Track / Stage 6 / 120 of 400 #1120 of 1964

Problem 1120

National Olympiad, first round

Number theory Difficulty 6.1 Prove it

Given sets $A, B, C, D$ with the property that any two of them, any three of them, and all four are disjoint, such that card $A=a^{5 n+3}, \operatorname{card} B=a^{5 n+2}, \operatorname{card} C=a^{5 n+1}, \operatorname{card} D=a^{5 n}, a, n \in N^{*}$ . Prove that card $(A \cup B \cup C \cup D)+4$ is an even number.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

## Solution and Grading Criteria:

Since $A \cap B=\emptyset, A \cap C=\emptyset, A \cap D=\emptyset, B \cap C=\emptyset, B \cap D=\emptyset, C \cap D=\emptyset, A \cap$ $B \cap C=\emptyset, A \cap B \cap D=\emptyset, A \cap C \cap D=\emptyset, B \cap C \cap D=\emptyset, A \cap B \cap C \cap D=$ $\emptyset$

It follows that: $\operatorname{card}(A \cup B \cup C \cup D)=\operatorname{card} A+\operatorname{card} B+\operatorname{card} C+\operatorname{card} D=a^{5 n+3}+a^{5 n+2}+$ $a^{5 n+1}+a^{5 n}=a^{5 n+2} \cdot(a+1)+a^{5 n} \cdot(a+1)=(a+1)\left(a^{5 n+2}+a^{5 n}\right)=(a+1) \cdot a^{5 n}$ . $\left(a^{2}+1\right)=a(a+1) \cdot a^{5 n-1} \cdot\left(a^{2}+1\right)=$ even, because $a(a+$

1) is an even number.

It follows that, $\operatorname{card}(A \cup B \cup C \cup D)+$ 4 is an even number.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.