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I tried to state the base using the underscore but it didn't work, so I used this notation

log(b, a)

which WA will interpret as

\(\log_{b}a\)

Happpyyy b-dayyy 2 uuuuuuuuuu

To Emerald wonder: You don't have neg points anymore yay

Maybe you typed it wrong?

Hectictar: The reason you got a different result is because of sqrt(2). The sqrt(2) is the BASE of log. In other words, it is meant to be : Log(base sqrt(2), k)). But you entered it in WA as:

k(ceil(log(sqrt(2), k))-floor(log(sqrt(2),k))). If you enter it the way he/she has it in LaTex, it should be entered like this:  k(ceil(log_(1.41421), k)) - floor(log_(1.41421,k)))= 500,499. See here:

https://www.wolframalpha.com/input/?i=%E2%88%91%5Bk*(ceil(log_(1.4142135,+k))+-+floor(log_(1.4142135,k))),+k,+1,+1000%5D

When  k  is a power of  2 ,   \(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 0\)

When  k  is not a power of 2 ,   \(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor\ =\ 1\)

So  N  is the sum of all numbers which are not powers of  2  in the interval  [1, 1000]

\(N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad \sum\limits_{n = 0}^{9}2^n\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad (1+2+4+8+16+32+64+128+256+512)\\~\\~\\ N \quad=\quad \sum\limits_{k = 1}^{1000}k\quad-\quad 1023\\~\\~\\ N \quad=\quad \Big(\frac{1000(1000+1)}{2}\Big)\quad-\quad1023\\~\\~\\ N \quad=\quad 500500\quad-\quad1023\\~\\~\\ N \quad=\quad 499477\)

Hopefully your teachers not on! 

Oh boy I better tell my teacher that this math dosent make any sense shes a idiot! 

Thanks guys for your help have a nice day! 

You know actually 

It's not the calculator's fault LOL. any decent calculator would give an error or not work for that question.

The question doesn't make any sense.....

Here is the answer I was about to post:

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This:   652314557445522288521/655852155255+441527514-^31524452

doesn't make sense because of the  -^

What do you mean by  -^   ?

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The - by itself is a valid operation.

The ^ by itself is a valid operation.

The credit  goes to hectictar for figuring this out she just sent this is a private message I told her I would post it!!! 

But -^ together is not a valid operation.

It's just like saying...

What is   5-x3 =?  

"What is  5 minus times 3 ?"

See? It doesn't make any sense!

THX nickolas 

Hey cubeman have you ever tried to copy then paste it onto this calculator! 

at this link >>>>>>> https://web2.0calc.com/ 

Please try it further on but the answer to this is 

1877661132.7735639929455461-^31524452-^31524452= 1436133618.7735639929455461   

Thats Long! 

Thanks 

Hey Guest thisis what I pulled up and seen! 

41 / 900 is already in the simplest form. It can be written as 0.045556 in decimal form (rounded to 6 decimal places).

I not sure on this though I hope it helps! 

This is incorrect

Ok!!!

1.  Let x  = p + a       let  floor (x)  =  p

So....we have the following series

(p + a) ,  p  ,  a

The common ratio, r,   =     a/p

So

(p + a) ( a/p)  =  p     ⇒   a + a^2/p  = p   ⇒   a^2 + pa - p^2 = 0

a =  -p +√[p^2 + 5p^2 ]          =        - p  + √5 p      =       p ( √5 - 1)    =    p * phi

       _______________                 __________            _________    

                 2                                             2                              2

Then  a/p  =  phi

Then

(p + a) (a/p)  =  p

(p  + p*phi) (phi)  =  p

p*phi + p*phi^2  =  p        if we let  p =1, then we have an identity for these powers of phi, namely :

phi + phi^2  =  1       

And  a  = p*phi  =  1*phi  =  phi ≈  .618

So x  =   (a + p)  =  ( phi + 1 )  =  Phi  ≈  1.618

So...this is the common series  :

x , floor(x)  , x - floor(x)  =  

 Phi, 1 , phi

2 -

sumfor(k, 1, 1000,  k* (ceil(logn(sqrt2,k)) -  floor(logn(sqrt2, k))) = 500,499 
This has this "closed form":
a(n) =1 / 2 [n -1]*[n + 2] =500,499 when n=1000

I think this problem is solved easily if we lay it out in this manner :

Let the tangent to the circles be the equation y  = -8

Let the center of circle O   be  (0, 0)

Let the center of circle P  be (a, - 6)

Let the equations of the two circles be :

x^2 + y^2  = 64   

And

(x - a)^2  + (y - 6)^2  = 4

And ....since the distance between the centers of the circle is the sum of the radi, i,e.,  10 units, we can find the x coordinate of the center  of circle P, a,   as follows :

(a - 0)^2 + (0 - 6)^2  = 10^2

a^2  + 36  =  100

a^2  =  64

a  = 8

So...the center of  P  is (8, -6)

So  S  must  have  the coordinates (8, -8)

So....OS    has a length of     sqrt  [ (8 - 0)^2 + (-8 - 0)^2 ]  =  sqrt [ 64 + 64]  =  sqrt (64 * 2) =

8 sqrt(2) units

Also since we know all the zeros are at   x = -2   and   x = 4 ,  we can say

y  =  a(x + 2)(x - 4)

Since the axis of symmetry on a parabola is always halfway between the roots, and since the maximum is 54....

vertex   =   ( \(\frac{4-2}{2}\), 54)   =   (1, 54)

Now that we know  (1, 54)  is a solution to  y = a(x + 2)(x - 4) , we can say

54  =  a(1 + 2)(1 - 4)

54  =  a( -9 )

-6  =  a

So the equation of the parabola is...

y  =  -6(x + 2)(x - 4)

y  =  -6( x² - 2x - 8 )

y  =  -6x² + 12x + 48

-6 + 12 + 48  =  54