Find the remainder when 2^1990 is divided by 1990.?

Question

Find the remainder when 2^1990 is divided by 1990.?

Answer 1

1990 = 10*199 = 2* 5* 199

prime factorisation

2^4 mod 5 = 1 as per fermat theore

so 2^1990 mod 5 = 2^2 mod 5 = 4 mod 5

2^ 198 = 1 mod 199

so 2^1990 mod 199 = 2^10 mod 199 = 1024 mod 199 = 29 mod 199

2^1990 mod 2 = 0

so 2^ 1990 mod 199 = 29

2^1990 mod 5 = 4

using chineese remainder theorem you can proceed

continuing further

2^ 1990 = 0 mod 2

2^ 1990 = 4 mod 5

this gives 2^1990 = 4 mod 10  

now using

a = 4  (mod 10)

a = 29 (mod 199)

Notice that 20*10+(-1)*199=1, thus 20*10≡1 (mod 199) and -199≡1 (mod 10)

let a= 29*(20•10)+4*(-199), then it is clear that

a = 29*(0)+4*(1)=4 (mod 10) and a=29•(1)+4•(0)=29 (mod 199) so this a works for what we want.

a= 5004=1024 (mod 1990)

so the remainder is 1024

Answer 2

I think it is zero.

Let

a / b = q

a = dividend

b = divisor

q = quotient

Ex.

1)  4/2 = 2 remainder 0

2) 5/3 = 1 remainder 2

We cay say that

dividend = quotient x divisor + remainder

For 4/2

dividend = 4

divisor = 2

quotient = 2

remainder = 0

4 = 2 x 2 + 0

4 = 4 + 0

4 = 4

For 5/3

dividend = 5

divisor = 3

quotient = 1

remainder = 2

5 =  1 x 3 + 2

5 = 3 + 2

5 = 5

============

For our problem

2^1990 / 1990

dividend = 2^1990

divisor = 1990

quotient = 2^1990/1990 (without going through the actual computation)

remainder = r (unknown)

Dividend = quotient x divisor + remainder

2^1990 = 2^1990/1990 x 1990 + r

2^1990 = 2^1990 + r

r = 2^1990 - 2^1990

r = 0

Can we say then

that

The remainder of x^n/n  = 0?

I think so.

Answer 3

I have forgotten most of the modular arithmetic I have learned but i can tell you that x^n / n does not necessarily have a remainder of 0

eg

let x = 2 and n = 3

2^3 / 3 = 8/3 = 2 remainder 2