A Pythagorean triplet is a set of three natural numbers, a b c, for which,
a^2 + b^2 = c^2
For example,
3^2 + 4^2 = 9 + 16 = 5^2 = 25.
There exists exactly one Pythagorean triplet for which a + b + c = N (the input).
Find the product abc.
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The description should be 9 + 16 = 5^2 and not 9+16 = 2^5 ?
Thanks Aurelien! Fixed it.
This is an interesting problem. Apparently everyone (including me) had to solve it by a basic brute force method with oly very slight variations. I did manage to use my MATLAB copy to use solve from the symbolic toolbox to get the answers, but the version of MATLAB on which Cody runs doesn't have access to a license for the symbolic toolbox.
@ Doug Hall, I have solved all the problems in the series. However, I have not received the badge and the associated scores on completion of the series. Can you please help me with this?
Wow, all the triangles are in the ratio of 8:15:17? Pretty huge oversight.
People have realized that all test cases are based on the primitive Pythagorean triplet (8, 15, 17) and the solutions are full of hacks. This is quite annoying as one cannot compare one's solution with other valid ones.
Test suite has been updated with additional test cases and leading solutions have been rescored.
Am I missing something, or are the problems with some of the test cases?
For test case 7:
x = 240
y_correct = 48e4
15^2 + 112^2 = 225 + 12544 = 12769 = 113^2
15 + 112 + 113 = 240, so this satisfies the requirement.
15 * 112 * 113 = 189840, which does not match the expected answer.
Similarly for test case 8:
x = 90
y_correct = 21060
9^2 + 40^2 = 81 + 1600 = 1681 = 41^2
9 + 40 + 41 = 90, so this satisfies the requirement.
9 * 40 * 41 = 14760 which does not match the expected answer.
Thanks for the heads up on the test cases. When I changed my code to spit out ALL the solutions, I was able to see that Test Case 7 has four solutions and Test Case 8 has two solutions. This contradicts the problem description that "There exists exactly one Pythagorean triplet for which a + b + c = N (the input)." If one's code assumes this, it is only happenstance whether it comes up with the same solution that the test case is looking for. In my code, it was always the last solution found, so for others who searched in the opposite direction, it may have been the first solution found.
Multiple solutions exist for multiple test cases
function ans = euler009(x)
k=[3,4,5;5,12,13;7, 24, 25;8, 15, 17;9, 40, 41;11, 60, 61;12, 35, 37;13, 84, 85;15,112,113;16, 63, 65;17,144,145;19,180,181;20, 21, 29;20, 99,101;21,220,221;23,264,265;24,143,145;25,312,313;27,364,365;28, 45, 53;28,195,197;29,420,421;31,480,481;32,255,257;33, 56, 65;33,544,545;35,612,613;36, 77, 85;36,323,325;37,684,685];
h=[primes(x)];
p=zeros(1,length(h));
for z=1:10
p(floor(x./h.^z)==x./h.^z )=p(floor(x./h.^z)==x./h.^z)+1;
end
h=h(p~=0)
p(p==0)=[]
x./sum(k,2)
t=k(ans==floor(ans),:)
ans=[t]
for z=1:length(p)
[ans;t.*(h(z).^p(z))]
end
ans(x==sum(ans,2),:)
prod(ans)
end