This is a simple addition activity that involves problem solving
at a level appropriate for young students (around 4th grade at least). It is based on a problem that appeared in the 1996 MATHCOUNTS competition exam materials. The problem stated:
|Charlene's age on her birthday in 1995 is equal to the sum of the digits in the year of her birth. How many years old is she on this birthday?|
The basic concept needed to solve this problem is that of the digit sum of a number. The digit sum of a number is nothing more nor less than the "sum of the digits of a given number." For example, the digit sum of 283 is 2 + 8 + 3, or 13.
To make an interesting and reasonably challenging activity for elementary students, we can proceed as follows:
|1. Pick any number.||283|
|2. Find the sum of its digits.||13|
|3. Add these two numbers.||296|
That's it. But that in itself is not much, you say. Of course, you're right. The challenge lies in reversing the whole situation. Ask the students to find a number that will produce a specific result or sum at Step 3. That is, given something like 296 for the "sum" in Step 3, the students are expected to come up with 283 as the "answer."
Let's illustrate what we mean. "Okay, kids, now that you know how to do the 3-step procedure, let's reverse things just a bit. Can you find a number that after you do the process on it you will have a sum of say, 152?"
[Draw a rectangular box to represent the top addend. Put a smaller box under it to represent the digit sum. Underline, and place the "+" sign. Write the number 152 under the line to form the vertical format that most students are accustomed to.]
Now say: "What could we put in the large box?" What do you think would be a reasonable guess for the number we're looking for?" Try to elicit something smaller than 152, of course, but not too small. (This is where good number sense comes in handy.) A possible line of reasoning might go as follows:
|1) Try 145||145 + 10 = 155 [Too high]|
|2) Try 144||144 + 9 = 153 [Still too high but closer]|
|3) Try 143||143 + 8 = 151 [Too low]|
|4) Try 139||139 + 13 = 152 [That's it!]|
Continue the investigation with other numbers, especially larger numbers like the one used in the MATHCOUNTS problem. Warning! Some "answer/sum numbers" can not be produced by the digit-sum addition process. Can you find some of them? Also, some "answer/sum numbers" can be produced by two different numbers. Can you find some of this type as well?
After the above material was originally written, I came across a nice idea in math: naming categories of numbers according to some particular characteristic they have. (See Keith Numbers.) So I have decided to call those numbers that do not have a solution in the procedure described above as CHARLENE NUMBERS.
After the above material was written, I was rumaging through some old notes and clipped magazine articles that I had gathered over past years (and were frankly gathering dust!). And lo and behold! The concept of the MATHCOUNTS problem goes back a lot further than I had remembered. It seems it has its genesis with that mathematician from India, named D. R. Kaprekar. (Click here to see another of his activities that I wrote about in this website and that is often mentioned in the literature.)
Now the situation is this: what I had called Charlene Numbers earlier already had a name, Kaprekar's self-numbers. Kaprekar called them by that name because they were "self-born", that is to say, not the result of adding the digit sum of a smaller number (which he called the generator of N) to produce that number. All numbers that are not self-numbers he called generated numbers because they were the results of the digit sum addition process, that is to say, "generated" by a smaller number.
My source of this and what will be explained below are two articles from Martin Gardner's Mathematical Games columns in the March and April 1975 issues of SCIENTIFIC AMERICAN magazine.
First, Kaprekar called this unusual algorithm "digitaddition". Sequences of numbers using this process can easily be constructed by any 3rd-grade schoolchild. For example, beginning with 47, one would have
Of course, such a sequence has no end, so there could be no sum of the terms of an infinite sequence of this type, as any schoolboy is supposed to learn in advanced mathematics. But, unlike that advanced math class, there is no "easy" (formula) way to find the sum of a finite series either. You just have to "add 'em up". For the partial sequence given above -- 47 to 95 -- the sum is 350.
But, Kaprekar did find an interesting short cut process for finding the sum of all the digits involved in such partial sequences. He said you merely subtract the first and last terms and add to that result the sum of the digits of the last term. For our example, we have
95 - 47 + (9 + 5) = 48 + 14 = 62.
As he himself said, "Is this not a wonderful new result? The Proof of all this rule is very easy and I have completely written it with me. But as soon as the proof is seen the charm of the whole process is lost, and so I do not wish to give it just now." (Gardner, March 1975, pp. 113-4)
As I discovered and mentioned earlier, some numbers can even be produced by more than one number or generator. Of course, Kaprekar knew all about this, too, back in 1949, in fact. He had found that many numbers could have more than one generator, calling such numbers by the name junction numbers. The smallest junction number, having 2 generators, is 101; its generators are 91 and 100. The smallest junction number having 3 generators is
its generators are 10,000,000,000,000, 9,999,999,999,901, and 9,999,999,999,892. That's getting pretty large, I'd say.
But in 1961, Kaprekar said he found the smallest junction number with four generators, a 25-digit number. It is
The generators, however, were not given in the article. Can you find them?
Gardner mentioned that numbers that were primes and self-numbers could be called self primes. (Sounds logical, I'd say.) 3, 5, 7, 31, and 53 are some small examples of this. And likewise for year numbers, mentioning that the years in this century prior to the time he wrote these articles that were "self-years" are the following:
[I found this interesting because my birth year is one of those self-years! Which one? Ah, I'll never tell.]
Gardner then comments that 11,111,111,111,111,111 (or repunit-17) and 3,333,333,333 are self-numbers.
Next, what about special sets of numbers. Gardner says that the powers of 10 produce some interesting results.
|Power of 10||Generator|
But one million is not there. Why? You guessed it; it's a self number! It has no generator. (Could we say that like many millionaire people are "self made", likewise the number 1,000,000 is "self born"?) The next self-power-of-ten is 1016.
The March article closes with this comment: "No one has yet discovered a nonrecursive formula that generates all self-numbers, but Kaprekar has a simple algorithm by which any number can be tested to determine whether it is self-born or generated." He challenges the reader to find it and find the next self-year after the one given in the list above. In the April issue he gave the answers, which I give below verbatim.
D. R. Kaprekar's method of testing a number, N, to see if it is a self-number is as follows. Obtain N's digital root by adding its digits, then adding the digits of the result and so on until only one digit remains. If the digital root is odd, add 9 to it and divide by 2. If it is even, simply divide by 2. In either case call the result C. Subtract C from N. Check the remainder to see if it generates N. If it does not, subtract 9 from the last result and check again. Continue subtracting 9's, each time checking the result to see if it generates N. If this fails to produce a generator of N in k steps, where k is the number of digits in N, then N is a self-number.
For example, we want to test the year 1975. Its digital root, 4, is even, so that we divide 4 by 2 to obtain C = 2. 1975 minus 2 is 1973, which fails to generate 1975. 1973 minus 9 is 1964. This also fails. But 1964 minus 9 is 1955, and 1955 plus the sum of its digits, 20, is 1975; therefore 1975 is a generated number. Since 1975 has four digits, we would have had only one more step to go to settle the matter. With this simple procedure it does not take long to determine that the next self-year is 1985. There will be only one more self-year in this century: 1996.
As a classroom math activity, I would hope that the teacher would not reveal all the secrets covered in the articles by Gardner, especially the shortcut for finding a given number's generator, if it indeed exists. Much math exploration can be provided if such concepts are revealed after the benefits of the activity are no longer important. Then the shortcut can be introduced to infuse new life into the problem.
And finally, if you really want to see some high-class math formulas related to self numbers, I recommend that you check out this page: Self Numbers.
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