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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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- Spring '08
- Ginsparg
- Rational function, Recurrence relation, partial sums