2015 Sum Sequences

Number Theory Level 2

We can represent the number $2014$ in various ways as the sum of positive consecutive integers:

$97+98+99+100+...+115=2014$

$502+503+504+505=2014$

The sum of consecutive integers with the highest number of terms that result in $2014$ is:

$12+13+14+15+16+17+...+61+62+63+64$

We need to add up $53$ consecutive numbers to get $2014$ .

What is the highest number of positive consecutive integers that we can add to get $2015$ .

62 60 103 2

1 solution

Nguyen Thanh Long
Dec 27, 2014

Suppose the sequence is: n, n+1, n+2, ..., n+m-1. So it sums to get the summary: $\frac{m^2}{2}+\frac{m(2n-1)}{2}-2015=0$ This equation has a positive root is: $\sqrt{\frac{(2n-1)^2}{4}+2 \times 2015}-\frac{2n-1}{2}$ By checking n=1,2,3,..., easy to get n=2. So $m=\boxed{62}$

2015 Sum Sequences

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