Problem 2

Let

n

be positive and let

A_1,\ldots,A_{2n+1}

be subsets of

B

. Suppose each

A_i

has exactly

2n

elements, every two distinct

A_i

have exactly one common element, and every element of

B

belongs to at least two

A_i

. For which

n

can one label each element 0 or 1 so that each

A_i

has 0 on exactly

n

elements?

Step 3 of 5: Step 3

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n(2n+1)/2\in\mathbb Z\Longrightarrow n\text{ even}

Detailed analysis

If the labeling exists, zero incidences total n(2n+1). Every element contributes two incidences, so n(2n+1)/2 elements are labeled zero. Hence n is even.