??anyone????

Hint

just use:$a\neq b$ then $gcd(a,b)<|a-b|$

nmd27082001 wrote: just use:$a\neq b$ then $gcd(a,b)<|a-b|$ Can you write full solution?

it has already been posted : https://artofproblemsolving.com/community/c6t177f6h1772446_a_number_theory_problem

$\gcd(a,b)\leq |a-b|$ my beloved The answer is $3$ only which clearly works. We now prove that we must have $N=3$. Since $\gcd(a,b) \leq b-a$ for all $b>a$, we have $$\gcd(d_1,d_2)+\cdots+\gcd(d_{s-1},d_s)\leq (d_2-d_1)+\cdots+(d_s-d_{s-1})=d_s-d_1=N-1.$$Thus for the sum of the written numbers is $N-2$, we should have equality everywhere, except for a single application of the inequality which is "off by one", i.e. $\gcd(d_i,d_{i+1})=d_{i+1}-d_i$ everywhere except for one index, where $\gcd(d_i,d_{i+1})=d_{i+1}-d_i-1$. In this case we should have $d_{i+1}-d_i-1 \mid d_{i+1}-d_i \implies d_{i+1}-d_i=2$, and $d_i,d_{i+1}$ are odd. Thus there are two "consecutive" factors of $N$ which can be written as $2n-1$ and $2n+1$. On the other hand, if $N \neq (2n-1)(2n+1)$, then the factors $\frac{N}{2n+1}$ and $\frac{N}{2n-1}$ are also consecutive and different from our original pair. But we have $$\gcd\left(\frac{N}{2n+1},\frac{N}{2n-1}\right)=\frac{N}{(2n-1)(2n+1)},$$which becomes clear if we multiply everything by $(2n-1)(2n+1)$, while $$\frac{N}{2n-1}-\frac{N}{2n+1}=\frac{2N}{(2n-1)(2n+1)},$$so equality does not hold here either. Thus we must have $N=(2n-1)(2n+1)$, which is odd. On the other hand, if $2n-1 \neq 1$, then $1$ and $p$ are consecutive and different from $2n-1$ and $2n+1$, where $2 \nmid p$ is the least prime divisor of $N$, and $\gcd(1,p)=1 \neq p-1$, so equality does not hold there either. Thus we are forced to have $2n-1=1$, which yields $N=1\cdot 3=3$ as desired. $\blacksquare$

Quite nice. The answer is $n=3$ only, which works. Notice that because $\gcd(d_1, d_2) \leq |d_1-d_2|$, we must have $\gcd(d_1, d_2) = d_1 - d_2$ except in one instance where $d_1, d_2$ are both odd and $d_1-d_2 = 2$. Claim. $2 \mid n$ unless $n=3$. Proof. Suppose otherwise. Then the smallest prime $p \mid n$ must be $p=3$; however, note that there are no valid possibilities for the next smallest divisor (as we have ``lost" our margin of $1$ already), hence contradiction. $\blacksquare$ Now set $(d_1, d_2) = (2k-1, 2k+1)$ and $N = c(2k-1)(2k+1)$. Notice that $$\frac N{2k-1} - \frac N{2k+1} = \frac{2N}{(2k+1)(2k-1)} = 2c.$$So $2c \mid c(2k + 1)$ which is an obvious contradiction as $2k+1$ is odd. Hence $c=1$. But then by the first claim, $n$ must be odd, contradiction.