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Any other ideas?

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5=Sqrt((.2)^(-2))=(.2)^(-0!)

48=((2+0!)!)!!

8=(2+2)!!

100=(.(0!))^(-2)

6=(2+0!)!

15=5!!

Remaining

67 to 69

71

74

77

79

82 to 84

86 to 89

91

93

Maybe we can use infinite square roots

84%done

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

YAYAYAYAYAY

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

Oopsies, my bad. sqrt(30), I must've missed a 0. 

And also, it it 6. :)))

Nice job. :DDDD

=^._.^=

"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

Thank you soo much!!!

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

What is a right triangle that has 22 as one of it's sides? It's sides must be integer lengths. 

If all sides are integers, that's called a Pythagorean Triple. 

If you look at the linked webpage, there are no Pythagorean Triples with a side 22.  

https://www.avc.edu/sites/default/files/studentservices/lc/math/pythagorean_triples.pdf  

If there had been one with a side of 11, we could have just doubled all the sides,  

but there isn't one of those either.  

If this is to be a Pythagorean Triiple, with one side 22, then at least one of the other  

sides would have be be greater than 50.  I don't know how to find the other sides,  

by a method other than brute force. 

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

By mass points, CD/BD = 1/3.

There are 14 such primes.

idiot it doesn't work