Question

What is the maximum possible value of k for which 2013 can be written as a sum of k con-secutive positive integers?

Answer 1

Let n be a positive integer.

Given  2013 = (n+1) +  (n+2)  + n+3 ...... + n+k
                    = (1+2+3+...+k) + (n+n+n... k times)
                    = k (k+1) / 2  +  n k

=> k² + (2n+1) k - 4026 = 0
=>   2 k = √[(2n+1)²+16104] - (2n+1)

Obviously, k is maximum when n is minimum possible. Because if n is smallest, then to get the sum of 2013, we need more terms. So k will be more.

Let n = 1 (minimum possible),  2 k = √16113 - 3,     k is not integer.
      n = 2 then    2 k = √16129 - 5   = 127 - 5 

so for n = 2,  k = 61

The numbers are :  3+4+5+6+.....+60+62+63   = 2013

 maximum value of k = 61

Answer 2

Answer:

Step-by-step explanation:

Let n be a positive integer.

Given  2013 = (n+1) +  (n+2)  + n+3 ...... + n+k

                   = (1+2+3+...+k) + (n+n+n... k times)

                   = k (k+1) / 2  +  n k

=> k² + (2n+1) k - 4026 = 0

=>   2 k = √[(2n+1)²+16104] - (2n+1)

Obviously, k is maximum when n is minimum possible. Because if n is smallest, then to get the sum of 2013, we need more terms. So k will be more.

Let n = 1 (minimum possible),  2 k = √16113 - 3,     k is not integer.

     n = 2 then    2 k = √16129 - 5   = 127 - 5  

so for n = 2,  k = 61

The numbers are :  3+4+5+6+.....+60+62+63   = 2013

maximum value of k = 61