Question

What is the maximum value of k for which 2013 can be written as a sum of k consecutive positive integers?

Answer 1

Let us assume that the k consecutive integers whose sum is 2013 has starting term as 'i'.

Sum of k terms starting from 'i',Sk=k2(2i+(k−1)∗1)=2013Sk=k2(2i+(k−1)∗1)=2013

k(2i+k−1)=4026k(2i+k−1)=4026

k2+(2i−1)k−4026=0k2+(2i−1)k−4026=0


This is a quadratic equation that we need to solve.

We know that solving a quadratic equation ax2+bx+c=0ax2+bx+c=0 by factoring or middle term splitting requires to list down the factor pairs of a×ca×c, which is -4026.

Factors of 4026 are: 2, 3, 11, 61

Factor pairs of - 4026 that add up to give a positive coefficient of 'k' are:

-2 + 2013 = 2011

-3 + 1342 = 1339

-6 + 671 = 665

-11 + 366 = 355

-22 + 183 = 161

-33 + 122 = 89

-61 + 66 = 5


Now, we also know that the coefficient of 'k' is 2i -1.

That means if we wish to have integer roots of the equation
k2+(2i−1)k−4026=0k2+(2i−1)k−4026=0, the coefficient of 'k' must be the sum of factor pairs listed above.

Also, we need to find the largest value of k.

That means we need to select such a factor pair that contains the largest negative value and it is -66 + 61 = 5

That means 2i - 1 = 5 or i = 3

Therefore, the quadratic equation becomes:

k2+5k−4026=0k2+5k−4026=0

(k−61)(k+66)=0(k−61)(k+66)=0

k = 61 or k = -66

So, k = 61 (Rejecting the negative value)

Thus, 2013 = 5 + 6 + 7 + ........ + 63 , a sum of 61 consecutive integers.