What is 32 to the 14th power, mod 7?

Answer 1

32^14 mod 7= (-3)^14 mod 7 = 4,782,969 mod 7 = 2 mod 7.

Answer 2

Above answer may get to the same answer as mine, but with no explanation of how to get there.

2^1 mod 7 = 2 mod 7 = 2
2^2 mod 7 = 4 mod 7 = 4
2^3 mod 7 = 8 mod 7 = 1
2^4 mod 7 = 16 mod 7 = 2
2^5 mod 7 = 32 mod 7 = 4
2^6 mod 7 = 64 mod 7 = 1
2^7 mod 7 = 128 mod 7 = 2
2^8 mod 7 = 256 mod 7 = 4
2^9 mod 7 = 512 mod 7 = 1
2^10 mod 7 = 1024 mod 7 = 2
2^11 mod 7 = 2048 mod 7 = 4
2^12 mod 7 = 4096 mod 7 = 1

etc etc etc. I think you can see the pattern.
32^14 = (2^5)^14 = 2^70

If the exponent is divisible evenly by 3 you get:
2^(3n) mod 7 = 1
2^(3n+1) mod 7 = 2
2^(3n+2) mod 7 = 4

Since 70 = 3*23 + 1
Therefore 2^70 mod 7 = 2

Answer 3

32 ^ 14 = (2^5)^14 = 2^70

now because 7 is prime
using format's little theorem

2^6 mod 7 = 1
so 2^66 mod 7 = 1
2^70 mod 7 = 2^ 4 mod 7 = 16 mod 7 = 2

that is the ans

Answer 4

By Fermat's little theorem,
32^14 = 4^14 = (4^7)² = 4^2 = 2(mod 7).