67 to d power of 101 is divided by 65, wht is d remainder?

Question

how do v solve it?? i have no idea

Answer 1

Where phi is Euler's phi-function,

lambda(65) = lcm(phi(5), phi(13)) = lcm(4, 12) = 12.

therefore a^12 = 65n + 1 for every a which has no common factor with 65. Try it and see!

By throwing away multiples of 65 and of 12 where we now see that they make no difference,

67^101 = 2^101 = 2^5 = 32.

Answer 2

we know 67 mod 65 = 2
67^101 mod 65 = 2^101 mod 65

now 2^6 mod 65 = 64 mod 65 = -1

2^96 mod 65 = (2^6)^16 mod 65 = 1

2^101 mod 65 = 2^5 mod 65 = 32 mod 65 = 32

so 67^101 mod 65 = 32

remainder = 32

Answer 3

i'm not sure what d(x) is. Do you advise the spinoff of f(x)? if so, then particular, this is accessible to divide f(x) by its spinoff to get a the remainder of 0, for particular purposes (which includes f(x) = e^x).