0
2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2
that is 2 mod 5 = 2.
On spreadsheets, the formula is usually "=MOD(2,5)"
2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2
that is 2 mod 5 = 2.
On spreadsheets, the formula is usually "=MOD(2,5)"
2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2
that is 2 mod 5 = 2.
On spreadsheets, the formula is usually "=MOD(2,5)"
2 = 5*0 + 2 so [the quotient is 0 and] the remainder is 2
that is 2 mod 5 = 2.
On spreadsheets, the formula is usually "=MOD(2,5)"
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What is remainder when 5 to the power of 30 is divided by 7?
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
What is 2 x 4 x 2 in the mod 5 system?
8
What is the remainder when the product of 100 5's are divided by 7?
The remainder is 2. Knowing how to do modular arithmetic makes the problem much easier to solve. We say a number x has a value of k "mod" 7 if dividing x by 7 leaves a remainder of k. Multiples of 7 are equal to 0 mod 7 (they leave no remainder), and 32 is equal to 4 mod 7, because 32 / 7 = 4 remainder 4. The product of 100 5s is equal to 5 to the 100th power, or 5^100 in shorthand. Let us look at what the first few powers of 5 are equal to mod 7: 5^1 = 5 = 5 mod 7 5^2 = 25 = 4 mod 7 5^3 = 125 = 6 mod 7 5^4 = 625 = 2 mod 7 5^5 = 3125 = 3 mod 7 5^6 = 15625 = 1 mod 7 5^7 = 78125 = 5 mod 7 5^8 = 390625 = 4 mod 7 It looks as if we might have a repeating pattern, and indeed the next few powers of 5 are equal to 6, 2, and 3, confirming that 5^k = 5^(k-6) mod 7. This means that 5^100 has the same remainder after dividing by 7 as 5^94, which has the same remainder as 5^88, etc. etc. until we reach 5^4, which leaves a remainder of 2. Therefore by induction 5^100 leaves a remainder of 2. Hope this all makes sense.
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.
Can you find a square number that when divided by 3 has a remainder of 2?
No because it is impossible. Let mod(x.3) denote the remainder when x is divided by 3. Let n be any integer. Then mod(n,3) = 0,1 or 2. When mod(n,3) = 0, mod(n2,3) = 0 When mod(n,3) = 1, mod(n2,3) = 1 When mod(n,3) = 2, mod(n2,3) = 4 and, equivalently mod(n2,3) = 1. So, there are no integers whose squar leaves a remainder of 2 when divided by 3.
