Henry Ernest Dudeney/Modern Puzzles/176 - The Five-Pointed Star/Solution

Modern Puzzles by Henry Ernest Dudeney: $176$

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.

Solution

Dudeney'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

• 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $176$. -- The Five-Pointed Star
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $393$. The Five-Pointed Star