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Question:

Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11, ..., and Y be the set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, .... Then, the number of elements in the set X∪Y is _______?

Solution:

X:1,6,11,_,10086Y:9,16,23,,14128X∩Y:16,51,86, _________Letm=n(X∩Y)∴16+(m𕒵)×35≤10086⇒m≤288.71⇒m=288∴n(X∪Y)=n(X)+n(Y)−n(X∩Y)=2018+2018𕒶88=3748.