This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
This homework help was uploaded on 02/01/2008 for the course CS 483 taught by Professor Ginsparg during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 Ginsparg

Click to edit the document details