CPhill

In a reduced form.....If we can express the denominator solely as a power of 2 or a power of 5 - or both - the fraction will terminate....else..it repeats.....

1/ 40  terminates  = 1 / [ 2^3 * 5]

1/ 24 does not =  1 / [ 2^3 * 3 ]

2/125 terminates  = 2 / 5^3

2/ 99 does not  =  2 / [ 3^2 * 11]  

7/ 300  does not =  7 / [ 2^2 * 5^2 * 3 ] 

7 / 200  terminates  =  7 / [ 2^3 * 5^2]

Here's the provision for the reduced form

3 / 300     appears to not terminate....however

3 / 300  =   1 / 100  =    1 /  [  2^2 * 5^2 ]    so this actually does terminate  !!!

And that's all there is to it  !!!!

Here's a fast way ....we multiply by the base and add the next digit.....etc.

Suppose we have

1011₂

Multiply the leading "1" by 2 and add the next digit  ⇒  2 * 1 + 0  = 2

Multiply this result by 2 and add the next digit ⇒   2 * 2 + 1  =  4 + 1  = 5

Multiply this result by 2  and add the final digit ⇒ 2 * 5 + 1  = 10 + 1  = 11  

Verify for yourself that  1011₂   =  11₁₀

BTW - This method will convert any base to a base 10 number......

ax^2 + bx + c     (1)

In this form, the x coordinate of the vertex is given by

 - b / [ 2* a ]         

To find the y coordinate, sub this result back into   (1)  

Examlple

2x^2 +  8x +  `10     ......        a = 2       b =  8

x coordinate of the vertex  =  -8/ [2 * 2 ]  =  -8 / [ 2 * 2 ]  =  -8 / 4  =  -2

y coordinate of the vertex =  2 (-2)^2 + 8(-2) + 10  =      2

(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8

Let's write this in a different manner

(t-1)^3  + 8  + 6(t-1)^2 + 12(t-1) 

The first two terms can be written as a sum of cubes thusly

[ (t - 1) + 2 ]  [ (t - 1)^2  - 2(t - 1)  + 4 ]  =

[ t + 1 ] [ t^2 - 2t + 1 - 2t + 2 + 4 ]  =

[ t + 1]  [  t^2 - 4t + 7 ]      (1)

The second pair of terms can be factored as

6 [ t - 1 ] [ t - 1 + 2]  =

6 [ t - 1] [t + 1]  =  [ t + 1] [6t - 6]    (2)

Putting (1) and (2) together, we have

[ t + 1 ]  [  t^2 - 4t + 7 + 6t - 6 ]  =

[ t + 1 ]  [ t^2 + 2t + 1 ]  =

[  t + 1 ] [t + 1] [t + 1 ]  =

[ t + 1]^3

They are indicating where the graphs cross a particular axis.....I have to admit, that threw me at first, too, ISG  !!!!......I don't even see why they are necessary in this problem !!!

F  is correct.....the graphs intersect in two places  [two x values ]

I can explain why the others are incorrect.....if you want me to.......!!!

The answer cannot be graphs   F , G, H  or  K

The reason is that some of the "shaded" portion  of each graph falls into the 3rd quadrant where x and y are both negative  which  would make   1  < x + y < 2     false

So...by default.....the answer is   J

Call the height of the triangle, y

Call the base length, x

And

sin ( theta)  =   y / hypotenuse   ⇒  hypotenuse * sin (theta)  =  y

cos (theta)  =  x / hypotenuse  ⇒ hypotenuse *cos (theta)  = x

So.....this must mean that the hypotenuse  = 10

Here's an graphical way of figuring this out

C (n , 4)   =  n! /  ( [ n - 4] !  * 4! )   .....so....we require that.......

C ( n , 4 )  ≥  365

n! /  ( [ n - 4] !  * 4! )  ≥  365

n! / [ n - 4]!  ≥   4! * 365

n! / [ n - 4]!  ≥  8760 

n * (n - 1) (n - 2) (n - 3)  ≥  8760

Look  at the graph here : 

https://www.desmos.com/calculator/j9rafrlbpn

Note that  the minimum n that makes this true ≈  11.3

So.....we need 12 friends

Proof  C(11, 4)  = 330    too few groups of 4

But  C( 12. 4)  = 495     .......we even have a few "leftover" groups of 4  !!!!!! 

Thanks, heureka....I am very intersted in this problem....your answer includes some combinatorial identities........as I am not too familiar with these, could you show me how these two are derived ??

[ n / ( k + 1) ]  =  [  (n - k ) / ( k + 1) ]   *  [ n / k ]

And

[ n / (k - 1) ]  =  [ k / (n - k + 1) ] * [ n / k ]

Also......what allows us to do this ??

2 [ (n - k) / (k + 1) ]  =  1

n - k   =  [ k + 1 ] / 2

Thanks....!!!!