Convince that zero is an even number   I am not quite sure on this question can anyone help?
3005
Some books say that even numbers start from $2$ but if you consider the number line concept, I think zero($0$) should be even because it is in between $-1$ and $+1$ (i.e in between two odd numbers). What is the real answer?
13311
Based on pure intuition, the root is 0. It’s very obvious. But what is the proper way to solve it? Also how do we know it’s the only possible root (the answer is 1 root only, but how can I confirm?)
26607
Is the concept of even/odd numbers applicable to decimal numbers? For e.g. - 4.222 is a even number?
113661
The way I have understood, $0/0$ is undefined or indeterminate because, if $c=0/0$ then $c\cdot 0=0$, where $c$ can be any finite number including $0$ itself.  If we also observe a fraction $F=a/b$ where $a,b$ are positive real numbers,  the value of $F$ increases with the decrements of $b$. Being $0$ the least non-negative integer, if $b$ tends $0$ then $F$ tends to $\infty$ which is greater than all finite numbers.    I have also heard that no number is equal to infinity, a variable can tend to infinity. Any ratio is also a variable or is resolved to a number.  Now,   $$\tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)}=\frac{1}{0},$$    is it undefined or infinity?  Also we know   $$\tan2x= \frac{2\tan x}{1-\tan^2x}$$  Is this formula valid for $x=\pi/2$?  Any rectification is more than welcome.  Some references:
78178
With given number N, how to determine first number after N with same amount of odd and even divisors? For example if we have N=1, then next number we are searching for is :   2 because divisors: odd : 1 even : 2   I figured out that this special number can't be odd and obviously it can't be prime. I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number. Cheers
59743
Let , $D$ be the set of tuples $(w_1,w_2,\cdots,w_{10})$ , where $w_i \in \{1,2,3\},1\le i\le10$ and $w_i+w_{i+1}$ is an even number for each $i$ with $1\le i\le 9$.   Then find the cardinality of $D$.   As , $w_i+w_{i+1}$ is an even number so either both even OR both odd. Again $w_i\in \{1,2,3\}$. So , both even possibility is exactly $1$ which is $(2,2,\cdots,2)$.  But I am unable to find the both odd possibility..
118105
I have a set of numbers $$ [n] = \{1,2,...,n\} $$ in my special case $n = 100$, and I have a subset of $[100]$ with the following specification $$ A\subseteq[100] $$ and $$ |A| &gt;= 55 $$ now I should prove, that this statement is true for some $$ a,b\in A: a-b=9 $$ I thought about the problem and I realised, that if I just take the numbers $1-55$ that there are a lot of pairs $a,b$ that match the condition. So I tried to build a set in which no pair matches the condition. Therefore I just used the even numbers from $2-100$. Because even-even=even. But there are only $50$ even numbers in $[100]$, so I have to add at least $5$ odd numbers. So as soon as I add one odd number my set matches the condition. Using the pigeonhole principle: $$ n,m \in \mathbb{N}, f: [n] \to [m], |f^{-1}(j)|, j \in [m], \exists j^{*} \in [m], |f^{-1}|&gt;=\lceil\frac{n}{m}\rceil $$ I get, that there is at least $$ \lceil\frac{100}{55}\rceil = 1 $$ solution to my problem. But I think that I have to specify the function for the projection to prove the problem. And I think that I can use the modulo operator to achieve my goal, but currently I am stuck. Could someone please help me?
111792
Possible Duplicate:             How many zeroes does $2012!$ end with?    My idea is:   402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625$   How can we "show" that this is true?
89159
What would the steps be for someone new trying to learn how to solve this? I feel like I understand the concepts and what they mean but not how to prove it. Thanks
26943
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.   I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:   The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.  Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.    For $n=k+1$, there are two ways the number of sixes are even:  a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n]$  b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  I have two questions   How do I get from $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$ Other than that, is my use of induction correct? Is it rigorous enough to prove the formula?
4042
Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.  Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.  If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.  And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
112337
Follow up to  question. Is $0$ a positive number?
72327
How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...  Thanks in advance...
9261
Hi I'm practicing some number theory and I've come across the following problem,  Suppose $n \geq 1$ is an odd integer that divides $a^2 - 1$ for some integer $a &gt; 1$. Prove that,  $$n = \gcd(a-1, n)\gcd(a+1, n)$$  Now I know that the $\gcd(a-1, a+1) = 1$ if $a$ is even and $2$ if $a$ is odd and I believe this is relevant but I can't solve the problem. I've not really any idea where to start so even just help with that would be great.  Thanks
62381
Can you please help me and tell, how should I move on?  Can this be proved by induction?     Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.   Thank you in advance
10523
I know how to do it normally, but I could swear that you could fill out out like a[0] = {0,0,0,0}; How do you do it that way? I did try Google, but I didn't get anything helpful.
25606
Can someone give me a hint how to evaluate $$ \binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{o(n)},$$ where $o(n)$ is $n$ if $n$ is even and $n-1$ otherwise ?