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The remainder of the division, by 4, is a number between 0 and 3. In the case of binary, this would maintain the last two bits of the original number.

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Q: What happens to a binary number n if you compute n mod 4?
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How do you find the multiplicative inverse of a number mod a number?

Formally, a number n, has an inverse mod p only if p is prime. The inverse of n, mod p, is one of the numbers {0, 1, 2, ... , k-1} such that n*(p-1) = 1 mod p If p is not a prime then: if n is a factor of p then there is no such "inverse"; and if n is not a factor of p then there may be several possible "inverses".


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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


What number divided by 3 has a remainder of 2 and divided by five has a remainder of three?

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None. And any attempts to say other wise are defeated in higher math. Even 0 is considered even by way of being 0 in mod(2). (similarly, 1 is odd by being 1 in mod(2) )

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