Saudi Arabia booklet 2024

It is easy to check that for $n=2,4,10$ the sequences are $(1),(1,3),(1,3,7,9)$ respectively. Thus these numbers are solution of the given problem. Considering $n\ge 3$, if $n$ is odd then $\text{gcd}(n,2)=1$ so clearly $a_1=1,a_2=2$ and their sum is divisible by $3$, which not satisfy. Hence, $n$ is even. It is easy to check that $n=6$ or $n=8$ also do not work. Now we consider $n\ge 12$ and separate in the following cases:

1. If $n=12k$ for $k\ge 1$ then consider $a=6k+1,b=6k-1$. Since $a$ is odd then $\text{gcd}(n,a)\mid \text{gcd}(n,2a)=\text{gcd}(12k,12k+2)=2$ so $\text{gcd}(n,a)=1$. Similarly, we have $\text{gcd}(n,b)=1$ so $a,b$ are two consecutive terms in the sequence of $n$, but $a+b=12k$ which is divisible by $3$. This case does not give any solution.

2. If $n=12k+2$ for $k\ge 1$ then consider $a=6k+3,b=6k+5$. Let denote $d=\text{gcd}(n,a)$ then $d|2a=12k+6$ so $d|4$, but $d$ is odd then $d=1$, thus $\text{gcd}(n,a)=1$. Similarly, denote $d^{\prime }=\text{gcd}(n,b)$ then $d^{\prime }|2b=12k+10$ so $d^{\prime }|8$, but $d^{\prime }$ is odd then $d^{\prime }=1$, thus $\text{gcd}(n,b)=1$. Now $a,b$ are consecutive terms in the sequence and $3\mid a+b$, not satisfies.

3. If $n=12k+4$ for $k\ge 1$ then consider $a=6k+1,b=6k-1$.

4. If $n=12k+6$ for $k\ge 1$ then consider $a=6k+1,b=6k-1$.

5. If $n=12k+8$ for $k\ge 1$ then consider $a=6k+3,b=6k+1$.

6. If $n=12k+10$ for $k\ge 1$ then consider $a=6k+1,b=6k-1$.