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This confuses me, please help me out:what is the remainder when 2014^2015 is divided by 9?
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Answer. 4.
Solution
1.
2014 is not divisible by 3. (The divisibility by 3 rule is divisibility by 3 the sum of the digits of a number.
The sum 2+ 0 + 1 + 4 = 7 in not divisible by 3.
See the lesson Divisibility by 3 rule in this site).
Hence, is not divisible by 9.
2.
The remainder of division 2014 by 9 is 7 (You can check it yourself by making long division.
By the way, it coincides with the number 7 above, and it is not occasional).
The remainder of division by 9 is the same as the remainder of division = 49 by 9 and is 4.
(I refer to elementary number theory, but you can check it yourself).
The remainder of division by 9 is the same as the remainder of division = 343 by 9 and is 1.
(Again, I refer to elementary number theory, but you can check it yourself).
After that the remainders of division by 9 will repeat again and again in the loop 7, 4, 1, 7, 4, 1, . . .
periodically/cyclically with the cycle length 3.
So, the sequence of remainders of divisions by 9 is
7, 4, 1, 7, 4, 1, . . . for n = 1, 2, 3, 4, 5, 6 . . . , (1)
i.e. is cyclical with the cycle length 3.
Thus when you get , the remainder will be WHAT?
Divide 2015 by 3 and you will get the remainder 2.
So, the number of interest is the second number in the sequence of remainders (1), i.e. 4.