15. The highest common factors of all the pairs chosen from the positive integers $Q, R$ and $S$ are three different primes.
What is the smallest possible value of $Q+R+S$ ?
A 41
B 31
C 30
D 21
E 10
Solution
$\mathbf{B}$
We use the notation $\operatorname{HCF}(X, Y)$ for the highest common factor of the two integers $X$ and $Y$. Suppose that $\operatorname{HCF}(Q, R)=a, \operatorname{HCF}(Q, S)=b$ and $\operatorname{HCF}(R, S)=c$, where $a, b$ and $c$ are three different primes.
It follows that both $a$ and $b$ are factors of $Q$. Therefore the smallest possible value of $Q$ is $a b$. Likewise, the smallest possible values of $R$ and $S$ are $a c$ and $b c$, respectively.
We seek the smallest possible value of $Q+R+S$, that is, of $a b+a c+b c$, where $a, b$ and $c$ are different primes. To do this we choose the values of $a, b$ and $c$ to be the three smallest primes, that is 2,3 and 5 , in some order.
Because $a b+a c+b c$ is symmetric in $a, b$ and $c$, the order does not matter. With $a=2, b=3$ and $c=5$, we have
$$
a b+a c+b c=2 \times 3+2 \times 5+3 \times 5=6+10+15=31 .
$$

We deduce that the smallest possible value of $Q+R+S$ is 31 .