1 mod 3 means the remainder after 1 has been divided by 3. Which is 1.
2 mod 3 = 2, 3 mod 3 = 0
There is a mod that removes blurs, i do not have it but type sims 3 nude mod or try mod the sims 3.
Garry's Mod 10. When people refer to Garry's Mod 11, they mean the up-to-date Garry's Mod 10
yes, but being a player mod doesnt mean u are a forum mod and vise verser. u need to be a mod in both different things
The Sims 3 censor blur can be removed by using a mod. A popular mod to use is the Mosaic-blur removal.
test how to mod dragon ball z budokai tenkaichi 3
If by "mod" you mean "modulo," then your question is meaningless, because "mod 1" is meaningless. For example, 18 mod 5 = 3, because you subtract the maximum number of multiples of 5 and the remainder is 3. But by definition any whole number modulo 1 would always be 0.
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.
This is very easy to prove using modulo arithmetic. Basically, what you do is to look at the remainder when a number (n) is divided by 3. Let k(mod 3) represent the remainder when a number is divided by 3. Since the divisor is 3, there are only 3 possible values for k, that is: n = 0(mod 3), 1(mod3) or 2(mod3). Suppose n = 0(mod 3) then n2 + 1 = 0 + 1(mod 3) = 1(mod 3) so that n2 + 1 leaves a remainder of 1 when divided by 3 and so is not divisible by 3. Suppose n = 1(mod 3) then n2 + 1 = 12 + 1(mod 3) = 2(mod 3) so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3. Suppose n = 2(mod 3) then n2 + 1 = 22 + 1(mod 3) = 5(mod 3) = 2(mod 3) so that n2 + 1 leaves a remainder of 2 when divided by 3 and so is not divisible by 3. Thus, for all possible values of n, division by 3 leaves a positive remainder. And so the result follows.
To find the units digit of 399 the question being asked is: What is (399) MOD 10? This does not necessitate evaluation of 399 before the modulus is done, as it can be done whenever it is possible during the multiplication as any multiple of 10 multiplied by 3 is still a multiple of 10. The first few powers of 3 modulus 10 are: 31 MOD 10 = 3 32 MOD 10 = (3 x 31) MOD 10 = (3 x 3) MOD 10 = 9 33 MOD 10 = (3 x 32) MOD 10 = (3 x 9) MOD 10 = 27 MOD 10 = 7 34 MOD 10 = (3 x 33) MOD 10 = (3 x 7) MOD 10 = 81 MOD 10 = 1 35 MOD 10 = (3 x 34) MOD 10 = (3 x 1) MOD 10 = 3 36 MOD 10 = (3 x 35) MOD 10 = (3 x 3) MOD 10 = 9 At this point, it can be seen that the answer is a repeating pattern of 3, 9, 7, 1, 3, 9, ... So we need the 99th element of this pattern. The pattern is a repeat of 4 digits, so we calculate 99 MOD 4 = 3. So the 3rd element of the repeating part is the answer: 7. (If the power MOD 4 had been 0, it would have been the 4th element of the pattern: 1)
1 + 1 = 0 in binary. Why does this happen?Note: Adding binary numbers is related to modulo 2 arithmetic.Let's review mod and modular arithmetic with addition.modulus 2 is the mathematical term that is the remainder from the quotient of any term and 2. For instance, if we have 3 mod 2, then we have 3 / 2 = 1 + ½. The remainder is 1. So 3 ≡ 1 mod 2.What if we want to add moduli?The general form is a mod n + b mod n ≡ (a + b) mod n.Now, for the given problem, 1 mod 2 + 1 mod 2 ≡ 2 mod 2. Then, 2 mod 2 ≡ 0 mod 2.Therefore, 1 + 1 = 0 in binary.
Mod is essentially the remainder when a given number is divided by the base (of the modulus).So10/3 has a remainder of 1 and so 10(mod 3) = 111/3 has a remainder of 2 and so 11(mod 3) = 2
69512578 mod 3 would be 1
The Mod Squad - 1968 My What a Pretty Bus 1-3 was released on: USA: 8 October 1968
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.
The Mod Squad - 1968 The Long Road Home 3-1 was released on: USA: 22 September 1970
The MOD function finds a modulus. That is the remainder when you divide one number into another. So if you divide 10 by 3, you would get a remainder of 1. To do that with the MOD function, you enter it as: =MOD(10,3)
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.