is a common measure for $AB$ and $CD$.
Suppose there were some magnitude $G$ which is greater than $AF$ which is also a common measure for $AB$ and $CD$.
Since:
:$G$ measures $AB$
and:
:$AB$ measures $ED$
it follows that:
:$G$ measures $ED$.
But $G$ also measures the whole of $CD$.
Therefore $G$ also measures the remainder $CE$.
But $CE$ measures $FB$.
Therefore $G$ also measures $FB$.
But $G$ also measures the whole of $AB$.
Therefore it also measures the remainder $AF$.
But $G$ is greater than $AF$, and so cannot measure it.
From this contradiction it follows that $G$ cannot be a common measure for $AB$ and $CD$.
That is, $AF$ is the greatest common measure of $AB$ and $CD$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Greatest Common Measure of Three Commensurable Magnitudes

== Theorem ==
=== Porism ===
== Proof ==
Let $A, B, C$ be three commensurable magnitudes.
From Greatest Common Measure of Commensurable Magnitudes, let the greatest common measure of $A$ and $B$ be $D$.
Either $D$ measures $C$ or it does not.
Let $D$ measure $C$.
Since:
:$D$ measures $C$
and:
:$D$ measures $A$ and $B$
then:
:$D$ is a common measure for $A, B$ and $C$.
As no magnitude can measure a smaller magnitude, it follows that $D$ is the greatest common measure of $A, B$ and $C$.
Let $D$ not measure $C$.
As $A$, $B$ and $C$ are commensurable, then some magnitude $E$ will measure them by definition.
$E$ will measure both $A$ and $B$.
From the porism to Greatest Common Measure of Commensurable Magnitudes, $E$ also measures their greatest common measure $D$.
But $E$ also measures $C$.
Therefore $E$ will measure $C$ and $D$.
Therefore $C$ and $D$ are commensurable.
So from Greatest Common Measure of Commensurable Magnitudes, let $E$ be the greatest common measure of $C$ and $D$.
Since:
:$E$ measures $D$
and:
:$D$ measures $A$ and $B$
then:
:$E$ measures $A$ and $B$.
But $E$ measures $C$ also.
So $E$ is a common measure for $A, B$ and $C$.
Suppose there were some magnitude $F$ which is greater than $E$ which is also a common measure for $A, B$ and $C$.
Since $F$ measures $A, B$ and $C$, it measures $A$ and $B$.
From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ also measures the greatest common measure of $A$ and $B$.
That is, $F$ measures $C$
But $F$ measures $D$.
From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ measures the greatest common measure of $D$ and $C$.
That is, $F$ measures $E$.
But $F$ is greater than $E$ and so cannot measure $E$.
From this contradiction it follows that no magnitude greater than $E$ can measure $A, B$ and $C$.
The result follows: $E$ is the greatest common measure of $A, B$ and $C$, unless $D$ measures $C$.
If $D$ measures $C$ then $D$ is the greatest common measure of $A, B$ and $C$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Greatest Element/Examples/Finite Subsets of Natural Numbers

== Examples of Greatest Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Consider the ordered set $\struct {\FF, \subseteq}$.
$\struct {\FF, \subseteq}$ has no greatest element.
== Proof ==
 $A \in \FF$ is the greatest element of $\struct {\FF, \subseteq}$.
From Greatest Element is Maximal, $A$ is a maximal element of $\struct {\FF, \subseteq}$.
But from Maximal Element: Finite Subsets of Natural Numbers, $\struct {\FF, \subseteq}$ has no maximal element.
Hence $A$ cannot be the greatest element of $\struct {\FF, \subseteq}$.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Greatest Elements
Greatest Element/Examples/Finite Subsets of Natural Numbers less Empty Set

== Examples of Greatest Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Let $\GG$ denote the set $\FF \setminus \O$, that is, $\FF$ with the empty set excluded.
Consider the ordered set $\struct {\GG, \subseteq}$.
$\struct {\FF, \subseteq}$ has no greatest element.
== Proof ==
 $A \in \GG$ is the greatest element of $\struct {\GG, \subseteq}$.
From Greatest Element is Maximal, $A$ is a maximal element of $\struct {\GG, \subseteq}$.
But from Maximal Element: Finite Subsets of Natural Numbers less Empty Set, $\struct {\GG, \subseteq}$ has no maximal element.
Hence $A$ cannot be the greatest element of $\struct {\GG, \subseteq}$.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Greatest Elements
Greatest Element is Dual to Smallest Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$.
The following are dual statements:
:$a$ is the greatest element of $S$
:$a$ is the smallest element of $S$
== Proof ==
By definition, $a$ is the greatest element of $S$ :
:$\forall b \in S: b \preceq a$
The dual of this statement is:
:$\forall b \in S: a \preceq b$
by Dual Pairs (Order Theory).
By definition, this means $a$ is the smallest element of $S$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Greatest Elements
Category:Smallest Elements
Category:Dual Pairs (Order Theory)
Greatest Element is Maximal

