huneynumb Posted May 17, 2006 Share Posted May 17, 2006 Express 1991 as a sum of consecutive positive integers and show that this is the only way to do it. Link to comment Share on other sites More sharing options...
GutZ Posted May 17, 2006 Share Posted May 17, 2006 Like A+B+C+D = 1991? How many numbers can we use? Link to comment Share on other sites More sharing options...
Severian Posted May 17, 2006 Share Posted May 17, 2006 1991 = 994 +995 but I am not sure how to prove that it is the only case (induction maybe) Link to comment Share on other sites More sharing options...
Tartaglia Posted May 17, 2006 Share Posted May 17, 2006 Don't the 22 consecutive numbers 80,81,82...101 add up to 1991? Link to comment Share on other sites More sharing options...
Tartaglia Posted May 18, 2006 Share Posted May 18, 2006 What about the 11 consecutive integers 176,177, 178,...186? Link to comment Share on other sites More sharing options...
Bignose Posted May 19, 2006 Share Posted May 19, 2006 1991 = 994 +995 but I am not sure how to prove that it is the only case (induction maybe) 1991=996 + 995 Link to comment Share on other sites More sharing options...
Tartaglia Posted May 19, 2006 Share Posted May 19, 2006 The proof or otherwise of this problem revolves around the factors of 1991 Since for an arithmetic progression S(n) = (n/2)*(2a+(n-1)d) either n/2 or n must be a factor of 1991. This means the string of digits can only be 1,11,181 long or 2,22,362. Link to comment Share on other sites More sharing options...
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