6) At a conference, 30 scientists participate, each of whom is a mathematician, a physicist, a chemist, or a biologist. The physicists and biologists together are half the number of mathematicians; the physicists and chemists together are twice the number of biologists. Moreover, there is at least one physicist. How many mathematicians are there?
(A) 6
(B) 10
(C) 12
(D) 15
(E) 18
6. The answer is $(\mathbf{E})$.

Let $M, F, C$ and $B$ denote the number of mathematicians, physicists, chemists, and biologists present at the conference, respectively. We know that

$$
\left\{\begin{array}{l}
M+F+C+B=30 \\
F+B=\frac{M}{2} \\
F+C=2 B \\
F \geq 1
\end{array}\right.
$$

From the second equation, it follows that $M$ must be even, and from the fourth equation, it follows that $M>0$. From the first and second equations, we obtain

$$
3 M+2 C=60
$$

and by subtracting the second equation from the third, we find

$$
C-B=2 B-\frac{M}{2} \Rightarrow 2 C+M=6 B
$$

Subtracting the last relation found from (1) yields

$$
M=30-3 B
$$

which tells us that $M$ must be divisible by 3. Therefore, $M$ is positive and divisible by 6. On the other hand, $M$ must be less than 30, so $M$ can only be one of the values: $6,12,18,24$. For $M=6$, we find $B=8$ and $F+B=F+8=M / 2=3$ which is impossible, as it implies $F<0$. For $M=12$, we find $B=6$ and $F+B=F+6=M / 2=6$ which is equally impossible because it implies $F=0$. If $M=24$, from (1) it follows that $C<0$ and thus the value 24 must also be discarded. The only possibility is $M=18$, which is consistent with the data, and indeed leads to: $B=4, F=5$, $C=3$.

[Problem proposed by S. Mongodi]