Question

2^1990 / 1990

Answer 1

Solution: Let N = 2¹⁹⁹⁰
Here, 1990 can be written as the product of two co-prime factors as 199 and 10.
Let R₁ ≡ MOD(2¹⁹⁹⁰, 199)
According the the Fermet’s Theorem, MOD(a^p, p) ≡ a .
∴ MOD(2¹⁹⁹, 199) ≡ 2.
∴ MOD((2¹⁹⁹)¹⁰, 199) ≡ MOD(2¹⁰, 199)
∴ MOD(2¹⁹⁹⁰, 199) ≡ MOD(1024, 199) ≡ 29 ≡ R₁.

Let R₂ ≡ MOD(2¹⁹⁹⁰, 10)

∴ R₂ ≡ 2 × MOD(2¹⁹⁸⁹, 5) Cancelling 2 from both sides.
Now, MOD(2¹⁹⁸⁹, 5) ≡ MOD(2 × 2¹⁹⁸⁸, 5) ≡ MOD(2, 5) × MOD((2²)⁹⁹⁴, 5)
Also MOD(4⁹⁹⁴, 5) ≡ (-1)⁹⁹⁴ = 1 & MOD(2, 5) ≡ 2
∴ MOD(2¹⁹⁸⁹, 5) ≡ 2 × 1
∴ R₂ ≡ 2 × 2 = 4.

∴ N leaves 29 as the remainder when divided by 199 and 4 as the remainder when divided by 10.
Let N₁ 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
∴ N₁ ≡ 199p + 29 = 10q + 4 (where, p and q are natural numbers)

∴	199p + 25	= q
10

Of course, the 5 is the least value of p at which the above equation is satisfied, Correspondingly, q = 102.
∴ N₁ = 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 = 2¹⁹⁹⁰ = 1024 + k × 1990
∴ MOD(2¹⁹⁹⁰, 1990) ≡ 1024.