28. It is given that $a, b, c$ and $d$ are four positive prime numbers such that the product of these four prime numbers is equal to the sum of 55 consecutive positive integers. Find the smallest possible value of $a+b+c+d$. (Remark: The four numbers $a, b, c, d$ are not necessarily distinct.)
28. Answer: 28
The sum of 55 positive consecutive integers is at least $(55 \times 56) / 2=1540$. Let the middle number of these consecutive positive integers be $x$. Then the product $a b c d=55 x=5 \cdot 11 \cdot x$. So we have $55 x \geq 1540$ and thus $x \geq 28$. The least value of $a+b+c+d$ is attained when $x=5(7)$. Thus the answer is $5+11+5+7=28$.