**A Digital Diversion**

Perform the following steps as indicated to see an interesting result.

- Form the smallest possible 4-place number using the four largest digits.
- Add that number to itself. The result is sum #1.
- Take sum #1 and add it to itself. The result is sum #2.
- Finally, take sum #2 and add it to itself. The result is sum #3.
- While all 3 sums are interesting & share a common property, sum #3 is probably the most unique of all, especially when compared with the original 4-place number.

What is that strange aspect?

Do you care for another similar trick? Then try this:

- Form the largest possible 4-place number using the four largest digits.
- Divide that number by 4.
- Multiply the quotient just obtained by 5.
- Observe the resulting product carefully.

Aren’t numbers marvelous?

**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}

**Kaprekar Revisited**

Above in another activity you were introduced to the set we call Kaprekar’s Digits. This time we will insert the digit 0 and call the set the augmented Kaprekar set.: {0, 1, 4, 6, 7}. You will be multiplying by numbers formed from that set: e.g. 10746, 41067, etc. As before, the set for the products will be {2, 3, 5, 8, 9}.

Now the number of permutations for 5 things is 120, which is a bit boring to list out. Therefore, we will change our approach this time. Your task will be in the form of a traditional matching quiz. This means, we will list the permutations that work in the left column and the other factor in the right column. Once a match has been made, those numbers can be set aside and considered as completed.

There is another difference this time. The second factor will not be a whole number. Instead it will be non-integral, greater than 1. An * improper fraction* if you prefer. This proved to be necessary due to the nature of the digits used.

So let’s go. Match them up!

permutation | factor |

67014 | 8/3 |

70146 | 6/5 |

67140 | 9/8 |

71460 | 16/7 |

74610 | 17/15 |

41067 | 9/4 |

14607 | 4/3 |

61470 | 17/4 |

17460 | 5/4 |

47016 | 4/3 |

41706 | 7/3 |

14076 | 19/4 |

46710 | 8/5 |

17460 | 5/4 |

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.

- 4187814
- 4870784
- 6097906
- 630036
- 6834386
- 4004
- 82466428
- 1993 (a prime year)
- 102013 (another prime)
- 30013 (a "lucky" number)

** **

**POSTSCRIPT**

The activities above are based on a common theme – using the digits from 1 to 9, once and only once, to produce a special effect. It is a popular theme that has many variations, sometimes including the 0 as well. Some of the other variations are in other pages of WTM, while still others in other books and websites. For the individual who may be interested, we will give the web links here.

From the WTM:

- Double/Triple
- Nine-Digits-&-Equal-Products
- Ticket Number Trivia
- More Strange Multiplication
- Five Distinct Digits
- S.T.T.H.W.N. (Part II.2; Part IV.1)
- Special Numbers II (99066)
- FUN for Figurenuts

From the World!OfNumbers:

- http://www.worldofnumbers.com/em107.htm
- http://www.worldofnumbers.com/em114.htm
- http://www.worldofnumbers.com/em137.htm
- http://www.worldofnumbers.com/ninedigits.htm
- http://www.worldofnumbers.com/em85.htmM