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set which has a greatest element.
Let $M$ be the greatest element of $\struct {S, \preceq}$.
Then $M$ is a maximal element.
== Proof ==
By definition of greatest element:
:$\forall y \in S: y \preceq M$
Suppose $M \preceq y$.
As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric.
Thus it follows by definition of antisymmetry that $M = y$.
That is:
:$M \preceq y \implies M = y$
which is precisely the definition of a maximal element.
== Also see ==
* Smallest Element is Minimal
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Greatest Elements
Category:Maximal Elements
Greatest Element is Supremum

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a greatest element $M$.
Then $M$ is the supremum of $T$ in $S$.
== Proof ==
Let $M$ be the greatest element of $T$.
Then by definition:
:$\forall x \in T: x \preceq M$
By definition of supremum, it is necessary to show that:
:$(1): \quad M$ is an upper bound of $T$ in $S$
:$(2): \quad M \preceq U$ for all upper bounds $U$ of $T$ in $S$.
By Greatest Element is Upper Bound, $M$ is an upper bound of $T$ in $S$.
It remains to be shown that:
:$M \preceq U$ for all upper bounds $U$ of $T$ in $S$.
Let $U \in S$ be an upper bound of $T$ in $S$.
By definition of upper bound:
:$\forall t \in T: t \preceq U$
We have that $M \in T$.
Therefore:
:$M \preceq U$
Hence the result.
== Examples ==
== Also see ==
* Smallest Element is Infimum
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum
Category:Suprema
Category:Greatest Elements
Greatest Element is Supremum/Examples/Arbitrary Example 1

== Example of Use of Greatest Element is Supremum ==
Let $S$ be the subset of the real numbers $\R$ defined as:
:$S = \set {1, 2, 3}$
Then the greatest element of $S$ is $3$.
From Greatest Element is Supremum it follows that:
:$\sup S = 3$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(i)}$
Category:Examples of Suprema
Greatest Element is Supremum/Examples/Arbitrary Example 2

== Example of Use of Greatest Element is Supremum ==
Let $V$ be the subset of the real numbers $\R$ defined as:
:$V := \set {x \in \R: x > 0}$
From Supremum of Subset of Real Numbers: Example 3, $V$ has no supremum.
It follows from Greatest Element is Supremum that $V$ has no greatest element either.
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(iii)}$
Category:Examples of Suprema
Greatest Element is Terminal Object

== Theorem ==
Let $\mathbf P$ be an order category.
Let $p$ be the greatest element of the objects $\mathbf P_0$ of $\mathbf P$, considered as a ordered set.
Then $p$ is a terminal object of $\mathbf P$.
== Proof ==
Since $p$ is the greatest element of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: q \le p$
that is, for every object $q$ of $\mathbf P$ there is a unique morphism $q \to p$.
That is, $p$ is terminal.
== Sources ==
* : $\S 2.2$: Example $2.11$: $5$
Category:Order Categories
Greatest Element is Unique

== Theorem ==
Let $\struct {S, \preceq}$ be a ordered set.
If $S$ has a greatest element, then it can have only one.
That is, if $a$ and $b$ are both greatest elements of $S$, then $a = b$.
=== Class-Theoretical Formulation ===
== Proof ==
Let $a$ and $b$ both be greatest elements of $S$.
Then by definition:
:$\forall y \in S: y \preceq a$
:$\forall y \in S: y \preceq b$
Thus it follows that:
:$b \preceq a$
:$a \preceq b$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence, by definition of antisymmetric relation, $a = b$.
== Also see ==
* Smallest Element is Unique
== Sources ==
* : $\S 14$: Order
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $5$
* : $1$: Set Theory and Logic: $\S 3$: Relations
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.11$
Category:Greatest Elements
Category:Greatest Element is Unique
Greatest Element is Unique/Class Theory

