[Math] Number puzzle, number arrangement

puzzle

Arrange all of the digits from the numbers 1 through 15 in such a order that the sum of any adjacent pair is a perfect square. No repeats. I can be staring at those number for days. But I wonder if there is a trick.

Best Answer

To perhaps simplify the suggestion I made above: Try writing each number from 1 through 15 in a column. Next to each, write all the numbers that are greater than that number and that can be added to that number to form a perfect square. For example:

1: 3, 8, 15

2: 7, 14

And so forth. For some of these numbers, you'll find your options are limited: There are very few other numbers they can be paired with. That will, in turn, restrict where in your string you can put them. With that in mind, draw a series of fifteen blanks and start putting in the numbers that must go in specific positions.

Let me know if you have questions.

Related Solutions

Recreational Mathematics – Two Numbers, Two Mathematicians Puzzle

There is in fact no solution to this puzzle. Martin Gardner realized this after presenting it in Scientific American:

'As hundreds of readers have pointed out, the "impossible problem" given in this department for December turned out to be literally impossible.' (For this quote and an explanation, see here.)

For the puzzle to have a solution, there would have to be an admissible sum such that no matter how it is split up into two admissible addends, the resulting product also has at least one other admissible factorization. The following short Java program checks that there is no such admissible sum.

public class SumAndProduct {
    final static int min = 2;
    final static int max = 20;

    public static void main (String [] args) {
        outer:
        for (int sum = min + min;sum <= max + max;sum++) {
            for (int i = min;i <= max;i++) {
                int j = sum - i;
                int product = i * j;
                int nfactorizations = 0;
                for (int a = min;a <= max;a++) {
                    int b = product / a;
                    if (min <= b && b <= max && a * b == product)
                        nfactorizations++;
                }
                nfactorizations = (nfactorizations + 1) / 2;
                if (nfactorizations == 1)
                    continue outer;
            }

            System.out.println ("sum " + sum + " can't be determined");
        }
    }
}

Note that the solution $(2,6)$ given on this page linked to by picakhu is incorrect. The sum is $8$, which could be the result of $(3,5)$, which has the unique factorization $3\cdot5$ that would allow B to deduce the sum $8$. The solution originally claimed by Martin Gardner was $(4,13)$.

[Math] Sliding puzzle disjoint sets proof

You’re working too hard: every state has an associated value of $N$, and that value is either even or odd.

Best Answer

Related Solutions

Recreational Mathematics – Two Numbers, Two Mathematicians Puzzle

[Math] Sliding puzzle disjoint sets proof

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