Tuesday, February 21, 2017

At least one of 3 consecutive odd numbers is a multiple of 3

The way to approach this problem is to show that there will always be a number among three consecutive odd numbers that is divisible by $3$. Let three consecutive odd numbers be $2n+1,2n+3,2n+5$. If $2n+1$ is a multiple of $3$ then we are done. Else the remainder when divided by $3$ is either $1$ or $2$. Suppose its $1$ then it means $2x$ is divisible by $3$ which means $2x+3$ is divisible by $3$. Or suppose the remainder is $2$, which means $2x-1$ is divisible by $3$ which means $2x+2$ and $2x+5$ are divisible by $3$. Hence proved.

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Variance of Uniform Distribution

Let the corresponding probabilities are $\frac{1}{n+1}$ for all at the points $\{0,1,2,\cdots,n+1\}$ The general formula is $\sigma^2 = E[x^2]-(E[X])^2$ Let's first calculate $E[X]=0\cdot \frac{1}{n+1}+1\cdot \frac{1}{n+1}+\cdots+n\cdot \frac{1}{n+1}=\frac{0+1+2+\cdots+n}{n+1}=\frac{n(n+1)}{2(n+1)}=\frac{n}{2}$ Now lets calculate $E[X]=0^2\cdot \frac{1}{n+1}+1^2\cdot \frac{1}{n+1}+\cdots+n^2\cdot \frac{1}{n+1}=\frac{0^2+1^2+2^2+\cdots+n^2}{n+1}=\frac{n(n+1)(2n+1)}{6(n+1)}=\frac{n(2n+1)}{6}$ Therefore $\sigma^2 = \frac{n(2n+1)}{6}-\left ( \frac{n}{2}\right )^2 $ $\Rightarrow \frac{n(2n+1)}{6}-\frac{n^2}{4}$ $\Rightarrow \frac{2n(2n+1)-3n^2}{12}$ $\Rightarrow \frac{4n^2+2n-3n^2}{12}$ $\Rightarrow \frac{n^2+2n}{12}=\frac{n(n+2)}{12}$ Now suppose we have same $n+1$ terms shifted from $a$ to $b$ in that case the variance becomes $\frac{(b-a)(b-a+2)}{12}$

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