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be an ordering.
Let $A$ be a subclass of the field of $\RR$.
Suppose $A$ has a greatest element $g$ $\RR$.
Then $g$ is unique.
That is, if $g$ and $h$ are both smallest elements of $A$, then $g = h$.
== Proof ==
Let $g$ and $h$ both be smallest elements of $A$.
Then by definition:
:$\forall y \in A: y \mathrel \RR g$
:$\forall y \in A: y \mathrel \RR h$
Thus it follows that:
:$g \preceq h$
:$h \preceq g$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $g = h$.
== Also see ==
* Smallest Element of Class is Unique
Category:Greatest Element is Unique
Greatest Element is Upper Bound

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a greatest element $M \in T$.
Then $M$ is an upper bound of $T$.
It follows by definition that $T$ is bounded above.
== Proof ==
Let $M \in T$ be a greatest element of $T$.
By definition:
:$\forall y \in T: y \preceq M$
But as $T \subseteq S$, it follows that $M \in S$.
Hence:
:$\exists M \in S: \forall y \in T: y \preceq M$
Thus $T$ is bounded above by the upper bound $M$.
== Also see ==
* Smallest Element is Lower Bound
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum
Category:Boundedness
Category:Greatest Elements
Greatest Lower Bound Property

== Theorem ==
Let $S \subset \R$ be a non-empty subset of the set of real numbers such that $S$ is bounded below.
Then $S$ admits an infimum in $\R$.
This is known as the greatest lower bound property of the real numbers.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Least Upper Bound Property
* Continuum Property
== Sources ==
* : Chapter $1$: Fourier Series: $1.1$ Basic Concepts: $1.1.2$ Definitions
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Proposition $1.1.5$
* : $1$: Review of some real analysis: Exercise $1.5: 6$
* : $\S 2$: Continuum Property: $\S 2.4$: The Continuum Property
* 
* 
Category:Continuum Property
Category:Named Theorems
Greatest Lower Bound Property/Proof 1

== Theorem ==
== Proof ==
Let $T = \set {x \in \R: -x \in S}$.
By Set of Real Numbers is Bounded Below iff Set of Negatives is Bounded Above:
:$T$ is bounded above.
Thus by the Least Upper Bound Property, $T$ admits a supremum in $\R$.
From Negative of Supremum is Infimum of Negatives:
:$\ds -\sup_{x \mathop \in T} x = \map {\inf_{x \mathop \in T} } {-x}$
That is, by definition of $T$:
:$\ds -\sup_{x \mathop \in T} x = \inf_{x \mathop \in S} x$
and so $S$ admits an infimum in $\R$.
== Sources ==
* : $\S 8$: Least Upper Bounds: Problem $2 \ \text {(a)}$
Category:Continuum Property
Greatest Lower Bound Property/Proof 2

== Theorem ==
== Proof ==
Let $T$ be the set of lower bounds of $S$:
:$T = \set {x \in \R: x \le \forall y \in S}$
Since $S$ is bounded below, $T$ is non-empty.
Now, every $x \in T$ and $y \in S$ satisfy $x \leq y$.
That is, $T$ is bounded above by every element of $S$.
By the Least Upper Bound Property, $T$ admits a supremum in $\R$.
Let $B = \sup T$.
Since every element of $S$ is an upper bound of $T$ it follows from the definition of supremum that:
:$\forall y \in S : y \ge B$
Thus $B$ is a lower bound of $S$.
Because $B$ is an upper bound of $T$:
:$\forall x \in T : x \le B$
and so $B$ is the infimum of $S$.
== Sources ==
* : $\S 8$: Least Upper Bounds: Problem $2 \ \text {(b)}$
Category:Continuum Property
Greatest Power of Two not Divisor

== Lemma ==
Let $S = \set {1, 2, 3, 4, \ldots, n}$ be a set of integers from $1$ through $n$.
Let $2^t$ be the greatest power of $2$ in $S$.
Then $2^t$ does not divide any other integer in $S$.
That is, no other member of $S$ is a multiple of $2^t$.
== Proof ==
Let $k$ be a multiple of $2^t$ in $S$.
Then $k = 2^t \times \ell$ for some $\ell \in \Z, \ell \ge 1$.
If $\ell \ge 2$, then we would have $k = \ell \times 2^t \ge 2\times 2^t = 2^{t + 1}$.
As $S$ contains all integers up to $k$, we would have $2^{t + 1} \in S$, contradicting $2^t$ being the highest power of $2$ in $S$.
So $\ell = 1$, that is, $2^t$ is the only multiple of $2^t$ in $S$.
 
