nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/167 - A General Election/Solution

== by : $167$ ==
;A General Election
:In how many different ways may a Parliament of $615$ members be elected if there are only $4$ parties:
::Conservatives, Liberals, Socialists, and Independents?
:You see you might have $\text C. 310$, $\text L. 152$, $\text S. 150$, $\text I. 3$;
::or $\text C. 0$, $\text L. 0$, $\text S. 0$, $\text I. 615$;
::or $\text C. 205$, $\text L. 205$, $\text S. 205$, $\text I. 0$; and so on.
:The candidates are indistinguishable, as we are only concerned with the party numbers.
== Solution ==
:$39 \, 147 \, 416$ different ways. 
== Proof ==
According to :
:The general solution is as follows.
:Let $p =$ parties and $m =$ members.
:Then $C^{p - 1}_{m + p - 1} =$ number of ways.
In the above, $C^{p - 1}_{m + p - 1}$ is the binomial coefficient $\dbinom {p - 1} {m + p - 1}$.
== Sources ==
* : Solutions: $167$. -- A General Election
* : Answers: $446$. A General Election
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench

== by : $168$ ==
;The Magisterial Bench
:A friend at Singapore asked me some time ago to give him my solution to this problem.
:A bench of magistrates consists of two Englishmen, two Scotsmen, two Welshmen, one Frenchman, one Italian, one Spaniard, and one American.
:The Englishmen will not sit beside one another, the Scotsmen will not sit beside one another, and the Welshmen also object to sitting together.
:Now, in how many different ways may the ten men sit in a straight line so that no two men of the same nationality shall ever be next to one another?
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $168$. -- The Magisterial Bench
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $447$. The Magisterial Bench
Category:Henry Ernest Dudeney/Modern Puzzles
Category: Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench
Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench/Solution

== by : $168$ ==
;The Magisterial Bench
== Solution ==
:$1 \, 895 \, 040$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Solutions: $168$. -- The Magisterial Bench
* : Answers: $447$. The Magisterial Bench
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category: Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench
Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench/Solution/Proof 1

== by : $168$ ==
;The Magisterial Bench
== Solution ==
== Proof ==
Without imposing any conditions, we can arrange $10$ men in one line in a total of $10! = 3 \, 628 \, 800$ different ways.
It remains to count how many of these are barred.
Let us regard two of a nationality together as one item.
:$(1): \quad$ We consider:
:$\paren {E \ E} \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$
as a set of $7$ objects that can be permuted in $7!$ ways, multiplied by $2^3$ to account for the fact that either of the two within the pairs can be arranged in $2$ different ways.
Hence we remove all such permutations, which is $7! \times 2^3 = 40 \, 320$.
:$(2): \quad$ Then we eliminate all permutations of the form:
:$\paren {E \ E} \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$
where the Welshmen are not together.
This gives $8! \times 2^2$, but we must deduct the above arrangements or we will count them twice.
This gives $8! \times 2^2 - 7! \times 2^3 = 120 \, 960$.
:$(3): \quad$ The same applies to permutations of the form:
:$E \ E \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$
This gives another deduction of $120 \, 960$.
:$(4): \quad$ The same applies to permutations of the form:
:$\paren {E \ E} \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.
Hence another deduction of $120 \, 960$.
:$(5): \quad$ Similarly we eliminate all permutations of the form $\paren {E \ E} \ S \ S \ W \ W \ F \ I \ S \ A$.
By similar reasoning, there are $9! \times 2$ cases, from which we need to deduct the results of $(1)$, $(2)$ and $(3)$ (but not $(4)$).
This gives $9! \times 2 - 40 \, 320 - 2 \times 120 \, 960 = 443 \, 520$.
:$(6): \quad$ Similarly we eliminate all permutations of the form $E \ E \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$.
This gives another $443 \, 520$.
:$(7): \quad$ Similarly we eliminate all permutations of the form $E \ E \ S \ S \ \paren {W \ W} \ F \ I \ S \ A$.
This gives another $443 \, 520$.
Add all these up together:
:$40 \, 320 + 3 \times 120 \, 960 + 3 \times 443 \, 520 = 1 \, 733 \, 760$
Hence we have:
:$3 \, 628 \, 800 - 1 \, 733 \, 760 = 1 \, 895 \, 040$
== Sources ==
* : Solutions: $168$. -- The Magisterial Bench
* : Answers: $447$. The Magisterial Bench
Category: Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench
Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench/Solution/Proof 2

