Problem 68. Kaprekar Steps
The behavior at x=6174 is artifically set to 0, which taints the pure-recursive solution, but hey it was cool problem!
The K constant is 495 for 3 digits.
So the test with 691 is wrong.
Why tests with only one digit ?
Did I miss something ?
I don't understand the Test Suite for x = 3 and x = 1 too. what should I do in this case?
For x=3, the steps are 3000-0003=2997, 9972-2799=7173, etc.
Problem description is confusing as there are different Kaprekar constants depending on the number of digits. [0 9 495 6174 for 1, 2,3 4 digits respectively.
getting this error:
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The problem should specify that any number with less than four digits should be filled up to four digits with leading zeros. (e.g. 3 -> 0003)
Very nice and interesting problem!
I like the recursion aspect of this problem.
There is a small correction needed in the problem statement. Not all natural numbers, but 4 digit numbers can be reduced to Kaprekar number by the mentioned method. Similarly 3 digit numbers can be reduced to 495
How it works x = 1？？？？？
For those confused with test cases 2,3 and 5, like myself before, do conversion to 4-digit integer. Here is an example:
x = 1:
1000-0001 = 999
9990-0999 = 8991
9981-1899 = 8082
8820-0288 = 8532
8532-2358 = 6174
Therefore, y_correct = 5
Very nice. Took some few minutes to crack this.
What I did is to convert x into string and then use sort function.
This problem statement is poorly given and leaves out a crucial element- the input number must always have 4 digits, so if 1 is the input, the first iteration should be 1000-0001. the second iteration should be 9990-0999. for the problem as stated, the test suite gives solutions that are incorrect.
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