Enter Modular Exponentiation
Solve 33 mod 7 using:
Modular exponentiation
Build an algorithm:
n is our exponent = 3
y = 1 and u ≡ 3 mod 7 = 3
See here
n = 3 is odd
Since 3 is odd, calculate (y)(u) mod p
(y)(u) mod p = (1)(3) mod 7
(y)(u) mod p = 3 mod 7
3 mod 7 = 3
Reset y to this value
Determine u2 mod p
u2 mod p = 32 mod 7
u2 mod p = 9 mod 7
9 mod 7 = 2
Reset u to this value
Cut n in half and take the integer
3 ÷ 2 = 1
n = 1 is odd
Since 1 is odd, calculate (y)(u) mod p
(y)(u) mod p = (3)(2) mod 7
(y)(u) mod p = 6 mod 7
6 mod 7 = 6
Reset y to this value
Determine u2 mod p
u2 mod p = 22 mod 7
u2 mod p = 4 mod 7
4 mod 7 = 4
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
33 mod 7 ≡ 6
What is the Answer?
33 mod 7 ≡ 6
How does the Modular Exponentiation and Successive Squaring Calculator work?
Free Modular Exponentiation and Successive Squaring Calculator - Solves xn mod p using the following methods:
* Modular Exponentiation
* Successive Squaring
This calculator has 1 input.
* Modular Exponentiation
* Successive Squaring
This calculator has 1 input.
What 1 formula is used for the Modular Exponentiation and Successive Squaring Calculator?
Successive Squaring I = number of digits in binary form of n. Run this many loops of a2 mod p
What 6 concepts are covered in the Modular Exponentiation and Successive Squaring Calculator?
- exponent
- The power to raise a number
- integer
- a whole number; a number that is not a fraction
...,-5,-4,-3,-2,-1,0,1,2,3,4,5,... - modular exponentiation
- the remainder when an integer b (the base) is raised to the power e (the exponent), and divided by a positive integer m (the modulus)
- modulus
- the remainder of a division, after one number is divided by another.
a mod b - remainder
- The portion of a division operation leftover after dividing two integers
- successive squaring
- an algorithm to compute in a finite field
