Solutions to Assignment 5

Solutions to Assignment 5 - stopping when one gets to a...

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April 6, 2006 Physics 681-481; CS 483: Discussion of #5 1. We know that r < 100 and that 11490 / 2 14 = 0 . 7012939453 . . . is within 1 2 2 - 14 < 1 2 1 100 2 of j/r for integers j and r . My ancient HP-20S calculator tells me that 11490 / 2 14 = 1 1 + 1 2 + 1 2 + 1 1 + 1 7 + 1 35 + · · · (1) If I drop what comes after the 35 and start forming the partial sums I immediately get to a denominator bigger than 100, so I can drop the last 1 35 as well. I then find that 11490 / 2 14 1 1 + 1 2 + 1 2 + 1 1 + 1 7 (2) which works out 1 to 54 77 . Since 77 is the only multiple of 77 less than 100, r = 77. And indeed, 2 14 × 54 77 = 11490 . 079 . . . which is within 1 2 of 11490. 2. My vintage 1987 calculator (still going on its original battery) tells me that 11343 / 2 14 = 1 1 + 1 2 + 1 3 + 1 1 + 1 419 + · · · (3) 1 Another way to get this is to use the recursion relation for the numerators p and denominators q of the partial sums: p n = a n p n - 1 + p n - 2 , and q n = a n q n - 1 + q n - 2 , with q 0 = a 0 , q 1 = 1 + a 0 a 1 and p 0 = 1 , p 1 = a 1 . One easily applies these to the sequence a 0 , a 1 , a 2 . . . = 1 , 2 , 2 , 1 , 1 , 7 , 35
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Unformatted text preview: , stopping when one gets to a denominator larger than 100. 1 Since 419 is already bigger than 100 we can drop the 1 419 to get 1 1 + 1 2 1 4 (4) which gives 9 13 , and indeed 2 14 × 9 13 = 11342 . 769 . . . (1) which is within 1 2 of 11343. The number r is thus a multiple of 13 less than 100, of which there are seven. But we also have 13653 / 2 14 = 1 1 + 1 4 + 1 1 + 1 1364 + ··· (5) Since 1364 is bigger than 100 we can drop the 1 1364 to get 1 1 + 1 5 (6) which gives 5 6 , and indeed 2 14 × 5 6 = 13653 . 333 . . . (2) which is within 1 2 of 13653. So r is also a multiple of 6 less than 100. Since 6 and 13 have no common factors the least multiple of both is 6 × 13 = 78. Since there is no multiple of 78 less than 100 other than 78 itself, r = 78. 2...
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