if n is a positive integer,how many of the ten digits from 0

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if n is a positive integer,how many of the ten digits from 0 through 9 could be the units digit of n^3?

如果N是正整数,当3^(8n+3) + 2 被5整除的余数是多少?
8n+3 when divided by 4 gives a remainder of 3.
Since powers of 3 make cycles of 4 i.e.3,9,27,81,..3,..9,..7,..1,..3
so 38n+3 will have 7 at the units place.
when 2 is added we get 9 at the units place.
Thus when divided by 5 we get 4 as remainder.
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