CPhill

Here's the way I see this........

If the number is a multiple of 5, it must end in a  "5"......and there are 5!  =  120  ways of arranging the other digits

Of the numbers of interest.....the "6" has to be the first digit if the number is > 500,000 and the "5" has to be the last

Thus....the "middle digits" in these numbers can be arranged in 4!  = 24 ways

So...the probability is  24 / 120   = 1 / 5

Edited to correct an earlier answer......

Nice, Alan  !!!!!

We have  50 integers that are multiples of 2  from 1 - 100

We have   96/6 + 1  = 16 + 1  =  17 additional  integers  that are divisble by 3  [and not 2 or 5 ] from 1 - 100

We  have  7   integers  from 1-100  divisible  by  5  [ and not 2 or 3] from 1-100

So...the probability  that  he  chooses  a card divisible by 2, 3 or 5  is   [ 50 + 17 + 7 ] / 100  = 74/100 = 37/50

Let the number of  red balls  = x     

So we have that

[x / 10] * [ (x - 1) / 9 ]  = 1/15     simplify

x(x - 1) / 90   = 1/15       multiply both sides by  90

x ( x - 1)   = 6   

x^2 - x -  6  = 0     factor

(x -3) ( x + 2)  = 0    set each facotr to 0 and solve for x and we get that

x = 3    or  x  = -2       reject the second answer 

So...there are 3  red balls

Proof :  First Draw Probability of  Red  * Second  Draw Probability of  Red  =

                   3 / 10   *  2 / 9   =  6 / 90    = 1 / 15

I think this is what you are  looking for, but I'm not sure... I'll give it my  best shot  using something known as  the "sum of differences "

4  16    42    88     160

  12   26   46   72

      14   20   26

           6     6

              0

Using the polynomial  form  p(x)  = ax^3 + bx^2 + cx + d , we can  solve this system  of equations  for a,b, c and  d

a  +  b +  c +  d  =  4         [ for x  = 1, the first term]

8a + 4b + 2c + d  = 16        [for x  = 2, the second term]

27a + 9b + 3c + d  = 42      [for  x  = 3, the third term] 

64a + 16b + 4c + d  = 88     [for x  = 4, the fourth term ]

 The solving process is a little lengthy [ but not difficult ] ...I used WolframAlpha to find the solutions

a = 1, b  = 1,  c  = 2  and  d  = 0

So....the generating polynomial p(x) =   [ a_n ]  =    x^3 +  x^2 + 2x

Note that the first  five  terms are as above .....https://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+%2B+2x,++for++x++%3D+1,2,+3,+4,5

I hope that helped....if this is not what you want.....check back.....our member "heureka"  is  pretty knowledgeable about this sort of thing

OK, Coldplay....!!!

This triangle is isoceles.....the shortest angle bisector  will  be the one that bisescts angle "A"....it will be the altitude of the triangle

We can find the length of this bisector using the Pythagorean Theorem

Length  =   √ [ 14^2 -  13^2 ]   =  √ [196 - 169 ] =  √27  =  3√3 units

Here's a pic with AD, BE  and CF  being the bisectors.....as you can see,  AD  is the shortest bisector

Actually....this could be done more simply

Draw  AD....now....we have ΔACD  with  angle ACD  = 150°

And since CD  = CA, then angle CAD  and And ADC  are equal

So angle CAD = [180 -150 ] / 2  = 30/2  =  15°

And if we draw, AE, by the same process,  in Δ ABE, angle EAB = 15°

So CAB  = 60°  =  angle CAD + angle DAE + angle EAB

 60   =  15  + angle DAE + 15

60  = 30 + angle DAE

30° =  angle DAE

That's pretty nifty, Melody... I don't think I am  familiar with this  !!!

x  = k / y   .....hold  k  constant

If y  is increased  by 25%   we have that

x  = k /  [ (5/4)y  ]     simplify

x = (4/5)k / y

Then  x  is 4/5 ths  =  80%  of its orginal value  ( it  decreases  by 20% )