== Also see ==
* Harmonic Number is not Integer
Category:Number Theory
Greatest Set is Unique

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the greatest set of $\TT$, if it exists, must be unique.
== Proof ==
Let $A, B \in \TT$ both be greatest sets of $\TT$.
Since $A$ is the greatest set:
:$B \subseteq A$
Since $B$ is the greatest set:
:$A \subseteq B$
Hence, by definition of set equality:
:$A = B$
Therefore the greatest set of $\TT$ is unique.
Category:Set Theory
Category:Greatest Elements
Greatest Set may not Exist

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The greatest set of $\TT$ may not exist.
== Proof ==
Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 1$ is not the greatest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 0$ is not the greatest set of $\TT$.
Therefore $\TT$ has no greatest set.
Category:Set Theory
Category:Greatest Elements
Greatest Term of Binomial Expansion/Examples/Arbitrary Example 1

== Theorem ==
Consider the expression:
:$E = \paren {1 + 2 x}^{10 \frac 1 2}$
Let $x = \dfrac 3 7$.
Then the greatest term in the power series expansion of $E$ by means of the General Binomial Theorem is:
:$\dfrac {21 \times 19 \times 17 \times 15 \times 13} {5!} \paren {\dfrac 3 7}^5$
== Proof ==
Let us perform the expansion:
:$\paren {1 + 2 x}^{\frac {21} 2} = 1 + \dfrac {21} 2 \paren {2 x} + \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} } {2!} \paren {2 x}^2 + \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \paren {\frac {17} 2} } {3!} \paren {2 x}^3 + \cdots$
Consider the $\paren {r + 1}$th term:
:$u_{r + 1} = \dfrac {\paren {\frac {21} 2} \paren {\frac {19} 2} \cdots \paren {\frac {23} 2 - r} } {r!} \paren {2 x}^r$
Hence $u_{r + 1} > u_r$ if:
So:
:if $r < 5 \tfrac 4 {13}$ then $u_{r + 1} > u_r$
but conversely:
:if $r > 5 \tfrac 4 {13}$ then $u_{r + 1} < u_r$
Hence the greatest value of $r$ for which $u_{r + 1} > u_r$ is $5$.
Then we have that:
:$u_6 > u_5$
but then:
:$u_7 < u_6$
Hence the $6$th term is greatest:
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: The greatest term
Category:Binomial Theorem
Greedy Algorithm guarantees Maximum Weight iff Matroid

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr I$ be a non-empty set of subsets of $S$.
Then $\mathscr I$ is the set of independent subsets of a matroid on $S$ :
:$(1) \quad \struct{S, \mathscr I}$ is an independence system
:$(2) \quad$ for all non-negative weight functions $w : S \to \R_{\ge 0}$, the Greedy Algorithm selects $A_w \in \mathscr I$:
:::::$\forall B \in \mathscr I: \map {w^+} {A_w} \ge \map {w^+} B$
:where $w^+$ denotes the extended weight function of $w$.
== Proof ==
== Sources ==
* Chapter $19.$ $\S 1.$ The greedy algorithm
Category:Maximization Problem for Independence Systems
Greedy Algorithm may not yield Maximum Weight

== Theorem ==
Let $\struct {S,\mathscr F}$ be an independence system.
Let $w : S \to \R_{\ge 0}$ be a weight function.
Then the maximal set $A_0 \in \mathscr F$ selected by the Greedy Algorithm may not have maximum weight.
== Proof ==
Category:Maximization Problem for Independence Systems
Greek Anthology Book XIV: 1. - Problem

== Problem ==
:Polycrates Speaks
::Blessed , Helliconian scion of the Muses, answer my question: How many in thy house are engaged in the contest for wisdom performing excellently?
: Answers
::I will tell thee the, Polycrates.
:::Half of them are occupied with belles lettres;
:::a quarter apply themselves to studying immortal nature;
:::a seventh are all intent on silence and the eternal discourse of their hearts.
:::There are also three women,
:::and above the rest is Theano.
::That is the number of interpreters of the Muses I gather round me.
== Solution ==
It is apparent from what follows that Theano is not actually counted with the others.
Let $n$ be the number of students.
Then:
:$\dfrac n 2$ are occupied with belles lettres
:$\dfrac n 4$ apply themselves to studying immortal nature
:$\dfrac n 7$ are all intent on silence and the eternal discourse of their hearts
:$3$ are women.
Thus:
Thus of those not women:
:$14$ are occupied with belles lettres
:$7$ apply themselves to studying immortal nature
:$4$ are all intent on silence and the eternal discourse of their hearts
and it is seen that:
:$14 + 7 + 4 + 3 = 28$
== Historical Note ==
== Sources ==
* : $1$. -- : 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 11. - Problem

