Find all sequence of consecutive positive numbers which the sum of them is equal with $2019$.
Problem
Source: Kosovo MO 2019 Grade 9, Problem 4
Tags: number theory
Mathsy123
02.03.2019 23:15
Solution:
Let $x_1+x_2+.....+x_n=2019$. Where $x_i$ and $x_j$ are number consecutive if $j-i=1$
So $nx_1+ n(n-1)/2=2019$
If $n$ is even. Let $n=2k$
$2kx_1+k(2k-1)=2019$
$k(2x_1+2k-1)=2019=3*673$ So $k=1,3,673,2019$
In Total: $(x_1,n)=(1009,2),(334,6)$
If $n$ is odd. Let $n=2k+1$
$(2k+1)(x_1+k)=2019=3*673$ So $2k+1=1,3,673,2019$ So $k=0,1,336,1009$
In Total: $(x_1,n)=(2019,1),(336,3)$
Edit: Say number positive
Rijadinho
28.10.2023 11:16
$a+(a+1)+(a+2)+...+(a+n-1)=2019$ $\frac{n(a+a+n-1)}{2} =2019$ $n(2a+n-1)=4038=2 \cdot 3 \cdot 673$ If $n$ is even, then $2a+n-1$ is odd, and if $n$ is odd, then $2a+n-1$ is even. $n=2 \Rightarrow 2a+n-1=2019 \Rightarrow a=1009 \Rightarrow (n,a)=(2,1009)$ $n=3 \Rightarrow 2a+n-1=1346 \Rightarrow a=672 \Rightarrow (n,a)=(3,672)$ $n=6 \Rightarrow 2a+n-1=673 \Rightarrow a=334 \Rightarrow (n,a)=(6,334)$ In the cases where $n>a$, we get that $a$ is a negative integer, which is contradictory, therefore: $(n,a)=(2,1009),(3,672),(6,334)$ $\blacksquare$