CPhill

It's still a pretty good "trick" ..........

3x/4-2x/3=1/2

Multiply through by the common denominator  of  4, 3 and 2  =  12

12 (3x/4) - 12(2x/3) = 12 (1/2)

(36/4) x  -  (24/3)x  =  6

9x  - 8x  =  6

x  =  6

Thanks, Omi and heureka.........I was unsure about this one  !!!

1 / [ 2m - 4n ]  - 1 / [ 2m + 4n ]  - m / [ m^2 - 4n^2]    factor the denominators

1 / [ 2 (m - 2n) ]  -   1  / [ 2 (m + 2n) ]  -  m  / [ (m + 2n) (m - 2n) ]

The common denominator  is    2 (m + 2n) (m - 2n).....so we have

 (  [  (m + 2n) ]   - [m - 2n] - [2m]  )  / [  2 (m + 2n) (m - 2n) ]

(  4n - 2m ]  / [ 2 (m + 2n) (m - 2n) ]

[ 2 (2n - m) ]  /  [  2 (m + 2n) (m - 2n) ]

( 2n - m)  /  [ (m + 2n) (m - 2n) ]  

- ( m - 2n) / [ (m + 2n) (m - 2n) ] 

- 1  / [ m + 2n ]

2√3-√3  =

Factor out  √3

√3 [ 2 - 1 ]   =  √3 [ 1 ]  =  √3

Percent  of females  =

Number of females /  Total number of participants

121  / [ 121 + 99]  =   121  / 220  =  11 / 20  =  .55  = 55%

6 / 800  =  3 / 400  = .0075  =  0.75 %

One question.....you must be able to use numbers that have 3 leading 0s as possible four digit numbers????

Because....had I picked 9999 originally and then 9998....you would have had to use 0001 to get the first sum of 9999

And if I had picked 9997 as my last number.....you would have had to use 0002   to get the second 9999 sum......

Is this correct ???

HAHAHA!!!!!...I know it will come to 29997  !!!

I think I have the trick....although I'm not sure

You ask the person to pick a 4 digit number

Then...you add  9999 * 2   =  19998  to that number

This will be the "future number"

When they pick the next number.....you make sure that the number you pick and  their second number sum to 9999

When they pick their final number......you do the same thing......you make sure that your final number and their final number sum to 9999

So....the future number is just   their first number +  9999 * 2

Am I close ???

I know the next number you're going to pick.....8888  !!!