Hard question!!!!!!!!!!?

Question

is it possible to do:

A^2 + 3B^2 + 3C^2 + 5D^5 = 1002
??????

all positive integers

all variables are different to eachother

Answer 1

I just corrected some mistakes. If you see any others, please excuse them. It is now 3:30 in the morning where I live, and I'm getting a bit weary.

Well, to start with . . .
D must be less than 3, because 5*3^5 > 1002.
Also, notice that 1002 is divisible by 3. So, the other side has to be divisible by 3 as well. 3B^2 and 3C^2 are already divisible by 3, so we have to focus on A^2 and 5D^5. As it turns out, 5D^5 will give a remainder of two when divided by three for any positive integer value of D, because it's multiplied by 5 (a number which has remainder 2 when divided by 3). So, (A^2)/3 must yield a remainder of 1.
That limits values of A^2 to the following:
1
4
16
25
49
64
100
121
169
196
256
289
361
400
484
529
625
676
784
841
961
(I used a program to generate these, because it would have taken a while otherwise).
That isn't so important right now. (A^2)/3 has remainder 1, and if you subtract it from the other side, you get some number with remainder 2. (5D^2)/3 will have to have remainder 2 to restore the balance among the remainders on both sides. D<3, so we try 2 first. (5*2^5) = 160; 160/3 has remainder 1, so it cannot be a possible solution. Then we try D=1: 5*1^5 = 5; 5/3 has remainder 2, so D=1.
Let's see what we have now . . .
A^2 + 3B^2 + 3C^2 = 997
And 1 is removed from the possible values for A^2.
Let us think about this for a moment . . .
The smallest possible values for B^2 and C^2 at the moment are 4 and 9. It doesn't matter which one is which because they both have the same coefficient. If you plug those into the equation, you get A^2 = 958, which is not one of the choices. This also means that A^2 must be smaller than 958. Let's try A^2 = 841.
If A^2 is 841, then 3B^2+3C^2 = 997-841 = 156.
Dividing both sides by 3, we get B^2+C^2 = 52.
Hmmm . . . that's 36+16! So B is 4 and C is 6 (or vice versa).

In summary, one (actually two, because 4 and 6 are interchangeable among B and C) solution that is in the set of positive integers is:
A = sqrt(841) = 29
B = 4 or 6
C = 6 or 4
D = 1
There may be others, but they have to satisfy the constraints I came up with above.
Does that answer your question or what? Have fun! ;-)

Edit: OK, on second thought I see that I made an error in my process above. The value of (5D^5)/3 can have remainder 0, 1, or 2. However, because the only values of D that satisfy the equation are 1 and 2, the only possible remainders are 1 and 2 (because 5/3 has remainder 2 and 160/3 has remainder 1). So, that might introduce another set of possible values for A^2: those perfect squares which yield a remainder of 2. Again, I generate a list with a program:
Hmmm . . . that's interesting . . . there are no perfect squares less than 1002 which yield a remainder of 2 upon division by 3. So my answer above is right even though my reason for eliminating 2 as a possible value for D was incorrect. Have fun!

Answer 2

A=31
So A^2=961

B=3
so 3B^2=27

C= Square root of 3
so 3C^2=9

D=1
so 5D^5=5

Adding all of them we get

961 + 27 + 9 + 5 = 1002

Answer 3

hmmm.. 4 variables.. I think you will need about 3 more equations to solve this.. .. or more restrictions of somekind.. like.. all integers... all positive.. etc... otherwise.. you can just pick some numbers.. say... A = sqrt(991), B = 1, C = 1, D = 1

Answer 4

Many, many solutions.

One would be a = 30
b = 5.83095
c = 0
d = 0

Answer 5

Possible to do what?