== by : $168$ ==
;The Magisterial Bench
== Solution ==
== Proof ==
Let $X$ be the set of all permutations of the bench.
Let $E, S, W$ be the sets of all permutations where the two Englishmen, Scotsmen and Welshmen are forced to sit next to each other, respectively.
Then $\map {\complement_X} E, \map {\complement_X} S, \map {\complement_X} W$ are the sets of all permutations where the two Englishmen, Scotsmen and Welshmen do not sit next to each other, respectively.
The number of permutations the question asks for is:
$E \cap S \cap W$ contains permutations of the form:
:$\paren {E \ E} \ \paren {S \ S} \ \paren {W \ W} \ F \ I \ S \ A$
which can be considered as a set of $7$ objects that can be permuted in $7!$ ways, multiplied by $2^3$ to account for the fact that either of the two within the pairs can be arranged in $2$ different ways.
Thus:
:$\size {E \cap S \cap W} = 7! \times 2^3 = 40 \, 320$
$E \cap S$ contains permutations of the form:
:$\paren {E \ E} \ \paren {S \ S} \ W \ W \ F \ I \ S \ A$
where no restrictions are placed on the Welshmen.
This gives:
:$\size {E \cap S} = 8! \times 2^2 = 161 \, 280$
Similarly:
:$\size {S \cap W} = \size {E \cap W} = 161 \, 280$
$E$ contains all permutations of the form:
:$\paren {E \ E} \ S \ S \ W \ W \ F \ I \ S \ A$
where the only restriction is the Englishmen must sit next to each other.
This gives:
:$\size E = 9! \times 2 = 725 \, 760$
Similarly:
:$\size S = \size W = 725 \, 760$
Finally $\size X = 10! = 3 \, 628 \, 800$.
Therefore:
is the number of permutations required.
Category: Henry Ernest Dudeney/Modern Puzzles/168 - The Magisterial Bench
Henry Ernest Dudeney/Modern Puzzles/169 - The Card Pentagon

== by : $169$ ==
;The Card Pentagon
:Make a rough pentagon on a large sheet of paper.
:Then throw down the ten non-court cards of a suit at the places indicated in the diagram,
::so that the pips on every row of three cards on the sides of the pentagon shall add up alike.
:
:The example will be found faulty.
:After you have found the rule you will be able to deal the cards into their places without any thought.
:And there are very few ways of placing them.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $169$. -- The Card Pentagon
* : Combinatorial & Topological Problems: Magic Star Puzzles: $398$. The Card Pentagon
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/169 - The Card Pentagon/Solution

== by : $169$ ==
;The Card Pentagon
== Solution ==
:Deal the cards $1$, $2$, $3$, $4$, $5$ in the manner indicated by the dotted lines
::(that is, drop one at every alternate angle in a clockwise direction round the pentagon),
::and then deal the $6$, $7$, $8$, $9$, $10$ in the opposite direction, as shown,
::taking care to start with the $6$ on the correct side of the $5$.
:The pips on every side add to $14$.
:
:If you deal the $6$, $7$, $8$, $9$, $10$ in the first manner and the $1$, $2$, $3$, $4$, $5$ in the second manner,
::you will get another solution, adding up to $19$.
:Now work with the two sets of numbers $1$, $3$, $5$, $7$, $9$ and $2$, $4$, $6$, $8$, $10$ in the same way
::and you will get two more solutions, adding, respectively, to $16$ and $17$.
:There are $6$ solutions in all.
:The last two are peculiar.
:Write in, in the same order, $1$, $4$, $7$, $10$, $13$ and $6$, $9$, $12$, $15$, $18$;
::also write in $8$, $11$, $14$, $17$, $20$ and $3$, $6$, $9$, $12$, $15$.
:Then deduct $10$ from every number greater than $10$.
== Sources ==
* : Solutions: $169$. -- The Card Pentagon
* : Answers: $398$. The Card Pentagon
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/16 - Digging a Ditch

== by : $16$ ==
;Digging a Ditch
:Here is a curious question that is more perplexing than it looks at first sight.
:Abraham, an infirm old man, undertook to dig a ditch for $2$ pounds.
:He engaged Benjamin, an able-bodied fellow, to assist him and share the money fairly according to their capacities.
:Abraham could dig as fast as Benjamin could shovel out the dirt,
::and Benjamin could dig four times as fast as Abraham could do the shovelling.
:How should they divide the money?
:Of course, we must assume their relative abilities for work to be the same in digging or shovelling.
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $16$. -- Digging a Ditch
* : Arithmetical and Algebraical Problems: Money Puzzles: $9$. Digging a Ditch
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/16 - Digging a Ditch/Solution

