The 3 integers I have in mind sum up to 35 and their product is 720.
Please find them.
(In reply to
Which did you have in mind? by Jer)
My program modified to allow negative integers supports completeness of Jer's set of solutions.
clearvars,clc
f1=divisors(720);
for i=1:length(f1)
n2=720/f1(i);
f2=divisors(n2);
f2=f2(f2>=f1(i));
for j=1:length(f2)
n3=n2/f2(j);
if n3>=f2(j)
for a=[f1(i) -f1(i)]
for b=[f2(j) -f2(j)]
for c=[n3 -n3]
set=[a b c];
if sum(set)==35 && prod(set)==720
disp([a b c sum(set) prod(set)])
end
end
end
end
end
end
end
integers sum prod
-2 -8 45 35 720
3 12 20 35 720
5 6 24 35 720
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Posted by Charlie
on 2023-02-27 08:38:01 |
re: Which did you have in mind?
