Melody

How many zeroes do we write when we write all the integers from 1 to 256 in binary?

I am going to look for a pattern.

256 = 2^8  so there is 8 zeros which is 9 digits

consider the numbers between 2^1 and 2^2-1          inclusive    10,11,                                           2/2=1   zeros

consider the numbers between 2^2 and 2^3-1          inclusive    100,101,110,111,                    2^2  *2 /2=4  zeros

consider the numbers between 2^3 and 2^4-1          inclusive    4 digits                                   2^3 *3/2 = 12 zeros

       ....

consider the numbers between 2^7 and 2^8-1          inclusive    8 digits                                   2^7 *7/2   zeros

    1+4 + 2^3*3/2 +   2^4*4/2    +   2^5*5/2  +  2^6*6/2   +  2^7*7/2  + 8

=  1+4+      4*3    +      8*4       +        16*5   +     32*6    +      64*7  + 8

=  1  + 4 +   12    +      32         +       80       +      192     +       448  +  8

=    777  base 10   

Just as guest already determined  :)

How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if each person can receive at most two bracelets?  Each bracelet must be given to someone.

1 each   4! ways                                          = 24 ways

1 pair two individuals    4C2=6     4*3*2*6 = 144  ways

2 pairs   4C2/2 = 3                        4*3   *3 = 24 ways

24+24+144 = 192 ways   

I think.

**73      9*10-1  =  89

*73*      9*10       = 90

73**      10*10-1  = 99

7373                        1

89+90+99+1   =  279 

Sorry,

It doesn't make much sense to me either.

Yes, I got the same answer.

Here is the really simple way to do it.

No algebra AT ALL!  You dont even have to work out the equation of any of the circles!

The shortest difference between two points is a straight line.

But in this case a part of the problem is that a 4 foot vertical drop must be included.

I have included it at the start but it can be included at the end if you want. I called the end point of it A'

Now use pythagoras to find distance A'D     5,12,13 is a pythagorean triad so  A'D=13

The rope is 7 foot

13 - 7 = 6 feet left!  

Easy peazy!

What asinus has said (sort of)  is that a part of the vertical drop is the 4feet.

Which on my number plane is the point (0,5)

THEN

the the shortest path to D is a straight line.  The rope will cover the first 7 feet.  then you have to work out what is left.

That is much easier than how I did it.  

EXCELLENT thinking asinus!

Thanks asinus,

I had not actually understood what you had said before but now this comment totally explaines it!

Yes that would be a good idea!

Guest asker, 

If you want some clarification, please ask. 

Thanks asinus,  

I got the same answser as you but my answer invlolved some calculus.

       To find the minimum distance between the points   \((12,0)\;\; and \;\;(x, \sqrt{49-x^2}+5)\)

I also graphed it to as a second way to solve it.

Here is my gaphical solution.  If you right click on the point C you can drag it and see what difference it makes to the distance CD.