== by : $16$ ==
;Digging a Ditch
== Solution ==
Abraham should receive $\tfrac 1 3$ of $\pounds 2$, and Benjamin should receive $\tfrac 2 3$.
== Proof ==
Let $A$ and $B$ denote Abraham and Benjamin respectively
Let $r$ denote the ratio of the ability of Abraham to work relative to Benjamin.
Suppose:
:$B$ can dig all in $r$ hours and shovel all in $s$ hours.
:$A$ can dig all in $s$ hours and shovel all in $4 r$ hours.
Then as their relative abilities for work is the same in digging and shovelling:
:$\dfrac r s = \dfrac s {4 r}$
then:
$\dfrac r s = \dfrac 1 2$
So $B$ accomplishes twice as much work as $A$.
Thus $B$ takes twice the pay that $A$ does.
The result follows.
== Sources ==
* : Solutions: $16$. -- Digging a Ditch
* : Answers: $9$. Digging a Ditch
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/170 - A Heptagon Puzzle

== by : $170$ ==
;A Heptagon Puzzle
:Using the fourteen numbers, $1$, $2$, $3$, up to $14$, place a different number in every circle
::so that the three numbers in every one of the seven sides add up to $19$.
:
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $170$. -- A Heptagon Puzzle
* : Combinatorial & Topological Problems: Magic Star Puzzles: $399$. A Heptagon Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/170 - A Heptagon Puzzle/Solution

== by : $170$ ==
;A Heptagon Puzzle
== Solution ==
:The diagram shows the solution.
:
:Starting at the highest point, write the numbers $1$ to $7$ in a clockwise direction at alternate points.
:Then, starting just above the $7$, write $8$ to $14$ successively in the opposite direction, taking every vacant circle in turn.
:If instead you write in $1$, $3$, $5$, $7$, $9$, $11$, $13$ and then $2$, $4$, $6$, $8$, $10$, $12$, $14$,
::you will get a solution with the sides adding to $22$ instead of $19$.
:If you substitute for every number in these solutions its difference from $15$ you will get the complementary solutions,
::adding respectively to $26$ and $23$ (the difference of $19$ and $22$ from $45$).
== Sources ==
* : Solutions: $170$. -- A Heptagon Puzzle
* : Answers: $399$. A Heptagon Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/171 - An Irregular Magic Square

== by : $171$ ==
;An Irregular Magic Square
:Here we have a perfect magic square composed of the numbers $1$ to $16$ inclusive.
$\qquad \begin{array}{|c|c|c|c|}
\hline 1 & 14 & 7 & 12 \\
\hline 15 & 4 & 9 & 6 \\
\hline 10 & 5 & 16 & 3 \\
\hline 8 & 11 & 2 & 13 \\
\hline
\end{array}$
:The rows, columns, and two long diagonals all add up to $34$.
:Now, supposing you were forbidden to use the two numbers $2$ and $15$, but allowed, in their place, to repeat any two numbers already used,
::how would you construct your square so that rows, columns, and diagonals should still add up to $34$?
:Your success will depend on which two numbers you select as substitutes for $2$ and $15$.
=== Click here for solution ===
== Sources ==
* : Magic Square Problems: $171$. -- An Irregular Magic Square
* : Combinatorial & Topological Problems: Magic Square Puzzles: $384$. An Irregular Magic Square
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/171 - An Irregular Magic Square/Solution

== by : $171$ ==
;An Irregular Magic Square
== Solution ==
Substitute $7$ and $10$ for $2$ and $15$, and we get:
$\qquad \begin{array}{|c|c|c|c|}
\hline 1 & 10 & 9 & 14 \\
\hline 13 & 10 & 5 & 6 \\
\hline 8 & 3 & 16 & 7 \\
\hline 12 & 11 & 4 & 7 \\
\hline
\end{array}$
== Sources ==
* : Solutions: $171$. -- An Irregular Magic Square
* : Answers: $384$. An Irregular Magic Square
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/172 - A Magic Square Delusion

== by : $172$ ==
;A Magic Square Delusion
:Here is a magic square of the fifth order.
$\qquad \begin{array}{|c|c|c|c|c|}
\hline 17 & 24 & 1 & 8 & 15 \\
\hline 23 & 5 & 7 & 14 & 16 \\
\hline 4 & 6 & 13 & 20 & 22 \\
\hline 10 & 12 & 19 & 21 & 3 \\
\hline 11 & 18 & 25 & 2 & 9 \\
\hline
\end{array}$
:I have found that a great many people who have not gone very profoundly into these things believe that the central number in all squares of this order must be $13$.
:One correspondent who had devoted years to amusing himself with this particular square was astounded when I told him that any number from $1$ to $25$ might be in the centre.
:I will show that this is so.
:Try to form such a magic square with $1$ in the central cell.
=== Click here for solution ===
== Sources ==
* : Magic Square Problems: $172$. -- A Magic Square Delusion
* : Combinatorial & Topological Problems: Magic Square Puzzles: $385$. A Magic Square Delusion
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/172 - A Magic Square Delusion/Solution

