The following fractions are written on the board 1/n, 2/(n-1), 3/(n-2), ... , n/1 where n is a natural number. Alice calculated the differences of the neighboring fractions in this row and found among them 10000 fractions of type 1/k (with natural k). Prove that he can find even 5000 more of such these differences.
When n, the sum of each numerator/denominator pair (1 more than the n used in the puzzle), is factored into primes, count the number of different primes used, k. The number of differences of successive fractions in the series that are unit fractions is 2^k - 2.
An example is n = 2*2*2*3*3*11*47 = 37224. There are 4 different primes in its factorization so there will be 2^4 - 2 = 14 pairs of adjacent fractions in the series whose difference is a unit fraction:
37224
2 2 2 3 3 11 47
1881/35343 1880/35344 19/357 5/94 1/33558
5688/31536 5687/31537 79/438 11/61 1/26718
8272/28952 8271/28953 2/7 919/3217 1/22519
10153/27071 10152/27072 923/2461 3/8 1/19688
13113/24111 13112/24112 31/57 149/274 1/15618
13960/23264 13959/23265 1745/2908 3/5 1/14540
15840/21384 15839/21385 20/27 337/455 1/12285
21385/15839 21384/15840 455/337 27/20 1/6740
23265/13959 23264/13960 5/3 2908/1745 1/5235
24112/13112 24111/13113 274/149 57/31 1/4619
27072/10152 27071/10153 8/3 2461/923 1/2769
28953/8271 28952/8272 3217/919 7/2 1/1838
31537/5687 31536/5688 61/11 438/79 1/869
35344/1880 35343/1881 94/5 357/19 1/95
Edited on February 21, 2021, 6:22 pm
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Posted by Charlie
on 2021-02-21 18:13:22 |
The current hypothesis
