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)"
5 mod 2 is a mathematical expression which is the REMAINDER of dividing 5 by 2 (i.e 1).
Mod is the abbreviation of Modulus or Modulo.
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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.
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.
Evaluate 2 x 4 x 2 in the mod 5 system?
evaluate 2x4x2 in the mod 5 system
What is the probability of getting exactly 5 boys and 5 girls if you have 10 kids?
Here the sample space is(s)=10, =>mod(s)=10 a=be the event of getting exactly 5 boys =>mod(a)=5 b=be the event of getting exactly 5 girls =>mod(b)=5 thus, p(a)=mod(a)/mod(s)=5/10=1/2 p(b)=mod(b)/mod(s)=5/10=1/2 p(5 boys and 5 girls)=p(a)*p(b)=1/2*1/2=1/4
What is 2 x 4 x 2 in the mod 5 system?
8
What are the release dates for The Mod Squad - 1968 To Linc - With Love 2-5?
The Mod Squad - 1968 To Linc - With Love 2-5 was released on: USA: 21 October 1969
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.
PROVE THAT nnn-n is divisible by 6?
A number is divisible by 6 if it is divisible by 2 and 3. Look at 333-3 which is 330 The sum of the digits is 6 and it is even so it is divisible by 6 Now consider 222-2 which I picked because unlike 333, 222 has even digits. 222-2=220, one again even number so divisible by 2 but NOT divisible by 3 so NOT divisible by 6 So it look like this is not true for all n For any odd n, we have the following 1. nnn-n ends in 0 so it is even if we can show it is divisible by 3 we are done. but 777-7 is 770 which is NOT divisible by 3 so it is NOT true. For some n it is true, but not for all n... Now when will nnn-n be divisible by 3. only when n+n is a multiple of 3, ie n=33,66, 99 an that is it! So we could easily prove that nnn-n is divisible by 6 if and only if n=3,6,or 9 ----------------------------- If by nnn, you mean n3, a proof is as follows: n=0,1,2,3,4, or 5 (mod 6) If n=0 (mod 6), we have (0 (mod 6))((0(mod 6))2-1)=0 (mod 6). [Since the first term is zero] If n=1 (mod 6), we have (1 (mod 6))((1(mod 6))2-1)=0 (mod 6) [Since 1-1=0]. If n=2 (mod 6), we have (2 (mod 6))((2(mod 6))2-1)=(2*3) (mod 6) = 6 (mod 6)=0 (mod 6). If n=3 (mod 6), we have (3 (mod 6))((3(mod 6))2-1)=(3*8) (mod 6) = 24 (mod 6) = 0 (mod 6). If n=4 (mod 6), we have (4 (mod 6))((4(mod 6))2-1)=(4*15) (mod 6) = 60 (mod 6) = 0 (mod 6). If n=5 (mod 6), we have (5 (mod 6))((5(mod 6))2-1)=(5*24) (mod 6) = 120 (mod 6) = 0 (mod 6). If you're not comfortable with the modular arethmetic, you can substitue 6m+_, where the blank is each of the numbers 0 through 5 (since every number can be expressed either as a multiple of six, or as a multiple of six plus some number between 1 and 5 --the remainder when the number is divided by six). Taking our example with 5, you would get: (n)(n2-1) can be written as (6m+5)((6m+5)2-1), where m is an integer. Simplifying this, you get: (6m+5)((6m+5)2-1) (6m+5)((6m2+60m+25-1) 6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*25+5(-1) 6m*6m2+6m*60m+6m*25-6m+5*6m2+5*60m+5*24 Since m is an integer and each term is divisible by 6, (n)(n2-1) is divisible by 6 for integers that can be expressed as 6m+5. You would then repeat the process for each of 0 through 4 to complete the proof. Clearly, if you are comfortable with it, modular arithmetic is the less cumbersome way to proceed.
What is the remainder if 97 is divided by 5?
19.4
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.
How do you get the 'mod' of a number?
"mod" is short for "modulo", and indicates the remainder after division of the first number by the second. For example, 11 mod 2 = 1 (11 / 2 has an integer quotient of 5, with 1 left over).
What are the release dates for The Mod Squad - 1968 Corbey 5-8?
The Mod Squad - 1968 Corbey 5-8 was released on: USA: 2 November 1972
Is the minecraft weeping angles a mod?
Yes it is Weaping Angles is a mod. Heres how to install. 1: Install The Mod You Wanted 2: Install Modloader Or Mcpatcher I would Prefer Mcpatcher 3: Click Mods And Add+ Go to Downloads Find your mod Click ok 4:Patch the Mod By Pressing "Patch" 2: Go on Minecraft 5: Enjoy Your Mod!
