what are all possibilities for a,b,c to be 72?

a = [0:1:20]; b = [0:1:20]; c = [0:1:20];

i want to know what all the possibilities are of the product of a*b*c to be 72

i tried with an if statement but this doesn't work.

[Adam]

What does 'a,b,c' mean?

[KSSV]

what is that possibility?

[John D'Errico]

a+b+c = 72?
a*b*c = 72? 
a+b*c = 72?
a*b-c = 72?

or something else completely?

[Tomas Hipolito]

I understand you want to find all the values of a, b and c for that a*b*c=72. To do that you can do three For cycles (one for each variable), generate a vector with the values of the variables each time they match 72 and save that in a cell. Something like that

 Final_set={};
  for i =1:length(a)
   for j=1:length(b)
     for k=1:length(c)
         Product=a(i)*b(j)*c(k);
         if Product == 72
         Solution={[a(i) b(j) c(k)]};
         Final_set={Final_set;Solution};
         end
      end
    end
   end

In the end Final_set has the set of combinations possible to your variables. There might be a more efficient way to do this, but this solves the problem.

[Quinten Lodewijks]

its a*b*c

[Quinten Lodewijks]

and Tomas, how do i have to use that? do i have to put a,b,c before the code and execute the whole section or how do i need to use your code?

anyway thanks for replying!

[Tomas Hipolito]

You need to define a,b and c before the code. a, b and c vary from 0 to 20, right? In Matlab you define them as vectors as you did in your question. I didn't try the code, there is the possibility to have a syntax error. So it would be

 a=[0:1:20];
 b=[0:1:20];
 c=[0:1:20];
 and the rest of the code written above.
 Tell me if it worked.

Answer 1

[Thorsten]

a = nchoosek(1:72, 3);
idx = prod(a, 2) == 72;
a(idx, :)

[Quinten Lodewijks]

thanks!

[Thorsten]

I think that the following is correct:

a = []; 
for i = 1:72, for j = 1:72, for k = 1:72, 
  if i*j*k == 72, a(end+1,:) = [i, j, k]; end
end, end, end

Answer 2

[Jan]

The allowed range is 1:20 for the 3 elements (ignoring the 0), not 1:72.

P   = nchoosek(1:20, 3)
idx = prod(P, 2) == 72;
P(idx, :)

Or equivalently for Torsten's comment: for i = 1:20, ...

The loops can be stopped prematurely:

Result = []; 
for i1 = 1:20
  for i2 = 1:20
    for i3 = 1:20,
      p = i1 * i2 * i3;
      if p == 72
        Result(end+1, :) = [i1, i2, i3];
        break;   % Former products for larger i3 are > 72
      elseif p > 72
        break;   % Former products for larger i3 are > 72 also
      end
    end
  end
end

We know the prime factors of 72:

factor(72)
% 2     2     2     3     3

This means that we do not have to check values, which cannot be created by numbers, which cannot be build as a product of these values (and the 1):

Pool   = [1, 2, 3, 4, 6, 8, 9, 12, 18];
Result = []; 
for i1 = Pool
  for i2 = Pool
    for i3 = Pool
      p = i1 * i2 * i3;
      if p == 72
        Result(end+1, :) = [i1, i2, i3];
        break;   % Former products for larger i3 are > 72
      elseif p > 72
        break;   % Former products for larger i3 are > 72 also
      end
    end
  end
end

[Quinten Lodewijks]

Thank you so much! this solved the problem for me. The code was longer than I expected it to be haha