A little maths problem for you

**dbaryl** · 05-20-2002

[EDIT]Highlited in red is incorrect!
I pick A, B, C, and D to be prices for the items being sold (in cents). The product will look like this:
A * B * C * D = 711
(A * B * C) * D = 711
>> 711 % (A * B * C) == 0

What I mean by above is that EACH of the items bust be an INTEGER FACTOR of 711, in order to divide 711 cents by either of the prices and get an integer value in cents again. This is based on the assumption that the prices are integer values in cents, can't have a dicimal of a cent in price.

Now, the integer factors of 711 are: 1, 3, 9, 79 and 237.

This is the same as:

$0.01, $0.03, $0.09, $0.79 or $2.37

Let's theorize that all items are priced below $2.37, that is either $0.01, $0.03, $0.09 or $0.79. This would mean that you must be able to get a sum of $7.11 out of 4 items that are $0.79 or below, which is impossible. If only 1 item is $2.37, then the other 3 must add to 7.11 = 2.37 = 4.74 >> impossible again with 3 items below $1. So, at least 2 items must be priced at $2.37 each, leaving other 2 to be: 7.11 - (2 * 2.37) = 2.37

As you can see, it's impossible to have 2 items priced $0.01, $0.03, $0.09, $0.79 or $2.37 to add to $2.37 without them being $0 and $2.37. Now, this again is impossible, because the product must be a non-zero value.

I see no solution for this probem.

**Jet_Master** · 05-20-2002

YES!!! Ha Ha Ha!!!

This IS getting INTERESTING!!! Ha Ha Ha !!!

I LOVE this thread!!!

**Jet_Master** · 05-20-2002

dbaryl, thanks a lot for the info you have provided. I WILL use all that to find any possible solution if possible. if not possible, i will not possibly find an answer that is possible and is correct. but if there is a possible answer, i possibly can find it! just wait for the possible; or for some, my accptance of the impossible.

(either way, i am sure that i have confused you by now)

**Dual-Catfish** · 05-20-2002

There is an absolute answer... I once had a program written in C which calculated the answer(s) but I lost it when my HD crashed.

I know the values, but the program would be damn cool to have now

**Jet_Master** · 05-20-2002

the closest i could get was:
the items cost $1.00, $2.91, $1.94, $1.26

the sum is $7.11
but the product is $7.113204 - when you round this up to two decimal points, i get: $7.11!!

but, i didnot use programming. just my mind and a calculator.

**Dual-Catfish** · 05-20-2002

Well, sadly, that's wrong. It's not an exact value..
I may be able to recreate the program which calculated the answer, but I doubt it :/

**Jet_Master** · 05-20-2002

>>Well, sadly, that's wrong. It's not an exact value..

i thought so. my program cannot calculate anything. this is how my program is:
i declared 4 variables - a,b,c,d

a+b+c+d=7.11
a*b*c*d=7.11

isn't that right. correct me if i am wrong.

also,
a=7.11-(b+c+d)
b=7.11-(a+c+d)
c=7.11-(a+b+d)
d=7.11-(a+b+c)

when i do that, it calculates the first statement first. so, it gives my a value=7.11 and all others 0. so it satisfies the first part of my function. but i dont know why it neglects the part (a*b*c*d=7.11)

i hope you understand what i am saying...

Unregistered · 05-20-2002

dbaryl is right

but here's a piece of code that'll prove it:

Code:

#include <stdio.h>

int main()
{
   int i,j,k;
   int a,b,c,d;

   for(i = 1, a = 1; i <= 711; i++, a++)
   {
      for(j = 1, b = 1; j <= 711; j++, b++)
      {
         for(k = 1, c = 1; k <= 711; k++, c++)
         {
            d = 711 - a - b - c;
            if(a * b * c * d == 711)
            {
               printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
               return 0;
            }
         }
      }
   }

   printf("No such answer\n");

   return 0;
}

Unregistered · 05-20-2002

oops my bad

here's the revised code:

Code:

#include <stdio.h>

int main()
{
   double a,b,c,d;

   for(a = 0; a <= 7.11; a += 0.01)
   {
      for(b = 0; b <= 7.11; b += 0.01)
      {
         for(c = 0; c <= 7.11; c += 0.01)
         {
            d = 7.11 - a - b - c;
            if(a * b * c * d == 7.11)
            {
               printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
               return 0;
            }
         }
      }
   }

   printf("No such answer\n");

   return 0;
}

**dbaryl** · 05-20-2002

Originally posted by Jet_Master
dbaryl, thanks a lot for the info you have provided. I WILL use all that to find any possible solution if possible. if not possible, i will not possibly find an answer that is possible and is correct. but if there is a possible answer, i possibly can find it! just wait for the possible; or for some, my accptance of the impossible.

(either way, i am sure that i have confused you by now)

No, you didn't.

**dbaryl** · 05-20-2002

Oh, a couple more things... those of you who say that there is a possible answer, the only way I see it is if the prices are some ugly "non-nice" decimal. If you check the first post on this page, that is where I show my thoughts on this... if you see anthing wrong, tell me, otherwise, my answer stands: no solution.

Unregistered, how about this one?

Code:

#include <stdio.h>

int main()
{
   printf("No such answer!\n");

   return 0;
}

**Dual-Catfish** · 05-20-2002

Unregistered is on the right track... that is somewhat what the program looked like, but the logic is still off.

If you check the first post on this page, that is where I show my thoughts on this... if you see anthing wrong, tell me, otherwise, my answer stands: no solution.

Well, there is a solution, I have it right here... so I guess you're wrong :/

**dbaryl** · 05-20-2002

Is my assumption incorrect? Or are the 3 values something other tahn int values of cents? Tell me that much.... pliz.

**Commander** · 05-20-2002

I worte a progra the second i saw thins q, to figure out the ans I'm attaching it, I ran the damn thing for 8 hours, still no hit!!!!!!!!!!!!

**Commander** · 05-20-2002

And here's the code

PHP Code:


#include <stdio.h>

#include <stdlib.h>

#include <conio.h>

#include <string.h>

#include <math.h>









void main(){



float a,b,c,d;



for(;;){



a= (rand() / 65654.356);    // I got these Numbers

c= (rand() / 3215.265);      //by the rand() function

b= (rand() / 3541.065);      // Therefore, they are 

d= (rand() / 12456.265);    //random numbers!



printf("\n%.2f + %.2f + %.2f + %.2f == %.2f", a,b,c,d, a+b+c+d);

printf("\n %.2f * %.2f * %.2f * %.2f == %.2f", a,b,c,d, a*b*c*d);



if( ((a+b+c+d)==7.11) && ((a*b*c*d)==7.11) ){



system("pause");}

}

}