Problem
ALG-B2-M01-P024 Integer values of products
Let \(x_1
Hint 1. First check \(p_1\).
Hint 2. For a construction, try \(x_i=i+1\).
The number \(p_1=x_1-\frac1{x_1}\) is not a positive integer: if \(x_1=1\), it is \(0\), and if \(x_1>1\), it is not an integer. Hence at most \(11\) of the \(p_i\) can be positive integers.
We show that \(11\) is attainable. Take \(x_i=i+1\). Then \[x_i-\frac1{x_i}=i+1-\frac1{i+1}=\frac{i(i+2)}{i+1}.\] Therefore \[p_i=\frac{1\cdot3}{2}\cdot\frac{2\cdot4}{3}\cdots\frac{i(i+2)}{i+1}=\frac{i!(i+2)}2.\] For \(i\ge2\), this is a positive integer. Thus \(11\) of \(p_1,\ldots,p_{12}\) can be positive integers, and more is impossible.
A. Source analysis. Main objects: inequalities, order, an extremal element, or an invariant. The obvious first move usually gives only a local estimate. The hidden observation is to choose the right nondecreasing quantity, or to add/multiply inequalities only after signs are controlled. The needed step is an ordering, an invariant, a product transformation, or a boundary case.
F. Difficulty justification. Regional level 7: both an upper bound and a construction are needed.
G. Why this is not a one-step exercise. The answer does not follow from an example alone; one must prove \(p_1\) never works.