Another Digital Diversion

     It is a well known, and easily proven fact, that using 4 distinct digits, such as 3, 7, 2, 4, and so on, you can form 24 different 4-place numbers. The same principle would apply using 4 different letters of the alphabet, namely you could form 24 different "words". The words need not make sense or even be pronounceable. Likewise, .you could arrange 4 persons in four chairs in a row in 24 different ways. Such an arrangement is called a permutation of the 4 items under consideration.

          Now to continue in the line of reasoning of our previous digital activity…

     Take the digits 2, 3, 7, & 9, and form all 24 possible 4-place numbers. It is our goal now to find which of those numbers will yield for us the remaining five digits (1, 4, 5, 6, & 8) when they are doubled.

     Extra 1: While you still have your list of 24 candidates handy, try this with those that did not work out: multiply by 5 in order to get products formed by the same five digits (1, 4, 5, 6, & 8).

     Extra 2: Now replace the 3 above with a 6 and repeat the above doubling & quintupling process on those 24 permutations. That is, use 2, 6, 7, & 9 to yield results containing only the digits of 1, 3, 4, 5, & 8.

A Digital Diversion a la Kaprekar

     A little known piece of number trivia concerns a mathematician from India, named D. R. Kaprekar. He discovered that if you use the digits 1, 4, 6, & 7 to make the largest and smallest 4-place numbers, that the difference between them is a number that is composed of those same four digits. This is easily shown as follows:

7641 - 1467 = 6174

     The number, 6174, is therefore called Kaprekar's Constant. Much has been written about this idea in various websites*, math books & periodicals.

     However, your task here is (1) to list all 24 permutations of those digits as 4-place numbers; then (2) use some other number(s) to multiply them by in order to obtain the remaining five digits (2, 3, 5, 8, & 9). [Hint: the factors you need are less than 10.]

     Extra: three of the permutations will yield good results when multiplied by 12, 16, & 53.

     Note: use of a spreadsheet is recommended for this problem.

     [*For my website, go to Kaprekar. There are other links there as well.]

DD's Two-by-Two

     By now, you should be good at listing the 24 permutations of 4 distinct digits to form 4-place numbers. Also you should be pretty good at finding which permutation(s) yield the remaining unused digits to form the final result.

     So without further ado, we will give you nine sets of 4 digits to work with. Find which permutations produce the other five digits in their products when multiplied by 8.

     (If you think about it for a moment, that's actually what you were doing in the problem task at the beginning of this collection of activities. That is, doubling something three times in sequence is equivalent to multiplying by 8: N * 2 * 2 * 2 = N * 8.]

     Here are the sets. Each set will produce 2 good products. Sorta like Noah putting the animals in the ark 2-by-2, isn't it?

{1, 2, 3, 7}    {1, 4, 5, 9}    {1, 4, 6, 9}
{1, 5, 6, 9}    {2, 3, 5, 9}    {3, 4, 8, 9}
{3, 5, 8, 9}    {4, 5, 8, 9}    {4, 6, 7, 9}

Pandigitals and Fibonacci

     Surely you are familiar with a famous category of numbers in the mathematical world, called the Fibonacci Numbers. If not, we recommend that you do a search of the web and you will be amazed with this fascinating topic. Suffice it to say here that they are those numbers that appear in this sequence

1, 1, 2, 3, 5, 8,13, 21, 34, 55, etc.

where each number after the first two is found by adding the 2 preceding numbers. Study that example to convince yourself that it's true.

     Now we plan to connect the idea of the Fibonacci addition procedure and pandigitals in a clever and interesting way. Here's how:

     We will start with a certain number - it may be special in its own way, like being a palindrome, a prime, etc. - then we will try to get a pandigital result sooner or later by applying the special method of addition. Watch this:

     Let's start with the palindrome 9530359. Then add 9530360, the next counting number after it. The result is 19060719. Now add 9530360 and 19060719, to obtain 28591079. Continuing like this, our next two sums are 47651798 and 76242877. When those two are finally added, -- voila! - we have our pandigital of 123894675. This only took us 5 steps of hard adding. Not bad, eh?

     To avoid unnecessary re-copying of these big numbers, we could condense our presentation like so:

9530359 + 9530360 ------------ 19060719 ------------ 28591079 ------------- 47651798 ------------- 76242877 ------------- 123894675

     Looks rather easy now, doesn't it? Well, we're just getting started. Here are some more starting numbers for you to play with. Some take more steps to arrive at the "pandigital 9", some require fewer. So be careful. One little slip and you'll be going off in the wrong direction.

1. 4187814
2. 4870784
3. 6097906
4. 630036
5. 6834386
6. 4004
7. 82466428
8. 1993  (a prime year)
9. 102013  (another prime)
10. 30013  (a "lucky" number)