Remainder when 2^31 is divided by 5
To find the remainder when 2^31 is divided by 5, we need to understand the concept of modular arithmetic. In modular arithmetic, we divide a number by a modulus and find the remainder. The remainder can only be from 0 to (modulus-1). For example, if we divide 7 by 3, the remainder is 1 because 7 divided by 3 is 2 with a remainder of 1.
Using Modular Exponentiation
We can use modular exponentiation to find the remainder when 2^31 is divided by 5. Modular exponentiation is a technique used to calculate large powers of a number modulo a modulus. We can use the following formula to calculate a^n mod m:
a^n mod m = (a^(n-1) mod m * a) mod m
Using this formula, we can calculate 2^31 mod 5 as follows:
2^31 mod 5 = (2^30 mod 5 * 2) mod 5
We can simplify this expression by calculating 2^30 mod 5 first:
2^30 mod 5 = (2^15 mod 5)^2 mod 5
Again, we can simplify this expression by calculating 2^15 mod 5:
2^15 mod 5 = (2^14 mod 5 * 2) mod 5
We can keep simplifying the expression in this manner until we get down to 2^1 mod 5, which is simply 2 mod 5:
2^1 mod 5 = 2
Using all the previous calculations, we can now calculate 2^31 mod 5 as follows:
2^31 mod 5 = (2^30 mod 5 * 2) mod 5
= ((2^15 mod 5)^2 mod 5 * 2) mod 5
= ((2^14 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5
= (((2^13 mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5
= (((2^12 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5
= (((2^11 mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5
= (((2^10 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5)^2 mod 5 * 2) mod 5
= (((2^9 mod 5)^2 mod 5 *