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What term in (7^2)+2 progression isn't divisible by 3?
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7 is congruent to 1 mod 3. Therefore all powers of 7 are congruent to 1 mod 3, meaning that all numbers of the form 7^n +2 are dibisible by 3.
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You could also do a proof by induction:

Basecase, n=0, 7^0 + 2 = 3, clearly divisible by 3.

Assume 7^n + 2 is divisible by 3, to show that 7^(n+1) + 2 is too.

7^(n+1) + 2 = 7β‹…7^n + 2 = 6β‹…7^n + 7^n + 2.

As we already assumed, 7^n + 2 is divisible by 3, and 6β‹…7^n is clearly divisible by 3 as well. Thus 7^n + 2 is divisible by three for all natural numbers n.
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You mean 7^n +2, right?

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