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hw05soln

# hw05soln - (3 Since 2339 is already bigger than 100 we can...

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April 8, 2004 Physics 681-481; CS 483: Discussion of #5 1. We know that r < 100 and that 11883 / 2 14 = 0 . 725280 . . . is within 1 2 2 - 14 < 1 2 1 100 2 of 1 /r . My HP-20S calculator tells me that 11883 / 2 14 = 1 1 + 1 2 + 1 1 + 1 1 + 1 1 + 1 3 + 1 1 + 1 1 + 1 19 + · · · (1) If I drop what comes after the 19 and start forming the partial sums I quickly get to a denominator bigger than 100, so I can drop the last 1 19 as well. I then find that 1 1 + 1 2 + 1 1 + 1 1 + 1 1 + 1 3 1 2 (2) which works out 1 to 66 91 . And indeed, 66 × 2 14 / 91 = 11882 . 9011 which is within 1 2 of 11883. 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 , 1 , 1 , 1 , 3 , 1 , 1 , 19 , . . . , stopping when one gets to a denominator larger than 100. 1

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2. We have 11703 / 2 14 = 1 1 + 1 2 + 1 1 + 1 1 + 1 2339 +
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Unformatted text preview: (3) Since 2339 is already bigger than 100 we can drop the 1 2339 to get 1 1 + 1 2 1 2 (4) which sums to 5 7 , and indeed 5 × 2 14 / 7 = 11702 . 8571 . . . (1) which is within 1 2 of 11703. The number r is thus a multiple of 7 less than 100. We also have 15019 / 2 14 = 1 1 + 1 11 + 1 341 + · · · (5) Since 341 is bigger than 100 we can drop the 1 341 to get 1 1 + 1 11 (6) which sums to 11 12 , and indeed 11 × 2 14 / 12 = 15018 . 666 . . . (2) which is within 1 2 of 15019. So r is a multiple of 12 less than 100. Since 7 and 12 have no common factors the least multiple of both is 7 × 12 = 84 and there is no other common multiple less than 100, so r = 84. 2...
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