Problem 1319. Leap years on other planets

Created by Tim

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>A year is a leap year if it is divisible by 4, but not if it is divisible by 100, unless it is also divisible by 400. This means the average length of a calendar year is</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ 365 + 1/4 - 1/100 + 1/400 = 365.2425 days]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>which approximately matches the length of an actual year (the number of days it takes the earth to go around the sun once).</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Given p, the length of a year on another planet in units of days on that planet, and given that normal years have N days and leap years have N+1 days, where N=floor(p), find integer vector m&gt;0 defining leap years [years divisible by m(1) are leap years, but years divisible by m(2) are not, unless they are divisible by m(3), but then not if they are divisible by m(4), etc.] so that</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"code\"/></w:pPr><w:r><w:t><![CDATA[ p = N + 1/m(1) - 1/m(2) + 1/m(3) - 1/m(4) + ...]]></w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Include enough terms so that this is correct to four decimal places. Note that m(i+1) should be an integer multiple of m(i). Results are not necessarily unique. For example, if p=365.2425 then m=[4 100 400] is an exact answer, but m=[4 132 13068] also could be used (and [4 132] gives 365.2424, a better approximation than [4 100], which gives 365.2400).</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>"}]}

A year is a leap year if it is divisible by 4, but not if it is divisible by 100, unless it is also divisible by 400. This means the average length of a calendar year is<pre> 365 + 1/4 - 1/100 + 1/400 = 365.2425 days</pre>which approximately matches the length of an actual year (the number of days it takes the earth to go around the sun once).Given p, the length of a year on another planet in units of days on that planet, and given that normal years have N days and leap years have N+1 days, where N=floor(p), find integer vector m>0 defining leap years [years divisible by m(1) are leap years, but years divisible by m(2) are not, unless they are divisible by m(3), but then not if they are divisible by m(4), etc.] so that<pre> p = N + 1/m(1) - 1/m(2) + 1/m(3) - 1/m(4) + ...</pre>Include enough terms so that this is correct to four decimal places. Note that m(i+1) should be an integer multiple of m(i). Results are not necessarily unique. For example, if p=365.2425 then m=[4 100 400] is an exact answer, but m=[4 132 13068] also could be used (and [4 132] gives 365.2424, a better approximation than [4 100], which gives 365.2400).

Solve

Solution Stats

38 Solutions
10 Solvers

Last Solution submitted on Jan 13, 2019

Last 200 Solutions

Problem Comments

1 Comment

1 Comment

Tim on 6 Mar 2013

The test cases were calculated from actual data (length of day and tropical orbit period) at www.heavens-above.com, but they are not necessarily as well-defined or accurately known as the number of decimal places used here might imply.

Problem Recent Solvers10

Problem Tags

astronomy leap year

Problem 1319. Leap years on other planets

Solution Stats

Last 200 Solutions

Problem Comments

Problem Recent Solvers10

Suggested Problems

More from this Author10

Problem Tags

Problem 1319. Leap years on other planets

Solution Stats

Last 200 Solutions

Problem Comments

Problem Recent Solvers10

Suggested Problems

More from this Author10

Problem Tags

Players