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Thanks Guest,

I am glad that we are now in agreement. That is always comforting.     

x =  4y - a

y  = 4x  + b

And we know that

1 = 4(2) - a

2 = 4(1) - b        simplify

1  = 8 - a

2  =  4  - b

a  =  7

b = 2

a + b  =    9

The x coordinate of the vertex of  f  =   -a/2

The x coordinate of the  vertex of g   = -c/2

So    (-c/2)  is a root of  f

And (-a/2)  is a root of g

So

(-c/2)^2  - a(c/2)  +  b   = 0

(-a/2)^2  - c(a/2)  +  d  = 0              simplify

c^2/4 - ac/2  + b  = 0

a^2/4   - ac/2  +  d  = 0           multiply both equations through by 4

c^2  -2ac  + 4b   =  0

a^2  - 2ac  + 4d  = 0            subtract the second equation  from the  first

c^2  - a^2  +  4 ( b - d)  =  0

(c + a) (c - a)  =     4 ( d - b)      (1)  

And since  f and  g intersect at  (100, -100)   then

100^2  + 100a  + b  =  100^2 + 100c + d

100a + b   =  100c + d         rearrange  as

100(a - c)  = (d - b)        multiply both sides  by 4

400 (a - c)  =  4 (d - b)      (2)

Sub (2)   into (1)

(c +a)(c - a)  =  400(a - c)        divide both sides by  ( c - a)

(c + a)  =   400( a - c)  / ( c - a)

(a + c)  =  -400 ( c - a)  / (c - a)

a + c   =  -400

Hi Melody: Bravo!!! You were right after all. I simply computed the permutations of the (3,4,5,6) =4 x 4 =16 and presto got the same number that you did. I believe my mistake was in using 6 C 4, which I believe now to be wrong, because 4 are 2 separate entities(1,1 and 2,2).
112233    = ........................................................90 permutations  
112234  112243    =..........................................360p
112235  112244  112253 =...............................450p
112236  112245  112254  112263=..................720p  
112246  112255  112264= ...............................450p  
112256  112265=..............................................360p  
112266=..............................................................90p
                                                GRAND TOTAL =2,520 PERMUTATIONS.  
 

By the graph here : https://www.desmos.com/calculator/zascepyruy

Note that :

The height  of the triangle  =   x^2

And the base of the triangle  =  2x

So   we have  that

(1/2) x^2  * ( 2x )   =  8

x^3  =  8

x = 2

So the minimum possible height =  2^2  = 4

So   y   =  3

And

(1/2) x^2  * (2x)  =  64

x^3  =  64

x  = 4

So....the max possible height =  4^2  = 16

So  y =  15

So

3  ≤  r  ≤  15

Thanks for all that information, I do appreciate your efforts.

But I do not think that is correct.

The way i have done it, 

Is to think of every dice as being distinguishable.  Like they are all different colours so

if x and y are different (3,4) is different from (4,3)

So what I think is,

You have found how many combinations there are that satisfy the conditions given that the dice are indistinguishable.

then

You have divided by by the total number of ways WHEN the dice are different from each other, (distinguishable)

So I think you have mixed identical dice in the numberator with different looking dice in the denominator.

So I still think my answer is more likely to be correct than your is.

Here's one more way :

Let    x   =    12 + a

Let  y  = 12 +  b

1 /  [ 12 + a]   +   1 / [12 + b]    =  1 /12

[ 24  + a + b ]  /  [  144 + 12a + 12b +ab ]  =   1/12

12* [ 24+ a + b ]   =   [ 144 + 12a + 12b + ab ]

288  + 12a + 12b  =  144 + 12a + 12b  + ab

144  = ab 

So

a      b

12   12

16    9

18    8

24    6

36    4

48    3

72    2

144  1

The  smallest value  of  x +  y  =  

12 + 16 ,  12 +  9   =

28  ,    21   =

49

i=1;p=0;a=1;m=13;cycle:b=a%m; if(b*i % m==1,goto6, goto8);printa," - ","The mmi = ",i;p=p+i;i++;if(i<=(m-1), goto cycle, 0); i=1;m=13;a++;if(a<=12, goto cycle, 0);print"Total of All mmis =",p
 

1  -  The mmi =  1
2  -  The mmi =  7
3  -  The mmi =  9
4  -  The mmi =  10
5  -  The mmi =  8
6  -  The mmi =  11
7  -  The mmi =  2
8  -  The mmi =  5
9  -  The mmi =  3
10  -  The mmi =  4
11  -  The mmi =  6
12  -  The mmi =  12
Total of All mmis = 78
78 mod 13 = 0

No. It is prime. -MPS101

I actually was working on this problem while you were, I also figured it out!!!

y = x^2 + a

y  = ax

Set the y's equal

x^2  +  a  =  ax

x^2  - ax  + a    =  0

If we only have one solution, the discrminant  must  =  0

So

a^2  -  4a  =   0      factor

a ( a - 4)  =   0

The solutions are    a  =  0    or a  =   4

So....the sum of all possible  a's    =   4

We have 2 of 13  ranks from which  to choose the first pair  = C(13,2)

And we  have C (4,2)  ways to choose the cards within each rank  = [ C(4,2) ]^2

And we have  44 ways to choose the last card....so

C(13,2) * [C(4,2)]^2  * 44  = 

123,552  ways

(2020) ( 2022)  =

(2021 - 1) ( 2021  + 1)  =

(2021)^2   -  1

Note  that   ( base  m )^2     =    100_m 

The sum of the digits  =   0

There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton.  So:  13*3*12*3*11*4 = 61776 in total.

I think that Guest noticed that this question looked similar to the link you posted, so he assumed it was the same one, since both solutions are the same.

Thank you!

thank you!!

thank you!!

If we let x  =   - b.......we have that

(-b)^2  + 2b(-b)  +  ( a  - b)    = 

b^2  - 2b^2  +  (b^2 + b - b)   =

b^2  - 2b^2   +  b^2   =

2b^2  - 2b^2  =

0

Possibly! If you know one of the advanced programming languages such as C++, you will have access to a vast library of all kinds of numbers, including primes.

Of course NOT !!! That is why it is a PRIME NUMBER, which means it cannot be broken any further.

1/ x   +   1/ y   =   1 / 12

[ x + y ]  / xy  =   1 / 12

12 [ x + y ]    = xy

12x + 12y  = xy

12y - xy  =  - 12x

y ( 12 - x)  =  - 12x

y  =        12x

            _______

              x - 12

We  will  have positive integers for x, y   when

x         y

13     156

14      84

15      60

16      48

18      36

20      30

24      24         

28      21

The smallest  sum   of  x , y    when   x ≠ y   is    49

Thanks to EP  for spotting my earlier error   !!!!

The side of the large square  =   sqrt (25)  =  5

The side of the small square =  sqrt (7)

The smaller  dimension of a rectangle, d, can be found as

2d  + sqrt (7)  = 5

2d =  5 - sqrt (7)

d  =  [ 5 - sqrt (7) ]  / 2  =    (5/2) - sqrt (7) / 2

So  the larger dimension, D  =  5 - [ 5/2 - sqrt(7) / 2 ]  =  5/2 + sqrt (7)  / 2

So.....the  length of the diagonal  of a rectangle  = 

sqrt [ d^2 + D^2 ]  = 

sqrt  [ ( 5/2 - sqrt(7) / 2 )^2   + ( 5/2 + sqrt (7)/2)^2 ]  =

sqrt [ 25/4  + 7/4  + 25/4  + 7/4]  =

sqrt [ 64/4]  =

sqrt [ 16]  =

4