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«Contents Ü Foreword Elwyn Berlekamp and Tom Rodgers ½ I Personal Magic ¿ Martin Gardner: A “Documentary” Dana Richards ½¿ Ambrose, Gardner, ...»

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In the case of the 123 hole, we can argue as follows: If n 999, then f (n) n. In other words, the new number that counts the digits is smaller than the original number. (It’s intuitively obvious, but try induction if you would like rigor.) Thus, starting at 1000 or above eventually pulls one down to under 1000. For n 1000, I’ve personally checked the iterates of f (n) for n = 1 to 999 by a computer program such as the one above. The direct proof is actually faster and easier, as a three-digit string for a number must have

one of these possibilities for (# even digits, # odd digits, total # digits):

–  –  –

(0, 3, 3) (1, 2, 3) (2, 1, 3) (3, 0, 3) So, if n 1000, within one iteration you must get one of these four triples.

Now apply the rule to each of these four and you’ll see that you always produce (1, 2, 3) — thus resulting in the claimed number of 123.

Words to Numbers: 4 Here is one that master recreationist Martin Gardner wrote to tell me about several years ago. Take any whole number and write out its numeral in English, such as FIVE for the usual 5. Count the number of characters in the spelling. In this case, it is 4 — or FOUR. So, work now with the 4 or FOUR. Repeat with 4 to get 4 again.

As another instance, try 163. To avoid ambiguity, I’ll arbitrarily say that we will include spaces and hyphens in our count. Then, 163 appears as ONE HUNDRED SIXTY-THREE for a total count of 23. In turn, this gives 12, then 6, then 3, then 5, and finally 4.

Though this result is clearly language-dependent, other natural languages may have a comparable property, but not necessarily with 4 as the black hole.

Narcissistic Numbers: 153

It is well known that, other than the trivial examples of 0 and 1, the only natural numbers that equal the sum of the cubes of their digits are 153, 370, 371, and 407. Of these, just one has a black-hole property.

To create a black hole, we need to define a universe (set U ) and a process (function f ). We start with any positive whole number that is a multiple of

3. Recall that there is a special shortcut to test whether you have a multiple of 3. Just add up the digits and see whether that sum is a multiple of 3. For instance, 111,111 (six ones) is a multiple of 3 because the sum of the digits, 6, is. However, 1,111,111 (seven ones) is not.

Since we are going to be doing some arithmetic, you may wish to take out a hand calculator and/or some paper. Write down your multiple of 3.

One at a time, take the cube of each digit. Add up the cubes to form a new


number. Now repeat the process. You must reach 153. And once you reach 153, one more iteration just gets you 153 again.

Let’s test just one initial instance. Using the sum of the cubes of the digits, if we start with 432 — a multiple of 3 — we get 99, which leads to 1458, then 702, which yields 351, finally leading to 153, at which point future iterations keep producing 153. Note also that this operation or process preserves divisibility by 3 in the successive numbers.

–  –  –

This program continues forever, so break out after you’ve grown weary.

One nice thing is that it is easy to edit this program to test for black holes using larger powers. (It is well known that none exists for the sum of the squares of the digits, as one gets cycles.) In more formal language, we obtain the 153 mathemagical black hole by letting U = 3 · = all positive integral multiples of 3 and f (n) = the sum of the cubes of the digits of n. Then b = 153 is the unique black-hole element.

(For a given universe, if a black hole exists, it is necessarily unique.) Not incidentally, this particular result, without the “black hole” terminology or perspective, gets discovered and re-discovered annually, with a paper or problem proposal in one of the smaller math journals every few years.

46 M. W. ECKER The argument for why it works is similar to the case with the 123 example. First of all, 1¿ + 3¿ + 5¿ = 153, so 153 is indeed a fixed point. Second, for the black-hole attraction, note that, for large numbers n, f (n) n. Then, for suitably small numbers, by cases or computer check, each value eventually is “pulled” into the black hole of 153. I’ll omit the proof.

To find an analagous black hole for larger powers (yes, there are some), you will need first to discover a number that equals the sum of the fourth (or higher) power of its digits, and then test to see whether other numbers are drawn to it.

