Henry Ernest Dudeney/Modern Puzzles/Magic Star Problems

Henry Ernest Dudeney: Modern Puzzles: Magic Star Problems

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.

In this case we can always use the twelve consecutive numbers $1$ to $12$ and the sum of the four numbers in every line will always be $26$.

The numbers at the six points of the star may add up to any even number from $24$ to $54$ inclusive, except $28$ and $50$, which are impossible.

It will be seen in the example that the six points add up to $24$.

If for every number in its present position you substitute its difference from $13$ you will get another solution, its complementary,

with the points adding up to $54$, which is $78$ less $24$.

The two complementary totals will always sum to $78$.

I will give the total number of different solutions and point out some of the pretty laws which govern the problem,

but I will leave the reader this puzzle to solve.

There are six arrangements, and six only, in which all the lines of four and the six points also add up to $26$.

Can you find one or all of them?

All you have to do is place the numbers $1$, $2$, $3$, up to $14$ in the fourteen discs so that every line of four disks shall add up to $30$.

The star may be formed in two different ways, as shown in our diagram, and the first example is a solution.

The numbers $1$ to $16$ are so placed that every straight line of four adds up to $34$.

If you substitute for every number its difference from $17$ you will get the complementary solution.

$\qquad$

Let the reader try to discover some of the other solutions, and he will find it a very hard nut, even with this one to help him.

But I will present the puzzle in an easy and entertaining form.

When you know how, every arrangement in the first star can be transferred to the second one automatically.

Every line of four numbers in the one case will appear in the other, only the order of the numbers will have to be changed.

Now, with this information given, it is not a difficult puzzle to find a solution for the second star.