8.5. 20 numbers: $1,2, \ldots, 20$ were divided into two groups. It turned out that the sum of the numbers in the first group is equal to the product of the numbers in the second group. a) What is the smallest and b) what is the largest number of numbers that can be in the second group?
Answer. a) 3; b) 5. Solution. a) The sum of all numbers from 1 to 20 is 210. It is obvious that the second group contains more than one number, since otherwise it would be equal to the sum of the remaining 19 numbers, i.e., no less than \(1+2+\ldots+19=190\). We will show that in fact the second group contains more than two numbers. Indeed, otherwise for the numbers \(a\) and \(b\) from the second group, we would have the equality \(210-a-b=ab\), which can be rewritten as \(211=(a+1)(b+1)\). However, the last equation has no solutions in integers from 1 to 20, since the number 211 is prime. Now let's give an example of the second group consisting of three numbers. We will look for such an example in the simplest form, assuming that one of the sought numbers is 1. Then the remaining two numbers will be found from the equality \(209-a-b=ab\), which can be rewritten as \(210=(a+1)(b+1)\). The last equation gives two examples that fit the condition of the problem: \((1,9,20)\) and \((1,13,14)\). b) The number of numbers in the second group is less than six: indeed, otherwise their product would be no less than \(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6=720>210\). Let's try to find an example of five numbers in the form of the set \((1,2,3,4,x)\) with an unknown \(x\) (from 5 to 20). The number \(x\) must satisfy the equation \(210-10-x=24x\), i.e., \(25x=200\). We get the solution \(x=8\), so the set \((1,2,3,4,8)\) satisfies the condition of the problem.