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21990 divided by 1990 is: 11 with a remainder of 100

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Q: Find the remainder when 21990 is divided by 1990?
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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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To find the remainder you can do the 'chunking' method. This would leave you with a remainder of 2 because 4x9=36 38-36=2


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The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.


What number divided by 7 get a remainder of 4?

A lot of numbers divided by seven get a remainder of four. Well, eleven, eighteen, twenty-five, and thirty-two are some. This is a very broad question that can be answered many ways. The way to get the remainder of four is to find a multiple of seven and add four then you divide and the remainder is four.

Related questions

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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35


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How do you divide unknown polynomials?

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What is the remainder of 38 divided by 4?

To find the remainder you can do the 'chunking' method. This would leave you with a remainder of 2 because 4x9=36 38-36=2


How do you find number is even or odd?

An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.


What is the remainder of 53 divded by 9?

The remainder is 8. To find the remainder when a number is divided by 9, add the digits together; if this sum has more than 1 digit, repeat until one digit remains. This digit is the remainder, unless it is 9 in which case the remainder is 0. examples: remainder 53 → 5 + 3 = 8 → remainder is 8 when 53 is divided by 9. remainder 126 → 1 + 2 + 6 = 9 → remainder is 0, ie no remainder, 126 is divisible by 9. remainder 258 → 2 + 5 + 8 = 15 → 1 + 5 = 6 → remainder is 6 when 258 is divided by 9.