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