Card Tricks, Even

Here’s an example that sounds a bit different, yet meets the two criteria for a black hole. It’s a classic card trick.

Remove 21 cards from an ordinary deck. Arrange them in seven horizontal rows and three vertical columns. Ask somebody to think of one of the cards without telling you which card he (or she) is thinking of.

Now ask him (or her) which of the three columns contains the card. Regroup the cards by picking up the cards by whole columns intact, but be sure to sandwich the column that contains the chosen card between the other two columns. Now re-lay out the cards by laying out by rows (i.e., laying out three across at a time). Repeat asking which column, regrouping cards with the designated column being in the middle, and re-dealing out by rows. Repeat one last time.

At the end, the card chosen must be in the center of the array, which is to say, card 11. This is the card in the fourth row and second column.

There are two ways to prove this, but the easier way is to draw a diagram that illustrates where a chosen card will end up next time. But for those who enjoy programs, try this one from one of my readers.

–  –  –

Perhaps it is not surprising, but this trick, as with the sisyphian strings, generalizes somewhat. ¾ Kaprekar’s Constant: What a Difference 6174 Makes!

Most black holes, nonetheless, involve numbers. Take any four-digit number except an integral multiple of 1111 (i.e., don’t take one of the nine numbers with four identical digits). Rearrange the digits of your number to form the largest and smallest strings possible. That is, write down the largest permutation of the number, the smallest permutation (allowing initial zeros as digits), and subtract. Apply this same process to the difference just obtained. Within the total of seven steps, you always reach 6174. At that point, further iteration with 6174 is pointless: 7641–1467 = 6174.

–  –  –

Example: Start with 8028. The largest permutation is 8820, the smallest is 0288, and the difference is 8532. Repeat with 8532 to calculate 8532–2358 =

6174. Your own example may take more steps, but you will reach 6174.

The Divisive Number 15 Take any natural number larger than 1 and write down its divisors, including 1 and the number itself. Now take the sum of the digits of these divisors.

Iterate until a number repeats.

The black-hole number this time is 15. Its divisors are 1, 3, 5, and 15, and these digits sum to 15. This one is a bit more tedious, but it is also that much more strange at the same time. This one may defy not only your ability to explain it, but your very equilibrium.

Fibonacci Numbers: Classic Results as Black Holes

–  –  –

Solving by the quadratic formula yields two solutions, one of which is phi. More about the second solution in a moment.

Notice that the above plausibility argument did not use the values F (1) and F (2). Indeed, more generally, if you take any additive sequence (any sequence — no matter what the first two terms — in which the third term and beyond are obtained by adding the preceding two), one gets the same result: convergence to phi. The first two numbers need not even be whole numbers or positive. This, too, is easy to test in a program that you can write yourself.

Thus, if we extend the definition of black hole to require only that the iterates get closer and closer to one number, we have a black hole once again.

But there is still another black hole one can derive from this. Consider the function ´Üµ ½·½ Ü for nonzero real numbers Ü. I selected (or rather stumbled across) this function because of the simple argument above that gave phi as a limit. Start with a seed number Ü and iterate function values ´Üµ ´ ´Üµµ, etc. One obtains convergence to phi — but still no appearance of the second number that is the solution to the quadratic equation above.

Is there a connection between the two solutions of the quadratic equation? First, each number is the negative reciprocal of the other. Each is also one minus the other. Second, had we formed the ratios ´Òµ ´Ò · ½µ, we would have obtained the absolute value of the second solution instead of the first solution.

´Üµ ½·½ Ü, could not use an input Third, note that our last function of 0 because division by zero is undefined. The solution to the equation ´Üµ ½ · ½ Ü ¼ is just Ü  ½. Thus, Ü may not equal  ½ either.

We have not finished. We must now avoid ´Üµ  ½ (otherwise, ´ ´Üµµ ¼ and then ´ ´ ´Üµµµ is undefined). Solving ½ · ½ Ü  ½ gives Ü  ½ ¾.

