SOLUTION: Could you please help me with this problem. Thanks in advance. What is the remainder when 3 to the power 777777 is divided by 7.

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Question 111888:
Could you please help me with this problem. Thanks in advance.

What is the remainder when 3 to the power 777777 is divided by 7.

Found 2 solutions by solver91311, scott8148:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
If you raise 3 to successive powers starting with 1, and then perform integer division on the result by 7, you get a repeating pattern of remainders that repeats every 6th time.

Use the modulo function which returns the remainder for an integer division:

3mod7=+++++++3
9mod7=%092
27mod7=%096
81mod7=%094
243mod7=%095
729mod7=%091
2187mod7=%093
6561mod7=%092
19683mod7=%096
59049mod7=%094
177147mod7=%095
16777216mod7=1

Now, 777777 mod 6 is 3 (mod 6 because it repeats every sixth time), so 3%5E%28777777%29+mod+7 should yield the third item in the pattern, namely 6

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
3^0 ___ 1 _____ R/7=1
3^1 ___ 3 _____ R/7=3
3^2 ___ 9 _____ R/7=2
3^3 ___ 27 _____ R/7=6
3^4 ___ 81 _____ R/7=4
3^5 ___ 243 _____ R/7=5
3^6 ___ 729 _____ R/7=1
3^7 ___ 2187 _____ R/7=3
3^8 ___ 6561 _____ R/7=2
3^9 ___ 19683 _____ R/7=6
3^10 ___ 59049 _____ R/7=4
3^11 ___ 177147 _____ R/7=5

the remainder pattern repeats every six powers

777777/6=129629 R3; so the remainder for (3^777777)/7 is 6