0
398
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Find the remainder when 21990 is divided by 1990?
11.0503
25 divided by 5?
5
What is K divided by 5?
k divided by 5
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.
How do you work out 41 divided by 5?
8.2
What is the remainder left when 21990 is divided by 1990?
11.0503
What is 1990 divided by 24 with remainders?
82.9167
Find the remainder when 21990 is divided by 1990?
11.0503
What country was divided by Iraq in 1990?
Iraq divided no country in 1990. It occupied Kuwait in that year, but it did not divide it as the entire country was annexed.
What is 1990 divided by 24?
82.9167
What would -x divided by -5 be?
-x divided by -5 is x/5.
25 divided by 5?
5
What is the distributive property for 425 divided by 5?
85
What is K divided by 5?
k divided by 5
What is the multiplicative inverse of 5 divided by 8?
0.625
What is 5 divided by 1 and 1 over 4?
5 divided by 1 is 5. 1 over 4 is a quarter. 5 divided by 1¼ is 5 divided by 5/4, which is 5 time 4/5, that is 4.
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.
