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1 mod 3 means the remainder after 1 has been divided by 3. Which is 1. 2 mod 3 = 2, 3 mod 3 = 0
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.
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)
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.
There is a mod that removes blurs, i do not have it but type sims 3 nude mod or try mod the sims 3.
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
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Add 5 in mod 8. (You can always subtract 3 in mod 8 as well, eg 5 + 3 = 0; 0 - 3 = 5.)
If 3 ≡ x mod 7, then x = 7k + 3 for some k.
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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.
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