Solution: Let N = 2^{1990}

Here, 1990 can be written as the product of two co-prime factors as 199 and 10.

Let R_{1} ≡ MOD(2^{1990}, 199)

According the the Fermet’s Theorem, MOD(a^{p}, p) ≡ a .

∴ MOD(2^{199}, 199) ≡ 2.

∴ MOD((2^{199})^{10}, 199) ≡ MOD(2^{10}, 199)

∴ MOD(2^{1990}, 199) ≡ MOD(1024, 199) ≡ 29 ≡ R_{1}.

Let R_{2} ≡ MOD(2^{1990}, 10)

∴ R_{2} ≡ 2 × MOD(2^{1989}, 5) Cancelling 2 from both sides.

Now, MOD(2^{1989}, 5) ≡ MOD(2 × 2^{1988}, 5) ≡ MOD(2, 5) × MOD((2^{2})^{994}, 5)

Also MOD(4^{994}, 5) ≡ (-1)^{994} = 1 & MOD(2, 5) ≡ 2

∴ MOD(2^{1989}, 5) ≡ 2 × 1

∴ R_{2} ≡ 2 × 2 = 4.

∴ N leaves 29 as the remainder when divided by 199 and 4 as the remainder when divided by 10.

Let N_{1} be the least such number which also follow these two properties i.e. leaves 29 as the remainder when divided by 199 and 4 as the remainder when divided by 10

∴ N_{1} ≡ 199p + 29 = 10q + 4 (where, p and q are natural numbers)

Of course, the 5 is the least value of p at which the above equation is satisfied, Correspondingly, q = 102.

∴ N_{1} = 1024.

∴ Family of the numbers which leaves 29 as the remainder when divided by 199 and 4 as the remainder when divided by 10 can be given by

f(k) = 1024 + k × LCM(199,10) = 1024 + k × 1990

N is also a member of the family.

∴ N = 2^{1990} = 1024 + k × 1990

∴ MOD(2^{1990}, 1990) ≡ 1024.

By: Sachin Kamath, 6 years ago