Bariz - Problem

We must have $a$ be a prime, so $k+1$ is prime. Since $k+1\in\{2, 3, 4, 5, 6, 7\}$ we must have $a=2, 3, 5$ or $7$. But $a\neq 2$, $a\neq 3$ because the first and second divisors are less than $b$, not greater. Also $a\neq 5$ because there are at least 4 divisors of $n$ less than $(1+a+b)b$, namely $1, a, b, 1+a+b$. So $a=7$. Thus $c=1+a+b=8+b$ and $(b, c)=(b, 8)=1$. Hence there are at least two prime factors of $n$. Then $n=7p^2$ or $49p$ for some prime $p>7$. Then the prime factors in order are $1<7<p<7p<p^2<7p^2$ or $1<7<p, 49<7p<49p$. The first is impossible because $(b, c)=1$. So $\{b, c\}=\{49, p\}$. Thus $p=49\pm 8=41$ or $57$. $57$ is not prime, so $p=41$ and $n=49\cdot41=2009$.