If we continue now working backward with function ’s preimages we find, in succession, that we must similarly rule out   ¿, then   ¿,  ,   ½¿, ¾ etc. Notice that these fractions are precisely the negatives of the ratios of consecutive Fibonacci numbers in the reverse order that we considered. All of these must be eliminated from the universe for, along with 0. The limiting ratio, the second golden ratio, must also be eliminated from the universe.

In closing out this brief connection of Fibonacci numbers to the topic, I would be remiss if I did not follow my own advice on getting black holes by looking for fixed points: values Ü such that ´Üµ Ü. For ´Üµ ½·½ Ü, we 50 M. W. ECKER obtain our two golden numbers. As things are set up, the number ½ ½ is an attractor with the black-hole property, while  ¼ ½ is a repeller.

All real inputs except zero, the numbers   ´Ò   ½µ ´Òµ, and the second golden number lead to the attractor, our black hole called phi.

Unsolved Problems as Black Holes Even unsolved problems sometimes fall into this black-hole scenario. Consider the Collatz Conjecture, dating back to the 1930s and still an open question (though sometimes also identified with the names of Hailstone, Ulam, and Syracuse). Start with a natural number. If odd, triple and add one. If even, take half. Keep iterating. Must you always reach 1?

If you start with 5, you get 16, then 8, then 4, then 2, then 1. Success!

In fact, this problem has the paradoxical property that, although one of the hardest to settle definitively, is among the easiest to program (a few lines).

If you do reach 1 — and nobody has either proved you must, nor shown any example that doesn’t — then you next get 4, then 2, then 1 again, a cycle that repeats ad infinitum. Hmm a cycle of length three. We’re interested now only in black holes, which really are cycles of length one. So, let’s be creative and fix this up by modifying the problem.

Define the process instead by taking the starting number and breaking it down completely into factors that are odd and even. For instance, 84 is ¾ ¢ ¾ ¢ ¿ ¢. Pick the largest odd factor. In the example that would be ¿¢ ¾½. (Just multiply all the odd prime factors together. The only non-odd prime is 2.) Now triple the largest odd factor and then add 1. This answer is the next iterate.

Now try some examples. You should find that you keep getting 4. Once you hit 4, you stay at 4, because the largest odd factor in 4 is 1, and ¿¢½·½. Anybody who proves the Collatz Conjecture will prove that my variation is a mathemagical black hole as well, and conversely.

Because this is so easy to program, I will omit a program here.

Close Other examples of mathemagical black holes arise in the study of stochastic processes. Under certain conditions, iteration of matrix powers draws one to a result that represents long-term stability independent of an input


vector. There are striking resemblances to convergence results in such disciplines as differential equations, too.

Quite apart from any utility, however, the teasers and problems here are intriguing in their own right. Moreover, the real value for me is in seeing the black-hole idea as the unifying theme of seemingly disparate recreations.

This is an ongoing pursuit in my own newsletter, Recreational and Eductional Computing, in which we’ve had additional ones. I invite readers to send me

other examples or correspondence:

Dr. Michael W. Ecker, Editor Recreational and Educational Computing 909 Violet Terrace Clarks Summit, Pennsylvania 18411-9206 USA Happy Black-Hole hunting during your salute to Martin Gardner, whom we are proud to make a tiny claim to as our Senior Contributing Mathematical Editor.

Puzzles from Around the World Richard I. Hess Introduction Most of the puzzles in this collection were presented in the Logigram, a company newsletter published at Logicon, Inc., where I worked for 27 years.

The regular problem column appeared from 1984 to 1994 and was called “Puzzles from Around the World”; it consisted of problems from a number of sources: other problem columns, other solvers through word of mouth, embellishments or adaptations of such problems, or entirely new problems of my own creation. These problems are for the enjoyment of the solver and should be passed on to others for their enjoyment as well.

The problems are arranged in three general categories of easy, medium, and hard ; the solution section gives an approach to answering each of the problems. A sources section provides information on the source of each problem as far as I know. I would be delighted to hear from anyone who can improve the solution approach, add information on the source of the problem, or offer more interesting problems for future enjoyment.

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