Math Is Fun Forum

7254    n = Post# = 558 (see top right)

                                           
                                                                  Pascal Triangle

So, Post #12 defined the game out of 5 rolls.       
The number to reach and overcome is Total = 82.

Until this moment  gave us the following rolls:
Roll #1 = 6: 1 + 3 + 5 + 7 + 9 + 11 = 6 * 6 in Post #13
Roll #2 = 3: 1 + 3 + 5 = 3 * 3 in Post #18
Roll #3 = 4: 1 + 3 + 5 + 7 = 4 * 4 in Post #19
Total = 36 + 9 + 16 = 61
                                                   We need two more rolls!

Today we will Roll #4 = ...

Ready, Set and Ready, Go: . . . 5

Thank you, !

Thus, Roll #4 = 5: 1 + 3 + 5 + 7 + 9 = 5 * 5
Total: 61 + 25 = 86

Can Anybody roll the dice for the last Roll #5?

So, Post #12 defined the game out of 5 rolls.       
The number to reach and overcome is = 82.

Until this moment gave us the following rolls:
6: 1 + 3 + 5 + 7 + 9 + 11 = 6 * 6 in Post #13
3: 1 + 3 + 5 = 3 * 3 in Post #18
4: 1 + 3 + 5 + 7 = 4 * 4 in Post #19

                                                   We need two more rolls!

          CAN anybody ROLL the DICE?.... ANYBODY ? . . . .             

              Let's do the deal: you will roll the dice and I will count the numbers! 

       
ok, ok. Let's do it together!

      
Find:
               1
Roll # = (∫x³dx)-¹
               0
                                             
Use the table to find it:
       1
for (∫x³dx)-¹:  (x^4/4)-¹ = 4/x^4  ???  Nope! See Mr. Franklin's corrective post below
       0
It is:
       1
for (∫x³dx)-¹ we plug 0 and 1 into formula as follows:
       0
                           (x^4/4)-¹ ol¹ = [(1^4)/4 - (0^4/4)]-¹ = (1/4 - 0/4)-¹ = (1/4)-¹ = 4

Therefore:
  1
(∫x³dx)-¹ = 4 and thus Roll # = 4
  0

      
PS I've got too much pre-occupied with placing correct math symbols as x^4 and ∫
to miss on the flow of the calculation itself.
It is interesting that we have x³ ready to use here, which is x^3, but not the same for x^4.
Thank you, John, for interaction.

For previous case of Roll # = 6, the equality is:

1 + 3 + 5 + 7 + 9 + 11 = 6 * 6

Next, let us say gives... Roll # = 3:

for Roll # = 3, write down the equality here:

1 + ... = 3 * 3
_______________________________________

      
Now let us have even more fun!

Let's say that gives...

                   1
... Roll # = (∫x³dx)-¹
                  0

First, count Roll# = __

Then, write down the equality here:   _ + _ + ... = _ * _

Have fun

Let's now play a little game: !!

Come back to our original formula:

                                       1 + 3 + 5 = 3*3

How many numbers are on the left side of the above equality? That's right: 3 !

Follow the same reasoning for other numbers as it is done below:

Example: roll a 6-sided dice to receive amount of number participants in the game:

e.g., Roll #1 = 5 and you make equality of 5 members. Always start from 1 and add 2 to get next number:
                                                                1 + 3 + 5 + 7 + 9 = 5*5
next: Roll #2 = 2, then 1 + 3 = 2*2
Roll #3 = 6, then 1 + 3 + 5 + 7 + 9 + 11 = 6*6
Roll #4 = 4, then 1 + 3 + 5 + 7 = 4*4
Roll #5 = 1, then 1 = 1*1
Game: roll 5 times and count the sum of all 5 rolls. The highest sum wins.

Let's count our 5 rolls: 25 + 4 + 36 + 16 + 1 = 82.

Roll your dice 5 times and try to reach that 82.

Your turn...give me your Roll #1 = ___

... breaking down the puzzle :

if 

Now it is your turn:
2 + 4 + 6  = _ * [( _ + __ ) / 2] = _ * _  ?

Next, may you bring an example with integers?

“The Lady Tasting Tea. How Statistics Revolutionized Science in the XXth Century” by David Salsburg. The book author described discovery of the five-year-old in a few sentences (p.138, Chapter 14. THE MOZART OF MATHEMATICS). It does deserve our playfull minds... let's have fun and may  even create a game on the numbers right here.