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what will be the remainder, when 2^1990is divided by 1990?(RMO 1990)

Posted by: eeshan banerjee in Olympiad{RMO+INMO+IMO} 6 years ago, Total Answer(s): 1

Answer(s)

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.

By: Sachin Kamath, 6 years ago

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