blover Posted October 5, 2008 Share Posted October 5, 2008 Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting http://i33.tinypic.com/14t6dsl.jpg Quote Link to comment Share on other sites More sharing options...
ibelieveican Posted October 5, 2008 Share Posted October 5, 2008 Good Q IMO E Here is how. 324700 to 458600 Let us consider intervals now 1. 324a13 what are the values a can take. 7, 8, 9 so 3C1 2. 32ab13 a can take values 5, 6, 7, 8, 9 ( 4 is already considered in 1) 5C1 b can take values from 0 to 9 so 10C1....5c1X10C1 3. 3abc13.. a takes values 3 to 9 so 7C1 b and c takes values 0 to 9 so... 7C1X10C1X10C1 now we need to do the same thing with 458600 and I get E Hope I made sense Quote Link to comment Share on other sites More sharing options...
saketbansal Posted October 6, 2008 Share Posted October 6, 2008 458513 - 324713 = 133800 133800 / 100, as only one digit per humdered 1338 +1 as the ending number is also included. 1339 E. Quote Link to comment Share on other sites More sharing options...
hawk007 Posted October 6, 2008 Share Posted October 6, 2008 The questions says we need to find no of XXXXX13 between 324700 to 458600 for every hundred we can only have one 13 so if we calculate no of hundreds between 324700 to 458600 we get the solution I.e is 458600-324700=133900 divding by 100 we 1339 ANS-E Quote Link to comment Share on other sites More sharing options...
ibelieveican Posted October 6, 2008 Share Posted October 6, 2008 Great approach guys.. Quote Link to comment Share on other sites More sharing options...
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