Melody

So how many numbers are there between 25 and 250 inclusive?

What have you done towards solving this for yourself?

Hi Guest, 

I do not know how good your explanation is, I have not looked through it.

BUT

The words and sentences, being all clumped together like that, are very uninviting.

Your answers are more likely to be used and/or appreciated if they are more visually friendly.

\(\frac{1}{\sqrt{2}} +\frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}\\ = \frac{\sqrt{32}}{\sqrt{32}} \times \left( \frac{1}{\sqrt{2}} +\frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}\right) \)

now expand the bracket and simplify.

\((50 + 72i)^2 - (50 - 72i)^2\\ =50^2+(72i)^2+(2*50*72i)\quad -[50^2+(-72i)^2-(2*50*72i)]\\ =2*(2*50*72i)\\ =14400i \)

Nice logic Guinevere  

You should give yourself a point  

The easiest way to do this is to just look at numbers that fit,  find a common one and then see what it is mod12

N=3K+2

so N = 2,5,8,11,14,17,20 etc

N=4G+1

so N = 1,5,  etc

find a number that is in both lists.

Attention:  Tinfoilhat

I know that you have given some good answers and I thank you for those...  but I think in this case you did not know how to answer this question. 

I encourage people to leave part answers but it is reasonable for the asker to assume that you do actually know how to finish it.

If you have not finished the question for yourself (maybe because of time restraints), or you do not know how to finish it, please make this very clear to the asker.

I think the main point here is that the polynomial has rational coefficients.

is  [x-8  - sqrt3 ] is one factor then [x-8  + sqrt3] would have to be another. 

So when they are multiplied the factor will be [(x-8)^2 - 3] or  [x^2-16x+61]  only rational coefficients are left.

 I like this question too.

\(25^\frac{1}{25}=5^{\frac{2}{5^2}}\\~\\ 125^\frac{1}{125}=5^{\frac{3}{5^3}}\\~\\ 625^\frac{1}{625}=5^{\frac{4}{5^4}}\\~\\ \)

\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\~\\ P = 5^{1/5} \cdot 5^{2/5^2} \cdot 5^{3/5^3} \cdot 5^{4/5^4} \dotsm\\~\\ P = 5^{{1/5}+2/5^2+{3/5^3} +{4/5^4} }\dotsm\\~\\ \)

So I need to find a sum to infinity.

\(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\ =\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\ =(\text{infinite sum of GP a=1/5, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\ =\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\ =\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\ =\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\ \text{Infinite sum of a GP a=1/4 r=1/5}\\ =\frac{1}{4}\div(1-\frac{1}{5})\\~\\ =\frac{1}{4}\times \frac{5}{4}\\~\\ =\frac{5}{16} \)

\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\ P= 5^\frac{5}{16}\)

NOTE: 

I wrote the latex and worked the logic at the same time so there is plenty of room for careless error.

You need to work through (understand) what I have done and find any errors that might be there.

Unless you only want the answer of course in which case you may or may not get lucky.

LaTex:

\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\
=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\
=(\text{infinite sum of GP a=1/5, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\
=\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\

=\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\

=\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\
\text{Infinite sum of a GP a=1/4   r=1/5}\\
=\frac{1}{4}\div(1-\frac{1}{5})\\~\\
=\frac{1}{4}\times \frac{5}{4}\\~\\
=\frac{5}{16}
 

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