10. Four positive integers $a_{1}<a_{2}<a_{3}<a_{4}$ are such that, given any two of them, their greatest common divisor is greater than $1$, but $\operatorname{gcd}\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=1$. What is the minimum value that $a_{4}$ can take?
(A) 10
(B) 12
(C) 15
(D) 30
(E) 105.
10. The answer is $(\mathbf{C})$. First, there is at least one number $a_{k}$ that is not divisible by 2, otherwise 2 would divide $\operatorname{gcd}\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$. We want to prove that $a_{k}$ must have at least two distinct prime factors: indeed, if this were not the case, we would have $a_{k}=p^{s}$ for some prime $p$ and natural number $s$. But in this case, for every other number $h=1,2,3,4, h \neq k$, we have that $\operatorname{gcd}\left(a_{h}, a_{k}\right)$ can only be a number of the form $p^{r}$, with $1<r \leq s$ (because it must be a divisor of $p^{s}$, and it is greater than 1 by the first hypothesis). But this would imply that all four numbers are multiples of $p$, which is impossible by the second hypothesis. Therefore, there is at least one number $a_{k}$ that has at least two prime factors different from two, and thus it is at least $3 \cdot 5=15$. On the other hand, it is verified that the quadruple $6,10,12,15$ satisfies all the hypotheses of the problem, so 15 is indeed the minimum we are looking for.