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4

$$a+e+2ae=a\underbrace{\Leftrightarrow}_{-a \, \mathrm{both}\,\mathrm{terms}}e+2ae=0 \underbrace{\Leftrightarrow}_{\mathrm{common}\,\mathrm{factor}} e(1+2a)=0 \\ \underbrace{\Leftrightarrow}_{e\ne 0}1+2a=0 \Leftrightarrow a=-\frac 12. $$


3

We need just 4 numbers. First, let's consider just two of those 4 numbers. One with all $2n+2$ bits 0 and one with all $2n+2$ bits 1. Any number of the length $2n+2$ will have either number of 1's or number of 0's greater than or equal to $n+1$. These two numbers will cover all the numbers with number of bits = $2n+2$ other than those which have exactly ...


3

Suppose we agree that we will always toss the coin twice, and then follow the schedule you described: Between $A$ and $B$: $A$ loses on heads, $B$ loses on tails Between $C$ and the loser of toss 1: $C$ loses on heads, the round 1 loser loses on tails There are four ways the coin could come up, all equally likely: $$\begin{array}{cc|c} \text{first toss} ...


2

It seems that you're trying to determine the identity of a binary operation $$a\star b:=a+b+2ab.$$ That is, you are trying to find some $e$ such that for every $a,$ we have $a\star e=a.$ Using the definition of $a\star e,$ we have $$a+e+2ae=a.$$ Subtracting $a$ from both sides and factoring out the $e,$ we have $$e(1+2a)=0.$$ At this point, context becomes ...


2

Going through a few examples, I finally understood it. The explanation under 2.3.5 in Example 1 actually explains the basics very well. The main idea is that is that you can reduce every coefficient on it's own. Instead of doing division you simply use the fact that the reduction of every power over your maximal degree $m$ can be computed by $z^a \equiv ...


1

00 starts with 00 which is non-decreasing so k=2 01 starts with 01 which is non-decreasing so k=2 10 starts with 1, but 10 is not non-decreasing, so k=1 11 starts which 11, so k=2 Three of them have k=2, one has k=1, so T(2,1)=1,T(2,2)=3


1

Ordered sets exist and are often called sequences. The entire space of six-bit numbers might be written as $\{0,1\}^6$. Even better, use $\Bbb Z_2^6$ to give you some implicitly understood operations you can exploit, with multiplication being bitwise and, and addition being bitwise xor.


1

Substitution works fine. Plug your second into the first to get $x_3=1$ and so on. Another ways is that the second gives you $x_1=x_2$ and you can plug that in to eliminate one of the variables.


1

Since you ask for any method, I will add that making an exhaustive search for a solution is perfectly viable here. For the first exercise, you only have $2^3=8$ cases to try and in the second $2^4=16$. It may not be as elegant or give as much insight and understanding, but sometimes it's the most practical way to go.



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