CPhill

https://web2.0calc.com/questions/help-cphill_6

Good job, NTS!!!!

Thanks, NTS!!!

Another way  using Euler's Totient Function

distinct prime  factors of 20 =   2, 5

ø(20)  =    20 ( 1 - 1/2) (1 -1/5)  =  20 (1/2)(4/5)  = 10 * 4/5  =  8

That's an interesting method, NTS!!!!....I haven't seen that before....

Here's another way....factor each number

36  = 2^2 * 3^2

90  =  2 * 3^2 * 5

385  = 5 * 7 * 11

Take each different  prime to its  highest power and  find their product

2^2 * 3^2 * 5 * 7 * 11  = 13860 sec  =  231 min

Here's my best attempt....

"Anchor"  the mathmaticians in  any two seats next to each other

And they  can sit  in 2 ways

The other 5 people can  be arranged in the other five seats in  5! = 120 ways

So  2 * 120  =   240 arrangements

https://web2.0calc.com/questions/geometry_55509

x^3 + y^3 = (x + y) (x^2 + y^2  - xy)

        5 =  (2) (x^2 + y^2 - xy)

       (5/2)  = x^2 + y^2  - xy

      (5/2) + xy  = x^2 + y^2

x + y = 2      square both sides

x^2 + y^2  + 2xy  = 4

x^2 + y^2  = 4  - 2xy

So

(5/2) + xy  =  4 - 2xy

3xy  =  4 - 5/2

3xy =  3/2

xy =  1/2

2xy= 1   

So

x^2 + y^2  = 4 - 1  =   3

Thx, Tottenham10  !!!

The solutions (per WolframAlpha) are

(  [ -3 + sqrt (29)]  / 2 , 4sqrt (29) - 17 )    and  ( [ -3 -sqrt (29)]/2 , -4sqrt (29) - 17 )

https://web2.0calc.com/questions/rates_57

\(\frac{4}{1 - \sqrt[3]{3}} + \sqrt[3]{9}. \)

We can write  (1 - 3)  as  ( 1 - 3^(1/3))  ( 1 + 3^(1/3) + 3^(2/3) )

Mulitiplying the numerator and  denominator of the first fraction by the last factor in red we get

4 [ 1 + 3^(1/3) + 3^(2/3) ]          -2 [ 1 + 3^(1/3) + 3^(2/3) ]

____________________  =  

               1 - 3

And  9^(1/3)  = ( 3^2)^(1/3)  =  3^(2/3)

So we have

 -2 [ 1 + 3^(1/3) + 3^(2/3) ] + 3^(2/3)   =

-2  - 2*3^(1/3) - 3^(2/3)