A little maths problem for you

**Dual-Catfish** · 05-20-2002

No, all four values are expressed to the cent.

Unregistered · 05-20-2002

Is one of the numbers over $7.11? And for those of you who think this is impossible, it isn't. You can have a negative number to bring down the number above $ 7.11-
ex: $10+($-5)=$5
However this creates a problem- when you multiply, no matter what you will come out a negative number ( a negative number times a positve comes out negative, negative * positive * positive * positve is still negative ) To solve the problem we just add another number.

**Driveway** · 05-20-2002

Dope, that was me, forgot to log in.

**dbaryl** · 05-20-2002

A couple of things:

First, I want to make sure that this is not one of those wording problems, where there is some hidden thing like: He multiplied the name of the story by ... Is it right to assume that if the 4 items are priced a, b, c, and d then a + b + c + d = 7.11 and a * b * c * d = 7.11?

Now, if that is the correct case, I seriously don't see any other solution to this... except for one: negative values. I'm not sure if it's possible to achieve the result that way, but you can try by having 2 of the numbers be negative [negative goes away that way].

Can anyone who knows the answer at least hint as to what I'm doing wrong in my guess at the top of page 4?

**Dual-Catfish** · 05-20-2002

There are no negitive values. No value is over $7.11
$7.11 = a + b + c + d
$7.11 = a * b * c * d

**novacain** · 05-21-2002

there is no solution, apart from the clerk is on crack.
as dbarry said
to go futher
sqrt(711) < 27
if more than one one object is priced at greater than 0.27 then the product will be greater than 711 ie 0.27 * 0.27 * 0.01 * 0.01 = 7.29

so no two values can be greater than 0.27

The largest factor of 7.11 is 2.37 so this is the largest value possible. (0.01*0.01*0.03*2.37=7.11)

but if 2.37 is one value

7.11 - 2.37 = a+b+c
4.74 = a+b+c

as we have proved only one value can be above 0.27 there is no possible integer solution.

**OxYgEn-22** · 05-21-2002

damnit, i'm gone for one day and i thought you people wouldn't know my problem.. heh, figures the first person to read it would get it

**Jet_Master** · 05-21-2002

I worked my ASS off.

Yesterday, i worked for 1.5 hours on paper and these are the results i got.

bcd=2.
so a=(7.11/2) but that gives 355.5 cents. so i guess that is not possible.

also i managed to eliminate a. but the equation i got further afterthat resembles this: (everything is in cents)
3555+2(b+c+d)=bcd=2
.: 3555+2(b+c+d)=2
.: 2(b+c+d)=(-3553)
.: b+c+d=(-1776.5)
.: a=711+1776.5=355.5
.: 1065.5=355.5 ?!?!

quite illogical isn't it???
does that prove my theory that 0=infinity??
c'mon, accept it. my theory is CORRECT!!!

YES!!!
0=infinty

Ha ha ha!!!

**Jet_Master** · 05-21-2002

I simply LOVE this post. its the greatest. MATH rules!!!

**Dual-Catfish** · 05-21-2002

Apparently everyone thinks its impossible... Well, it's not, and I shall show you.
$3.16 $1.20 $1.25 $1.50 seem to work out to $7.11
http://mathforum.org/library/drmath/view/55897.html
http://www.mozart-oz.org/documentation/fdt/node21.html#section.elimination.grocery

There are many ways to do it... I used the now non-existant program.

**dbaryl** · 05-21-2002

Originally posted by OxYgEn-22
damnit, i'm gone for one day and i thought you people wouldn't know my problem.. heh, figures the first person to read it would get it

O2, if you are talking about thee 3-lightbulb problem, that one is a bit overused and I think evewryone knows it by now

**dbaryl** · 05-21-2002

>>0.27 * 0.27 * 0.01 * 0.01 = 7.29
I'm not sure that this is correct

Now, for Jet_Master: where did you come up with the fact that one of the values is 2? Is that a random guess?

One thing I'm sticking to is that the values have to be either $0.01, $0.03, $0.09, $0.79 or $2.37... those are the only integer factors of 711. I simply don't see any other way.

Now, about my reasoning a couple of pages back, I made a mistake in one of the assumptions, but this above does stay true, as far as I know.

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.

This is what I said, and I thinkit's correct. See if anything is wrong with this reasoning.

Unregistered · 05-21-2002

Anyone know why the code before with the floats didn't work? I converted the code to cents so all 4 items were multiplied by 100. So when multiplying the four items I just compared it with
$7.11 *100 ^ 4

Here's the code and it surprisingly works:

Code:

#include <stdio.h>

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

   for(a = 0; a <= 711; a++)
   {
      for(b = 0; b <= 711; b++)
      {
         if(a + b > 711)
            break;
         for(c = 0; c <= 711; c++)
         {
            if(a + b + c > 711)
               break;
            d = 711 - a - b - c;
            if(a * b * c * d == 711000000)
            {
               printf("a = %d\nb = %d\nc = %d\nd = %d\n", a, b, c, d);
               return 0;
            }
         }
      }
   }

   printf("No such answer\n");

   return 0;
}

**Driveway** · 05-21-2002

Here's a possibility, there is an answer:
all numbers are null set. In math, null set doesn't equal zero, it means no real numbers. Therefore, I say all four values are null set