Express 1991 as a sum of consecutive positive integers and show that this is the only way to do it.

Like A+B+C+D = 1991?

 

How many numbers can we use?

1991 = 994 +995 but I am not sure how to prove that it is the only case (induction maybe)

Don't the 22 consecutive numbers 80,81,82...101 add up to 1991?

What about the 11 consecutive integers 176,177, 178,...186?

1991 = 994 +995 but I am not sure how to prove that it is the only case (induction maybe)

 

1991=996 + 995

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.