ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
61 lines
2.3 KiB
Markdown
61 lines
2.3 KiB
Markdown
---
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id: 5900f50b1000cf542c51001d
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title: 'Problem 414: Kaprekar constant'
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challengeType: 5
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forumTopicId: 302083
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dashedName: problem-414-kaprekar-constant
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---
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# --description--
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6174 is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get 7641-1467=6174.
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Even more remarkable is that if we start from any 4 digit number and repeat this process of sorting and subtracting, we'll eventually end up with 6174 or immediately with 0 if all digits are equal.
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This also works with numbers that have less than 4 digits if we pad the number with leading zeroes until we have 4 digits.
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E.g. let's start with the number 0837:
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8730-0378=8352
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8532-2358=6174
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6174 is called the Kaprekar constant. The process of sorting and subtracting and repeating this until either 0 or the Kaprekar constant is reached is called the Kaprekar routine.
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We can consider the Kaprekar routine for other bases and number of digits. Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers. However, it can be shown that for 5 digits and a base b = 6t+3≠9, a Kaprekar constant exists. E.g. base 15: (10,4,14,9,5)15 base 21: (14,6,20,13,7)21
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Define Cb to be the Kaprekar constant in base b for 5 digits. Define the function sb(i) to be 0 if i = Cb or if i written in base b consists of 5 identical digits the number of iterations it takes the Kaprekar routine in base b to arrive at Cb, otherwise
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Note that we can define sb(i) for all integers i < b5. If i written in base b takes less than 5 digits, the number is padded with leading zero digits until we have 5 digits before applying the Kaprekar routine.
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Define S(b) as the sum of sb(i) for 0 < i < b5. E.g. S(15) = 5274369 S(111) = 400668930299
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Find the sum of S(6k+3) for 2 ≤ k ≤ 300. Give the last 18 digits as your answer.
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# --hints--
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`euler414()` should return 552506775824935500.
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```js
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assert.strictEqual(euler414(), 552506775824935500);
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```
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# --seed--
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## --seed-contents--
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```js
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function euler414() {
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return true;
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}
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euler414();
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```
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# --solutions--
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```js
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// solution required
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```
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