== by : $172$ ==
;A Magic Square Delusion
== Solution ==
Make $9$ squares like this one and put them together to form a larger square:
$\qquad \begin{array}{|c|c|c|c|c|}
\hline 9 & 11 & 18 & 5 & 22 \\
\hline 3 & 25 & 7 & 14 & 16 \\
\hline 12 & 19 & 1 & 23 & 10 \\
\hline 21 & 8 & 15 & 17 & 4 \\
\hline 20 & 2 & 24 & 6 & 13 \\
\hline
\end{array}$
Then you can pick out a $5 \times 5$ square in any position and it will be a magic square.
Hence it is seen that you can arrange for any number you like to be in the centre.
This is what is known as a Nasik square (named by after the place in India where he was living).
== Sources ==
* : Solutions: $172$. -- A Magic Square Delusion
* : Answers: $385$. A Magic Square Delusion
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/173 - Difference Squares

== by : $173$ ==
;Difference Squares
:Can you rearrange the nine digits in the square so that in all the eight directions the difference between one of the digits and the sum of the remaining two shall always be the same?
$\qquad \begin{array}{|c|c|c|}
\hline 4 & 3 & 2 \\
\hline 7 & 1 & 9 \\
\hline 6 & 5 & 8 \\
\hline
\end{array}$
:In the example shown it will be found that all the rows and columns give the difference $3$:
::(thus $4 + 2 - 3$, and $1 + 9 - 7$, and $6 + 5 - 8$, etc.),
::but the two diagonals are wrong, because $8 - \paren {4 + 1}$ and $6 - \paren {1 + 2}$ is not allowed:
::the sum of the two must not be taken from the single digit, but the single digit from the sum.
:How many solutions are there?
=== Click here for solution ===
== Sources ==
* : Magic Square Problems: $173$. -- Difference Squares
* : Combinatorial & Topological Problems: Magic Square Puzzles: $386$. Difference Squares
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/173 - Difference Squares/Solution

== by : $173$ ==
;Difference Squares
== Solution ==
These may be the only ones:
$\qquad \begin{array}{|c|c|c|}
\hline 2 & 1 & 4 \\
\hline 3 & 5 & 7 \\
\hline 6 & 9 & 8 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 8 & 1 & 4 \\
\hline 3 & 5 & 7 \\
\hline 6 & 9 & 2 \\
\hline
\end{array} \qquad \begin{array}{|c|c|c|}
\hline 2 & 1 & 6 \\
\hline 3 & 5 & 7 \\
\hline 4 & 9 & 8 \\
\hline
\end{array}$
The difference throughout is $5$.
== Sources ==
* : Solutions: $173$. -- Difference Squares
* : Answers: $386$. Difference Squares
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/174 - Swastika Magic Square

== by : $174$ ==
;Swastika Magic Square
:A correspondent sent me this little curiosity.
:It is a magic square, the rows, columns, and two diagonals all adding up to $65$,
::and all prime numbers that occur between $1$ and $25$
::(viz. $1$, $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$)
::are to be found in the swastika except $11$.
:
:"This number," he says, "in occult lore is ominous and is associated with the eleven Curses of Ebal,
::so it is just as well it does not come into this potent charm of good fortune."
:He is clearly under the impression that $11$ cannot be got into the swastika with the other primes.
:But in this he is wrong, and the reader may like to try to reconstruct the square
::so that the swastika contains all the prime numbers 
::and yet forms a correct magic square, for it is quite possible.
=== Click here for solution ===
== Sources ==
* : Magic Square Problems: $174$. -- Swastika Magic Square
* : Combinatorial & Topological Problems: Magic Square Puzzles: $387$. Swastika Magic Square
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/175 - Is it Very Easy

