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If you have a number in a base $b$, then the 'allowed' digits are normally $0,1,2,\ldots,b-1$. If you have a digit bigger or equal as $b$, then you could rewrite the number with out those digits. In your example the 'allowed' digits are $0,1,2,3,4,5,6,7$, so $8_8 = 10_8$, therefore $3738_8 = 3740_8$. But your calculation seems absolutely fine.
$\begin{matrix} & & & & & 1&1&0&1\\ & & & &\times & \color{red}{\rm 1} & \color{purple}{\rm 0} & \color{blue}{\rm 1} & \color{green}{\rm 1} \end{matrix}$ $\begin{matrix} & & & & 1 & 1 & 0 & 1 & (1101 \times \color{green}{\rm 1}) \\ ~ & & & 1 & 1 ... 4 I think the Hamming code with extra parity provides a solution. If there are$2^n$logcians, this requires sacrificing$n$of them (each with chance$1/2$). Let the rear$n$be the parity bits of the code, called out by those logicians. All the rest can run the error detection scheme. If no bits are wrong yet, either because the blind logician has not ... 4 Easiest answer as I see it: flip$a_1$. This is a bijection as the inverse function is given by flipping$a_1$. 0 Hint: Case1:$n$is odd. Let's say you have$k$number of$1's$where$k$is even. Then send this string to the string obtained by "flipping"$0's$and$1's$. This has$n-k$(which is odd) number os$1's$. Check that this is a bijection. Case 2:$n$is even. Let's say you have$k$number of$1's$, with$k$even. If the left most$a$is$1$then consider ... 0 Hint: Start with the case that$n$is odd, and think about what happens if you simply flip all the 0's to 1's and flip all the 1's to 0's. 1 One object can have$12$states-$6$for the operator times$2$for on/off. As you have$10$objects and the state of each object is independent, the number of total states is$12^{10}$0 This is what I ended-up writing... (by Josh Thomson) Mathematical Relations in Computing Computing is reliant upon countless mathematical operations, among which are core numeric systems such as ‘Unary’, ‘Binary’ and ‘Ternary’. These different numeric systems have their own uses in computing, but each share a common difference. The Unary number system has ... 2 Some hints: Suppose the upper$k$bits of$A,B$match. Then for any$x$satisfying$A \le x \le B$, the upper$k$bits of$x$will match those of$A,B$. You can do nothing about those except count them. So, suppose the top bit of$A$is 0 and the corresponding bit of$B$is 1 (and both have$n$bits). Then you should find a number in the range$A,...,B$... 0 To answer the first question: Suppose the initial string$A$is not all$0$or all$1$. Then after one operation it's easy to see that you get all$0$. Suppose the initial string$A$is not all$0$or all$1$. Then there is some$i$such that$A_i \neq A_{i+1}$. After doing this operation as you specified, we have a binary string$A'$. By the above ... 0 The algorithm will always be the same. Note two things: in notation the sign will always be kept, i.e.$-16_{10} = -10_{16}, -15_{10} = -\text F_{16}$(base subscripted) computers work differently: they store the sign by letting the first stored bit to equal$-2^k$where$k$is the bit-length of the data structure such that$\text{FF} = -1$while ... 0 For big numbers, you try to find the largest power of$16$that is still smaller than your number, and see how many times that one divides your number. Then you proceed with the second largest power, and so on. For negative numbers, you can calculate the hexadecimal of the absolute value of the number, and then stick a minus sign in front of it. 1 For X=00110101 and Y=10110101, Binary Addition X + Y = 11101010 Binary Subtraction X - Y = -10000000 Binary Multiplication X * Y = 0010010101111001 Binary Division X / Y = 0000 Check results @ Binary Arithmetic Calculator 0 I'm not sure what you personally mean by "binary". Maybe the binary numeral system for the natural numbers?... In any case, here, a binary relation between two sets of objects, say$A$and$B$, is a collection of pairs$(a_1, b_1), (a_2, b_2), \ldots, (a_m, b_m), \ldots$(finite or infinite), where every$a_i$belongs to$A$and every$b_i$to$B$. Note that ... 0 Write an$n+1$-digit binary number using leading zeros (so there are always$n+1$digits). Let each digit correspond to a vertex of an$n$-dimensional simplex. The binary number then represents a subset of the vertices of the simplex: if the value of a digit is$1$then the corresponding vertex is a member of the subset, and if the digit is$0$then the ... 1 It seems the following. Let$n$be a natural number,$X=\{x_0,x_1,\dots, x_n\}$be a set of vertices of an$n$-dimensional simplex and$B$be the set of binary sequences of length$n+1$. Then$|B|=2^{n+1}$. For each sequence$b=(b_0,\dots, b_n)$let$c(b)=\operatorname{conv}\{x_i: b_i=1\}$be a convex hull of a set$\{x_i: b_i=1\}\$ of vertices of the ...