Enter Modular Exponentiation
Solve 35 mod 11 using:
Modular exponentiation
Build an algorithm:
n is our exponent = 5
y = 1 and u ≡ 3 mod 11 = 3
See here
n = 5 is odd
Since 5 is odd, calculate (y)(u) mod p
(y)(u) mod p = (1)(3) mod 11
(y)(u) mod p = 3 mod 11
3 mod 11 = 3
Reset y to this value
Determine u2 mod p
u2 mod p = 32 mod 11
u2 mod p = 9 mod 11
9 mod 11 = 9
Reset u to this value
Cut n in half and take the integer
5 ÷ 2 = 2
n = 2 is even
Since 2 is even, we keep y = 3
Determine u2 mod p
u2 mod p = 92 mod 11
u2 mod p = 81 mod 11
81 mod 11 = 4
Reset u to this value
Cut n in half and take the integer
2 ÷ 2 = 1
n = 1 is odd
Since 1 is odd, calculate (y)(u) mod p
(y)(u) mod p = (3)(4) mod 11
(y)(u) mod p = 12 mod 11
12 mod 11 = 1
Reset y to this value
Determine u2 mod p
u2 mod p = 42 mod 11
u2 mod p = 16 mod 11
16 mod 11 = 5
Reset u to this value
Cut n in half and take the integer
1 ÷ 2 = 0
Because n = 0, we stop
We have our answer
Final Answer
35 mod 11 ≡ 1