== by : $175$ ==
;Is it Very Easy?
:Here is a simple magic square, the three columns, three rows, and two diagonals adding up to $72$.
$\qquad \begin{array}{|c|c|c|}
\hline 27 & 20 & 25\\
\hline 22 & 24 & 26 \\
\hline 23 & 28 & 21 \\
\hline
\end{array}$
:The puzzle is to convert it into a multiplying magic square,
::in which the numbers in all the eight lines if multiplied together give the same product in every case.
:You are not allowed to change, or add to, any of the figures in a cell
::or use any arithmetical sign whatever!
:But you may shift the two figures within a cell.
:Thus, you may write $27$ as $72$, if you like.
:These simple conditions make the puzzle absurdly easy, if you once hit on the idea;
::if you miss it, it will appear to be an utter impossibility.
=== Click here for solution ===
== Sources ==
* : Magic Square Problems: $175$. -- Is it Very Easy?
* : Combinatorial & Topological Problems: Magic Square Puzzles: $388$. Is it Very Easy?
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/175 - Is it Very Easy/Solution

== by : $175$ ==
;Is it Very Easy?
== Solution ==
All $8$ rows give the same product $4096$:
$\qquad \begin{array}{|c|c|c|}
\hline 2^7 & 2^0 & 2^5\\
\hline 2^2 & 2^4 & 2^6 \\
\hline 2^3 & 2^8 & 2^1 \\
\hline
\end{array}$
As reminds us:
:Of course, every arithmetician knows that $2^0$ equals $1$.
== Sources ==
* : Solutions: $175$. -- Is it Very Easy?
* : Answers: $388$. Is it Very Easy?
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/176 - The Five-Pointed Star

== by : $176$ ==
;The Five-Pointed Star
:There is something very fascinating about star puzzles.
:I give an example, taking the case of the five-pointed star.
:
:It is required to place a different number in every circle so that the four circles in a line shall add up to $24$ in all the five directions.
:No solution is possible with $10$ consecutive numbers, but you can use any whole numbers you like.
=== Click here for solution ===
== Sources ==
* : Magic Star Problems: $176$. -- The Five-Pointed Star
* : Combinatorial & Topological Problems: Magic Star Puzzles: $393$. The Five-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/176 - The Five-Pointed Star/Solution

== by : $176$ ==
;The Five-Pointed Star
== Solution ==
's unsatisfyingly unstructured and prolix answer is given below.
A full and coherent analysis is welcomed.
:Referring to Figure $\text I$, we will call $A$, $B$, $C$, $D$, $E$ the "pentagon", and $F$, $G$, $H$, $J$, $K$ the "points".
:
:::Figure $\text I$
:Write in the numbers $1$, $2$, $3$, $4$, $5$ in the pentagon in the order shown in Figure $\text {II}$, where you go round in a clockwise direction,
::starting with $1$ and jumping over a disc to the place for $2$, jumping over another for $3$, and so on.
:Now to complete the star for the constant summation of $24$, as required, use this simple rule.
:To find $H$ subtract the sum of $B$ and $C$ from half the constant plus $E$.
:That is, subtract $6$ from $15$.
:We thus get $9$ as the required number for $H$.
:Now you are able to write in successively $10$ at $F$ (to make $24$), $6$ at $J$, $12$ at $G$, and $8$ at $K$.
:There is your solution.
:
:::Figure $\text {II}$
:You can write any five numbers you like in the pentagon, in any order, and with any constant summation you wish,
::and you will always get, by the rule shown, the only possible solution for that pentagon and constant.
:But that solution may require the use of repeated numbers and even negative numbers.
:Suppose, for example, I make the pentagon $1$, $3$, $11$, $7$, $4$, and the constant $26$, as in Figure $\text {III}$,
::then I shall find the $3$ is repeated, and the repeated $4$ is negative and must be deducted instead of added.
:You will also find that if we had written our pentagon numbers in Figure $\text {II}$ in any other order we should always get repeated numbers.
:
:::Figure $\text {III}$
:Let us confine our attention to solutions with ten different positive whole numbers.
:Then $24$ is the smallest possible constant.
:A solution for any higher constant can be derived from it.
:Thus, if we want $26$, add $1$ at each of the points;
::if we want $28$, add $2$ at every point or $1$ at every place in both points and pentagon.
:Odd constants are impossible unless we use fractions.
:Every solution can be "turned inside out".
:Thus Figure $\text {IV}$ is simply a different arrangement of Figure $\text {II}$.
:Also the four numbers in $G$, $K$, $D$, $J$ may always be changed, if repetitions do not occur.
:For example, in Figure $\text {II}$ substitute $13$, $7$, $6$, $5$ for $12$, $8$, $5$, $6$ respectively.
:Finally, in any solution the constant will be two-fifths of the sum of all the ten numbers.
:So, if we are given a particular set of numbers we at once know the constant,
::and for any constant we can determine the sum of the numbers to be used.
:
:::Figure $\text {IV}$
== Sources ==
* : Solutions: $176$. -- The Five-Pointed Star
* : Answers: $393$. The Five-Pointed Star
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions