CPhill

D  - 45   =  4  [ L - 60 ]

But,at first,  D  = L     ...so....

D - 45  =  4 [ D  - 60 ] 

D  - 45  = 4D -  240

240  - 45 =  4D - D

195  = 3D

195  / 3   =  D   =  65

Each started with 65

This will be  a circle....BUT....we  have to set  it  =  to  something

Is it supposed to be set to  0  ???

Allowing for  the left factorial

! [ 2 * 2 ] ^2  +  0!   =   ! [ 4]^2 + 1  =   [9]^2  + 1  =   82

! [ 2  + 2  +  0! ] *  2   = ! [ 5 ] *2  =   [44] * 2  =   88

([ 2 + 0! ] !) !!   +  20  =  [ [ 3! ]!!  + 20  =  6!! + 20  = 48 + 20  = 68

C ( !(2 + 2) , 2 + 0!)  =  C ( 9,3)  = 84

When Claire  gets to the outpost......Megan will be 5/6 of the way there....so....she  will have jogged  (5/6)16  = 40/3  mi

So....she  is  16 - 40/3 =     48/3  -40/3  = 8/3  miles  from  the  outpost

The  time it will take  for  them to meet  (in hrs)   =

(8/3 mi)  / their combined rates  = 11mi/hr

So

(8/3) m  / 11 m /hr  =  8/33  hrs

So..... from  Claire's  point of  view   she jogs  back   6 m/h ( 8/33)h  =  48/33  mi

So.....they  meet   48//33 mi  from the outpost  =  1 + 15/33  miles from the outpost

sin  (x/2)   =      sqrt   [ ( 1 - cos x)  / 2 ]   =  [  (1   - (-1/3) ) / 2 ]  =  sqrt  [ (4/3) / 2 ]  =

sqrt [ 4/6]  =  sqrt (4)  /sqrt (6)   =  2  sqrt 6  =  (2/6) sqrt 6  =  sqrt 6 / 3

cos (x/2)  = sqrt [ (1 + cos x) /2 ) ]  =  sqrt [ ( 1   + - 1/3) / 2] =  sqrt  [  (2/3)/2] = sqrt [ 1/3]  =

sqrt ( 1 )/ sqrt (3)   =    sqrt (3)/3

"Anchor"  the  boys in any 2 chairs......there  are  2 ways they can sit

And foreach of these,  the  girls can be seated in 31 = 6 ways

So

2 * 6 =    12  possible arrangements

2n^2  +  2n  +  38  =

2 ( n^2  + n +  19)

Since  this  is a multiple of 2  the  only possibilities  are

2222,   4444,   6666  or 8888

There  are no integer  n's  such  that

2n^2  + 2n +   38  =    any  of these

Possible    digits

3  5   7

Arranging  these  in 3!  ways  we  get  6   M's

Also

1 3  5  7

Arrranging  these  in  4!  ways  we  get  24 more

So

24 + 6   = 

30 possible M's

Note  that sides of  11, 60 , 61  form a right  triangle

Any integer  multiple of these also forms a right triangle

So

2(11)   ,  2(60)  ,  2 (61)   =

22   ,    120  ,   122    form  a right triangle

(2/5)E  =  (6/7)N

(7/6) (2/5)E  =  N

(14/30)E  =  (7/15)E   = N

E - 36  =  N  + 20

E - 36  =  (7/15)E  +  20

E  -  (7/15)E  =  20  +  36

(8/15)E   =   56

E  =  (15/8)(56)

E  =  15  ( 56/8)

E  =  105   =   what  he started with

So.....he  ended  up  with    105 - 36  =   $69