Divisibility chain?

Let's call your favorite integer $N$ .

$8$ divides $N$ and $4$ divides $8$ , so $4$ divides $N$ .

$9$ divides $N$ and $3$ divides $9$ , so $3$ divides $N$ .

Now $4$ and $3$ are coprimes, so their product $3*4 = 12$ divides $N$ .

11,13 are primes so no possibility to divide the number divisible by 6,7,8,9. To be divisible by 10 there should 0 at end which isnot possible. Therefore, 12 is only the answer.

---

Since the $\text{l.c.m}(6,7,8,9)$ is the product of all primes less than equal to 9 that appears in given integers .ie $\text{l.c.m} (6,8,7,9) =\text{l.c.m}(6×2,7,2^3,3^2) = 2^3 . 7. 3^2 = 12(7.3.2)$ Hence, 6,7,8,9 are always divisible by 12.

---

A number divisible by $6$ and $8$ is also divisible by its least common multiple $24$ , and since $24 = 2 \cdot 12$ , a number divisible by $24$ must also be divisible by $\boxed{12}$ .

Just take a look at the prime factorization of each number and the divisor....

6 · 7 · 8 · 9 = 3,024 and 3,024 / 12 = 252