# Progressions| Existence of 1980 in the given progression

Given three infinite arithmetic progressions of natural numbers such that each of the numbers 1,2,3,4,5,6,7 and 8 belongs to at least one of them, prove that the number 1980 also belongs to at least one of them.

I may begin saying that 2,3,4 belong to the 3 progressions indivudually.

edit: my working for above statement

let a1,a2,a3 be the 3 progressions now if a) all 3 belong to only one of them, then 1980 obviously exists in it b) if only two of 2,3,4 belong to one of them then also it becomes a series with common difference given so which means 1980 belongs to one of them

so without loss of generality, i can safely assume 2,3,4 to belong to each one of them indivudually i.e 2 to a1, 3 to a2 and 4 to a3. now how to proceed?

well i got this far due to a clue that was associated with the question.

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I see what you mean, but can you include a reasoning for this claim. In general, it is a nice idea to show your work so that others can see where you are stuck. (Also, please avoid posting questions in the imperative mode.) –  Srivatsan Sep 3 '11 at 14:08

So: we may assume no progression has difference 1. You show that 2,3,4 belong to three different progressions. Similarly 6,7,8 belong to three different progressions. How do these match up? (Each of 2,3,4 matches one of 6,7,8.) If 4,8 are in the same progression, then 1980 is also in it, so assume not. If 4,6 are in the same progression, then 1980 is also in it, so assume not. Therefore 4,7 are in the same progression. This has difference 3 (since we assume no progression has difference 1). Next: if 3,6 are in the same progression, then 1980 is also in it, so assume not. Therefore 3,8 are in the same progression. This has difference 5. But now 5 and 6 are both missing from these two progressions, so they are both in the third progression, impossible.

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