Few days back I posted an article based on the interesting properties of 153. Lot of people got very excited to know about similar other numbers with such interesting properties. Today I will be discussing about another such interesting number: 6174
Kaprekar’s Constant
6174 is known as Kaprekar’s constant named after the Indian recreational mathematician D. R. Kaprekar. I’ve written about D.R.Kaprekar and contribution of other Indian mathematicians. 6174 has got a very interesting property. To know what that mysterious property is take any four-digit number. Arrange the digits in ascending and then in descending order to get two four-digit numbers. Then subtract the bigger number from the smaller number. If we keep on repeating this process we will end up in 6174. This process is called Kaprekar’s routine. All the numbers will yield 6174 in 7 or less than 7 iterations.
Example
Let’s randomly choose any number, say 4518:
Now, arranging the digits in ascending and then in descending order to get two four-digit numbers.
8541-1458 = 7083
8730-0378 = 8352
8532-2358 = 6174
Hence we get 6174 in 3 iterations.
4651 reaches 6174 after 7 iterations
6541-1456 = 5085
8550-558 = 7992
9972-2799 = 7173
7731-1377 = 6354
6543-3456 = 3087
8730-378 = 8352
8532-2358 = 6174
Try it for any 4-digit number yourself and see if it works.
Questions
- For a specific set of numbers Kaprekar’s routine will not work. Can you tell me what numbers will those be?
- If you follow Kaprekar routine with any 3 digit number it will also result in one specific number. Can you find out that 3 digit equivalent constant?
- The result of each iteration of Kaprekar’s routine is a multiple of 9. Can you explain why?
Hint: you have seen the application of similar mathematical logic in the earlier post – mind reading trick.
Leave your answers in the comments below.
Its realy good to b here. Well i found
3724 is a no. which follows the kaprekar’s observation in 8th iteration.
Hi Goyalk,
I was surprised to see someone here has found something path-breaking, but sad luck it takes 7 iteration only. Here you go –
7432 – 2347 = 5085
8550 – 558 = 7992
9972 – 2799 = 7173
7731 – 1377 = 6354
6543 – 3456 = 3087
8730 – 378 = 8352
8532 – 2358 = 6174
I’m a big fan of ur site
and the number 2588 is getting 6174 at 10th iteration
Hi Rajasekhar,
Thanks for your kind words. Check your calculations again. 2588 reaches 6174 in just 3 iterations. Here you go –
8852-2588=6264
6642-2466=4176
7641-1467=6174
Answer for the second qtn is 594
Answer for 2nd question is 495
1) The answer to 1st question is that all the numbers with repeating digits are an exception like 3333, 4444, 5555 etc.
2) The number should be 297
3) This is because whenever one subtracts a number from its reverse number it is always divisible by 9
a four digit number can be written as (1000x + 100y + 10z + w) – (1000w + 100z + 10y + x) = 999x + 90y – 90z – 999x which will always be divisible by 9. The same rule applies to the 3 digit number above.