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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

Q: What does 1 mod 3 mean?

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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 + Â½. The remainder is 1. So 3 â‰¡ 1 mod 2.What if we want to add moduli?The general form is a mod n + b mod n â‰¡ (a + b) mod n.Now, for the given problem, 1 mod 2 + 1 mod 2 â‰¡ 2 mod 2. Then, 2 mod 2 â‰¡ 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

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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.

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.

The Mod Squad - 1968 The Long Road Home 3-1 was released on: USA: 22 September 1970