CPhill

Put 6 people on one side of the see-saw and 6 on the other

Obviously....the heavy person will be on the lower side

Divide this side into two groups of 3......again...the heavy person will be on the lower side

Choose any 2 of these 3 people to weigh

We have two possibilities for the third weighing :

1) They balance.....the heavy person is the one not weighed

2) The don't balance....again....the heavy person is on the lower side

Note, YEEEEEET.....we are summing  99 terms   [ 100 - 2] + 1  = 99

So....

(1/6) (99) ( 2*(99)^2 + 9*(99) + 13)    =    338,349

Check this against    sum (r + 1)^2  from r = 1 to r = 99  = 338,349

Excellent, MG  !!!

>>>>>>>Sorry... I didn't see "even".....look at the guest's answer.....!!!

Look  at the grid below

7    8      9

4    5     6

1    2     3

First.....note that only  the corner points  and the middlle point can be reached in 4  moves from 1....so points 2, 4 , 6 and 8 cannot be reached in any set of 4 moves from 1

To reach 1 from 1 we have the following sets of moves

(1, 2, 1, 2, 1)  ( 1, 2, 5, 4, 1) (1, 4, 5, 2, 1) (1, 2, 3, 2, 1) (1, 4, 1, 4, 1) (1, 4, 7 , 4, 1)

To reach 3 from 1, we have these sets of moves

(1, 2, 3, 2, 3) ( 1, 2, 1, 2, 3 ) (1, 2, 3, 6, 3) (1, 2, 5, 6, 3) (1, 2, 3, 2, 1) ( 1, 4, 5, 6, 3) 

To reach 7 from 1 we have the same number of sets of moves as that of reaching 3 from 1

To reach the middle point 5 from 1 we have these sets of moves

(1, 2, 3, 6 , 5) (1, 2, 1, 2 , 5) (1, 2, 3, 2, 5) ( 1, 2, 5, 4 , 5) (1, 2, 5, 8, 5) (1, 2, 5, 2, 5) (1, 2, 5, 6, 5)

(1, 4, 7, 4, 5) ( 1, 4, 7, 8, 5) (1, 4, 5, 6, 5) (1, 4, 5, 8, 5) ( 1, 4, 1, 4 , 5)

To reach 9 from 1 we have

(1, 2, 3, 6, 9) (1, 2, 5, 6, 9) ( 1, 2, 5, 8, 9) (1, 4, 5, 6 , 9) (1, 4, 5, 8, 9) (1, 4, 7 , 8 , 9)

So.....we have a total of  4(6) + 12  =  36 possible moves

And 6 of these reach 9 from 1   -  B from A -.....so....the probability is  6 / 36   =  1 / 6

Well, PM....just thank the Lord that you're not that tree in your backyard....!!!

Good job, PM...!!!!

Good, synth...hope you are OK, too....!!!!!

Welcome, EP.....

[ Also note that -12 / 8 = - 3 /2 ]

For  f( f(x) to = 5

We have several possibilities

First

(x + 3) + 3 = 5

x + 6 = 5

x = -1

But.....note that  f(f(-1)) = f (-3)  = 5   so....-1  works

Also

(x^2 - 4)^2 - 4 = 5

x^4 - 8x^2 + 16 - 9 = 0

x^4 - 8x^2 + 7 = 0        factor

(x^2 - 7) ( x ^2 - 1)

We have 4 possible solutions

x = -√7, √7, -1 , 1

 x = - √7  will produce  because f(f(-√7)) = f (3) = 6

We have seen that x = - 1 works

x = 1 will work because  f(f(1)) = f(-3) = 5

x = √7  will work because   f(f(√7)) =  f(3)  =  5

Another possiibility is  that

(x +3)^2 - 4 = 5

x^2 + 6x +  9 - 9 = 0

x^2 + 6x = 0

x(x + 6) = 0

x =  0   or x = -6

f(f(0)) = f(-4) =  12

f(f(-6)) = f(-3) = 5

The last possibility is that

(x^2 - 4) + 3 = 5

x^2 - 6 = 0

x = -√6  or x = √6

f(f(-√6)) = f (2) = 0

f(f(√6)) = f(2) = 0

So....the x values  where f(f(x) ) = 5  are

x = -1, 1, -√7, √7, -6