一个自然数,最后一位数值移到最前一位.得到一个新的自然数.
新的自然数刚好是这个自然数的两倍.
求此自然数为?
如:
旧 新
x=12345 y=51234
二进制:01,10 |
105263157894736842 *2 = 210526315789473684 |
157894736842105263 ） 210526315789473684 （转自：顶顶华闻 www.TopChineseNews.com ） 263157894736842105 315789473684210526 368421052631578947 （转自：顶顶华闻 www.TopChineseNews.com ）421052631578947368 ） 473684210526315789（转 It is not fair! You beat me on this one! But I have more! |
You always win, I have never beaten you! |
You did. I cheated to get the rest. Just cut and pasted from yours. They are periodical and you found the period before I did. As for the 思路, you just start with a digit less than 5 and start to multiply by 2 and to see when it repeats itself. You will get one of the above numbers. It looks like they and their repeats are the only such numbers. A repeat is something like 105263157894736842105263157894736842 |
Let the original number be 10x + y, where y is a single digit positive integer, and x is any positive integer. We have 2 * (10x + y) = x + 10^m * y, and [log x] = m-1. We use [x] to denote the largest integer that is less than or equal to x. Collect the terms, we have 19 * x = (10^m – 2) * y. Since y is only single digit, 10^m -2 has to be divisible by 19. Let 10^m – 2 = 19 * k. The left hand side is divisible by 2, so k is divisible by 2. Let k = 2n. Now we have 5 * 10^(m-1) – 1 = 19 * n. Our task is now to find 499….9 such that it's divisible by 19. This may take a long slip of paper but it is not hard. Next we will find 99…9 such that it's divisible by 19. This can be simplified as finding 11…1 such that it's divisible by 19. Another long slip of paper will do it. You can append any multiples of chunks of 9's you found to the 499…9 you found initially. |
This number 105263157894736842 has an interesting property: multiplying it by any of 2,3,4,5,6,7,8,9 gets a such repeating number. However, of course, not by 10. My method: Let x be the old number, y be the last diget, then (x-y)/10+y*10^n=2x we got x =(99...9/19)*y =52631578947368421*y (instead of multiplication we can use division which I think easier) Since the first diget must be less than 5 (otherwise it multiplied by 5 will be more diget), y not= 1 when y=2, we have x= 105263157894736842 when y=3, we have 157894736842105263 when y=4... look like all the results for x are such repeating numbers. |
Let the number be 10x+y, we have (10^n)*y+x=2(10x+y), 19x=(10^n-2)y. so 10^n-2 is divisible by 19. 17 is the smallest number satisfying this condition. all those numbers are of the form n=17+18k. Divide 1 by 19, we get an infinite sequence of digits with period 18. Begin with 1,2,3 or 4, cut a segment of length n=17+18k, we get the number. |
1/19=0.052631578947368421052631578947368421....... Starting at any place with digit 1,2,3 or 4, cut a segment of this sequence of length 17+18k(k=1,2,3,...), we get the number. All those numbers can be obtained by this way.
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