What is the remainder when 2 rest to 1990 divided by 1990?

Answer 1

4. Find the remainder when 21990 is divided by 1990.

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(ap, 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.