Problem 42935. Sums of cubes and squares of sums

Created by John D'Errico

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Given the positive integers 1:n, can you:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[1. Compute twice the sum of the cubes of those numbers.\n2. Subtract the square of the sum of those numbers.\n3. Divide that result by n/2.]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>So, for n = 3, we might compute a result like this:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[((1^3 + 2^3 + 3^3)*2 - (1 + 2 + 3)^2)/(3/2)\nans =\n    24]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Yes, you probably can do all of this, but be careful on this problem, as n may be somewhat large, and I am expecting to see the correct result, not just an approximate value. Remember there are always different ways one may solve a problem.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>I point out the Project Euler reference because PE problem 6 is what made me think of this problem, and because the test cases will push the limits of what you can do if you are not careful.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

Solve

Solution Stats

1399 Solutions
180 Solvers

Last Solution submitted on Sep 27, 2022

Last 200 Solutions

Problem Comments

2 Comments

2 Comments

John D'Errico on 29 Aug 2016

I did not originally look to see exactly how far one could go. It looks like, IF one is careful in the important expression, you should be able to get to 3329020. I'd expect that the code I'd write that would solve this to go as far as 3329020 would employ an initial test to know if n is even or odd, changing the expression I'd write depending on the parity of n. Of course that branch would also increase the complexity, so raising the Cody score.

A C on 16 Jan 2021

Can someone point me in the right direction?
n = 1e5;
y_correct = uint64(500010000050000);
I get 500010000050005. I do not know how to go beyond double precision.

function y = cubesLessSquare(n)
% Formula
N = (((sum((1:n).^3)*2 - sum(1:n)^2)*2/n))
% double to char
C = sprintf('%f',N)
% char to vector
Y = C - '0'
% take out the dot and the numbers after the dot
Dot = find(Y<0)
Y([Dot:end]) = []
% join the the numbers
Z = sprintf('%d',Y)
% make to double
y = str2num(Z)
end

Solution Comments

Solution 2810326

1 Comment

1 Comment

Rafael S.T. Vieira on 8 Aug 2020

This appears to be true only for sequences of numbers starting from 1 and some combinations of these sequences.

Solution 1532083

1 Comment

1 Comment

Cai Cai on 21 Nov 2020

unbelievable!get it

Solution 1532027

2 Comments

2 Comments

bainhome on 17 May 2018

it is exactly the right way to solve this one.

Jiawei Gong on 9 Apr 2020

Could you please explain a bit about how the problem was converted to a polynomial function?

Solution 944984

1 Comment

1 Comment

Jan Orwat on 29 Aug 2016

works for n up to 3329020

Solution 944708

1 Comment

1 Comment

Jan Orwat on 29 Aug 2016

works for n up to 2642245, so it doesn't use the full potential of the uint64

Solution 944705

1 Comment

1 Comment

Peng Liu on 28 Aug 2016

isequal ignores the data type of the values in determining whether they are equal.

isequal(uint64(15503197751395200),15503197751395199,15503197751395201,15503197751395200)

flintmax('double') = 2^53 < 15503197751395200

eps(15503197751395200) = 2

Solution 944703

1 Comment

1 Comment

Jan Orwat on 28 Aug 2016

Big letters, small letters and a bit of luck.

Problem Recent Solvers180

Problem Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Problem 42935. Sums of cubes and squares of sums

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers180

Suggested Problems

More from this Author4

Problem Tags

Community Treasure Hunt

Problem 42935. Sums of cubes and squares of sums

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers180

Suggested Problems

More from this Author4

Problem Tags

Community Treasure Hunt

Players