Problem
ALG-B2-M01-P027 Partition into two groups
There are \(101\) nonzero integers. For each number, the sum of this number and the product of all the other numbers is negative. Prove that for any partition of the numbers into two nonempty groups, the sum of the products of the numbers in the two groups is negative.
Hint 1. First determine the parity of the number of negative elements.
Hint 2. In the negative product group, isolate one negative factor.
If the number of negative entries were even, there would be a positive number \(a\), and the product of all the other numbers would be positive. Then \(a+\prod_{\ne a}>0\), a contradiction. Hence the number of negative entries is odd.
Partition the numbers into two groups. Exactly one of the two products is negative. Let \(X=x_1x_2\cdots x_r<0\), \(Y=y_1y_2\cdots y_s>0\), and choose \(x_1<0\). Then \(x_2\cdots x_r>0\), and since the numbers are integers, \(x_2\cdots x_r\ge1\). Therefore \[X+Y=x_1(x_2\cdots x_r)+Y\le x_1+Y\le x_1+Y(x_2\cdots x_r).\] The last expression is negative by the condition applied to \(x_1\). Hence \(X+Y<0\).
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: two key steps are sign parity and using integrality to bound a product.
G. Why this is not a one-step exercise. Knowing the sign of the total product is not enough; every partition must be bounded.