what are all possibilities for a,b,c to be 72?
6 views (last 30 days)
Show older comments
Quinten Lodewijks
on 28 Mar 2017
Commented: Quinten Lodewijks
on 29 Mar 2017
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.
7 Comments
Tomas Hipolito
on 28 Mar 2017
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.
Accepted Answer
More Answers (1)
Jan
on 29 Mar 2017
Edited: Jan
on 29 Mar 2017
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
See Also
Categories
Find more on Logical in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!