Melody

Great,

It is always good to get feedback.  :)

Thanks Voldemort, that makes sense.

Here is a youtube clip, the first example is quite similar to yours

Incidentally, this has nothing to do with calculus.  

ok, I have not answered this yet but I will share my thoughts with you, maybe we can nut it out together.

The first thing I thought of it that I want the highest possible value of a_1.

Each number is different so the smallest factor possible is 2

so   if I let the the first one be A then the numbers will be

A,2A,4A,8A,16A,32A,64A

If A is 1 that is ok

If A is 2 then 64A is too big so the first term has to be1

The multiple could be 2 every time giving one possible sequence.

What about if one of the muliples was 3 rather than 2 then it would end in 96A,  So there are more possibilities there.

What about if one of the multiples 4 rather than 2 then it would end in 128A which is too big

so now you have a lot less possibilities to consider.

Can you find the answer and tell us what it is ?   

Hi Nerdiest.  

I just worked out each bit seperately and then put it altogether.

I have not presented it becasue it is a bit long and I no doubt made careless errors 

But I will walk you through it

\(z=a+bi\\ z^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=etc\\ |z|^2=a^2+b^2\)

Do the addition

The real part is 0 and the imaginary part is 0

Solve simultaneously.

p(-1)=5            p(-5)=-7

there is probably some super short cut now but I don't know it.

p(x)=Q(x)(x+1)(x+5)+Ax+B

So when divided by (x+1)(x+5) the remainder will be   Ax+B

p(-1)=A(-1)+B=-A+B=5

p(-5)=A(-5)+B=-5A+B=-7

solve simultaneously

-A+B=5

-5A+B=-7

I do ot understand the question.

The SQUARE ROOT of that is   |5x-3|

How about proofreading your question?

If you are too lazy to do that, it suggests to me that you will be too lazy to try and understand anybodies answer.

I have 10 distinguishable socks in my drawer: 4 white, 4 brown, and 2 blue. In how many ways can I choose a pair of socks, provided that I get two brown socks or two white socks?

There are 4*3 = 12 ways to get two brown socks and

there are 4*3 = 12 ways to get two white socks

so that is 24 ways.