## Task B-1.3.

In how many ways can the number 455 be written as the sum of an increasing sequence of two or more consecutive natural numbers?
## Solution.

Let $n$ be the first number in a sequence of $k+1$ consecutive numbers, $k \geqslant 1$. Then,

$$
\begin{aligned}
455=n+(n+1)+(n+2)+\cdots+(n+k) & =(k+1) n+(1+2+3+\cdots+k) \\
& =(k+1) n+\frac{k(k+1)}{2}=\frac{(k+1)(2 n+k)}{2}
\end{aligned}
$$

which implies $910=(2 n+k)(k+1)$.

The prime factorization of 910 is $910=2 \cdot 5 \cdot 7 \cdot 13$.

Notice that the numbers $(2 n+k)$ and $(k+1)$ have different parities, meaning that one of the parentheses must be equal to an even divisor of 910, which are the numbers 2, 10, 14, 26, 70, 130, 182.

Given that $2 n+k > k+1$, we get 7 linear systems whose solutions yield the numbers $n$ and $k$:

$$
\begin{aligned}
& \begin{array}{lllllll}
k+1=2 & k+1=10 & k+1=14 & k+1=26 & k+1=13 & k+1=7 & k+1=5
\end{array} \\
& 2 n+k=455 \quad 2 n+k=91 \quad 2 n+k=65 \quad 2 n+k=35 \quad 2 n+k=70 \quad 2 n+k=130 \quad 2 n+k=182
\end{aligned}
$$

Each of these systems has a unique solution, so there are a total of 7 ways to represent the number 455 as the desired sum.