== Problem ==
:I desire my two sons to receive the thousand of which I am possessed,
:but let the fifth part of the legitimate one's share exceed by ten the fourth part of what falls to the illegitimate one.
== Solution ==
Let $s$ be the share received by the legitimate heir.
Then:
Thus the legitimate heir inherits $577 \frac 7 9$ , while the illegitimate one receives $1000 - 577 \frac 7 9 = 422 \frac 2 9$.
== Sources ==
* : $11$. -- 
* : Metrodorus and the Greek Anthology: $36$
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 12. - Problem

== Problem ==
:Croesus the king dedicated six bowls weighing six ,
:each one heavier than the other.
== Solution ==
It is known that there are $100$ to the .
The weights of the bowls form an arithmetic sequence.
Let $a$ denote the weight of the lightest bowl in .
Let $n$ denote the number of bowls.
Let $d$ denote the common difference between the weights in of consecutive bowls arranged in order of weight.
Let $t$ denote the total weight of all bowls in .
We are given:
So:
So the lightest bowl weighs $97 \frac 1 2$ , and it follows that the weights of each bowl in is:
:$97 \frac 1 2, 98 \frac 1 2, 99 \frac 1 2, 100 \frac 1 2, 101 \frac 1 2, 102 \frac 1 2$
== Sources ==
* : $12$. -- 
Category:Greek Anthology Book XIV
Category:Examples of Sum of Arithmetic Sequence
Greek Anthology Book XIV: 13. - Problem

== Problem ==
:We both of us together weigh twenty , I, Zethus, and my brother;
:and if you take the third part of me and the fourth part of Amphion here, you will find it makes six,
:and you will have found the weight of our mother.
== Solution ==
Let $z$ be the weight in of Zethus.
Let $a$ be the weight in of Amphion.
We have:
So:
:Zethus weighs $12$ 
:Amphion weighs $8$ 
and (presumably):
:their mother weighs $6$ .
== Sources ==
* : $13$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 147

== Epigram ==
:Answer of Homer to Hesiod when he asked the Number of the Greeks who took part in the War against Troy
::There were seven hearths of fierce fire, and in each were fifty spits and fifty joints on them.
::About each joint were nine hundred Achaeans.
== Solution ==
Let $n$ be the number of Achaeans (that is, Greeks).
So there were $315 \, 000$ Achaeans in total.
== Sources ==
* : $147$
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 2. - Problem

== Problem ==
:On a Statue of Pallas
::I, Pallas, am of beaten gold, but the gold is the gift of lusty poets.
:::Charisius gave half the gold,
:::Thespis one-eighth,
:::Solon one-tenth,
:::and Themison one-twentieth,
::but the remaining nine talents and the workmanship are the gift of Aristodicius.
== Solution ==
Let $n$ talents be the weight or worth of the statue of Pallas (it is not clear which is meant, but this detail is immaterial).
Then:
:$\dfrac n 2$ was given by Charisius
:$\dfrac n 8$ was given by Thespis
:$\dfrac n {10}$ was given by Solon
:$\dfrac n {20}$ was given by Themison
:$9$ talents were given by Aristodicius.
Thus:
Thus
:$20$ talents were provided by Charisius
:$5$ talents were provided by Thespis
:$4$ talents were provided by Solon
:$2$ talents were provided by Themison
and the remaining $9$ talents were provided by Aristodicius.
and it is seen that:
:$20 + 5 + 4 + 2 + 9 = 40$
== Sources ==
* : $2$. -- 
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 20. - Enigma

== Enigma ==
:If you put one hundred in the middle of a burning fire, you will find the son and slayer of a virgin.
== Solution ==
Pyrrhus, son of Deidamia, and slayer of Polyxena.
The Greek letter $\rho$ (rho) is the symbol for $100$ in the classical number system of ancient Greece
The word for fire in ancient Greek was $\pi \upsilon \rho \acute \omicron \varsigma$ (that is, pyros).
Inserting $\rho$ into $\pi \upsilon \rho \acute \omicron \varsigma$ gives $\pi \upsilon \rho \rho \acute \omicron \varsigma$, which is a rendition of Pyrrhus.
== Historical Note ==
== Sources ==
* : $20$. -- 
* : Metrodorus and the Greek Anthology
Category:Greek Anthology Book XIV
Greek Anthology Book XIV: 20. - Enigma/Historical Note

== Historical Note on Greek Anthology Book $\text {XIV}$: $20$ - Enigma ==
Pyrrhus in this context is , who took as a concubine after the death of her husband in the .
He it was who murdered , according to some legends (or sacrificed, which is tantamount to the same thing).
 was indeed his mother.
Category:Historical Notes