Birth Year Magic

Everybody currently living was born in a year that contains four digits. I was born in 1949. My granddaughter was born in 2002. Let's take her year of birth and arrange the digits into ascending and descending order to form two new numbers: 2200 and 0022 (or simply 22). Let's subtract the smaller from the larger. The result is 2178. Let's repeat that process to from 8721 and 1278. Subtraction yields 7443. Continuing this process we get 3996, 6264, 4176 and finally 6174. Why finally? Because subtracting 7641 and 1467 leads us back to 6174! So for 2002, the progression is 2178, 7443, 3996, 6264, 4176, 6174.

Remarkably, all birth years lead to this very same number. Let's take my year of birth: the progression is 1949, 8442, 5994, 5355, 1998, 8082, 8532, 6174. The number 6174 is known as Kaprekar's constant and because it involves four digit numbers with at least two digits different, it's ideally suited to current birth years. You're not going to find someone who was born in 1111 or 2222 who will spoil the trick and so it's an ideal party trick. Everyone has their smartphone near at hand and, if not, one can always be borrowed. Using a calculator app or simply the Chrome browser, the necessary calculations can be quickly and easily carried out.

Figure 1: Google Chrome's Calculator

So Kaprekar's constant connects everybody alive today and in fact everybody born, or still to be born, between the years 1111 and 2222 with those two limits excluded.

There is no other four digit number that returns itself under the operation of subtract digits in ascending order from digits in descending order. There is a nice proof of why this is so on this site that also alerted me to the fact that for three digit numbers there is a constant that is reached as well, namely 495.

Suppose we have four digits \(a,b,c,d\) such that \(9 \geq a>b>c>d \geq 0\). If we can get the same digits (in any order) arising after the subtraction of \(dcba\) from \(abcd\), then we have an unchanging constant. It turns out that that only \(abcd-dcba=bdac\) satisfies and when this is the case \(a=7, b=6, c=4, d=1\). It's the same for the three digit situation \(abc\) and \(cba\) where \(9 \geq a>b>c \geq 0\). Here only \(abc-cba=cab\) satisfies and \(a=9,b=5,c=4\).

A table is also provided showing the situation for numbers of lengths other than 3 and 4. It can be seen that only three and four digit numbers lead to a unique constant. See Figure 1.

Figure 1: source

The site also shows the frequencies of the number of iterations required for four digit numbers to reach 6174. See Figure 2.

Figure 2: source

Here a permalink to some SageMath code that will display, for four digit numbers, the progression in reaching 6174 and count the number of iterations required. Here is another permalink, this time for three digit numbers.

Kaprekar is an interesting mathematician (biography link) and he is also associated with:

• Kaprekar numbers (Wikipedia link)
• Harshad numbers (see my blog posts: here and here)
• Self numbers and Junction numbers (see my blog post here)

Finally we should say a little about the number 6174 and some of its properties. One interesting property is that it can be written as the sum of the first three degrees of 18:$$18^3 + 18^2 + 18^1 = 5832 + 324 + 18 = 6174$$This relationship also means that 6174 is a Harshad number because it is divisible by the sum of its digits (18).