For any integral value of n, $$3^{2n}$$ + 9n + 5 when divided by 3 will leave the remainder.
Expression = $$3^{2n}$$ + 9n + 5
= $$3^{2n}$$ + 9n + 3 + 2
Taking 3 common from each term, we get :
=> 3 ($$3^{2n-1}$$ + 3n + 1) + 2
Now, if we divide the above term by 3, remainder will be 2.
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