Question

What is the maximum possible value of k for which 4026 can be written as a sum of ‘k’ consecutive positive integers?

Answer 1

Step-by-step explanation:

And indeed, 2013 is the sum of the 61 consecutive integers 3+4+5+… +63. If k is even, then the average is a half-integral submultiple of 2013, so that k must be one of 2, 6, 22, 66 … 4026; but it must again, for the same reason, be <63.5, so that we get no new maximum solutions that way.

Answer 2

Answer:

$\: \: \: \: \: \: \: \: \: \:$