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Why does -24 mod 7 equal 4 but 24 mod 7 equal 3?
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
What does 3 mod 0 equal?
The "mod" operator is the remainder from a division. Since you can't divide by zero, "mod zero" doesn't make sense, either (i.e., it is undefined).
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 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.
How many thirds are equal to six sixths?
There are: 3/3 = 6/6 because they both are equal to one
Why does -24 mod 7 equal 4 but 24 mod 7 equal 3?
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
What will be assigned to the Integer answer variable when the answer equals 45 Mod 6 statement is processed?
The value assigned to the Integer answer variable will be 3. This is because 45 modulo 6 equals 3, so the value of the statement 45 Mod 6 is 3.
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 does 3 mod 0 equal?
The "mod" operator is the remainder from a division. Since you can't divide by zero, "mod zero" doesn't make sense, either (i.e., it is undefined).
What are the release dates for The Mod Squad - 1968 Search and Destroy 3-6?
The Mod Squad - 1968 Search and Destroy 3-6 was released on: USA: 27 October 1970
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 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 are the release dates for 24 Hour Design - 2006 Mod Foyer 3-6?
24 Hour Design - 2006 Mod Foyer 3-6 was released on: USA: 21 August 2007
Is the ratio 6 to 3 the same as 3 to 6?
No. A ratio can be expressed as a fraction. 6 to 3 can be written as 6/3 which is equal to 2. 3 to 6 can be written as 3/6, which is equal to 1/2.
Why does 3 over 4 equal 6?
3 over 4 isn't equal to 6.
If pi is equal to 6 then what is 3?
Pi is not equal to 6. It is equal to 3.14159...
What does -6 over-3 equal?
-3
