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

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Quinten Lodewijks
Quinten Lodewijks 2017-3-28
评论： Quinten Lodewijks 2017-3-29
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.
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Quinten Lodewijks
Quinten Lodewijks 2017-3-28
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
Tomas Hipolito 2017-3-28
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.

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采纳的回答

Thorsten
Thorsten 2017-3-28
编辑：Thorsten 2017-3-28
a = nchoosek(1:72, 3);
idx = prod(a, 2) == 72;
a(idx, :)
  2 个评论
Thorsten
Thorsten 2017-3-29
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

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Jan
Jan 2017-3-29
编辑：Jan 2017-3-29
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
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Quinten Lodewijks
Quinten Lodewijks 2017-3-29
Thank you so much! this solved the problem for me. The code was longer than I expected it to be haha

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