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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).
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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 6 equal?
yo moms
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.
Why is n squared plus one never divisible by three?
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.
What is 0 to the power of 3 equal to?
03 = 0
What is the lowest common multiple for 2 3 and 9?
We see that we must find a number n such that it satisfies the condition: n ≡ 0 (mod 2) ≡ 0 (mod 3) ≡ 0 (mod 9) Since 9 is a multiple of 3, we can forget about the 0 (mod 3). Since 2 and 9 are relatively prime, the Chinese Remainder Theorem states that there indeed exists a number n such that it satisfies n ≡ 0 (mod 2) ≡ 0 (mod 9). Now let 2K represent some multiple of 2, and set it congruent to 0 (mod 9): 2K ≡ 0 (mod 9) This is a particularly easy case; 2K would have to equal some multiple of 9 for it to satisfy this expression. Therefore, K = 9 and n must = 18c, where c is an arbitrary multiplier. This is your new modulus: n ≡ 0 (mod 18) Any n that satisfies this condition will also satisfy n ≡ 0 (mod 2) ≡ 0 (mod 3) ≡ 0 (mod 9).
How many numbers between 1 to 1000 are divisible by 3 but not divisible by 5?
267 This question can easily be answered with some knowledge of programming, where the counts every case where x mod 3 = 0, AND X mod 5 (not equal to) 0
What is the additive inverse of 3 in mod 8?
Add 5 in mod 8. (You can always subtract 3 in mod 8 as well, eg 5 + 3 = 0; 0 - 3 = 5.)
What does 1 mod 3 mean?
1 mod 3 means the remainder after 1 has been divided by 3. Which is 1. 2 mod 3 = 2, 3 mod 3 = 0
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 6 equal?
yo moms
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 number containing the numbers 1 through 9 with all numbers only used once is divisible by all the numbers from 1 through 9?
36288 WRONG!(1*2*3*4*5*6*7*8*9)/10 WRONG!I obviously misreads the question. after writing a simple visual basic program, see below. I am changing my answer to "does not exist." as the program failed to return a value.Private Sub Command0_Click()Dim i As Longi = 1Do While i Mod 9 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 8 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 7 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 6 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 5 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 4 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 3 > 0 Or i Mod 10 = 0i = i + 1Do While i Mod 2 > 0 Or i Mod 10 = 0i = i + 1LoopLoopLoopLoopLoopLoopLoopLoopMsgBox iEnd Sub
Why does 1 plus 1 equals 0 in binary?
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.
Why is n squared plus one never divisible by three?
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.
Relational operator string vs numarical?
QBASIC relational operators are... = equal > greater than < lesser than >=greater than/OR, equal to <=lesser than/OR, equal to EXAMPLE QBASIC CODE:- IF 1 > 2 THEN PRINT "1 greater than 2" IF 1 < 2 THEN PRINT "1 lesser than 2" IF 1 = 2 THEN PRINT "1 equal to 2" IF 1 >= 2 THEN PRINT "1 greater than/OR, equal to 2" IF 1 <= 2 THEN PRINT "1 lesser than/OR, equal to 2" ...Output... 1 lesser than 2 1 lesser than/OR, equal to 2 Press any key to continue... QBASIC mathematical operators are... + plus - minus / divide * multiply MOD modulus (remainder) ^ raise to the power EXAMPLE QBASIC CODE:- PRINT "3 + 3 = "; 3 + 3 PRINT "3 - 3 = "; 3 - 3 PRINT "3 / 3 = "; 3 / 3 PRINT "3 * 3 = "; 3 * 3 PRINT "3 MOD 3 = "; 3 MOD 3 PRINT "3 ^ 3 = "; 3 ^ 3 ...Output... 6 0 1 9 0 27 Press any key to continue...
What is a fraction of 3 over 0 equal?
It is not equal to anything. Division by zero is not a valid operation.
