Problem 1672. Leftovers? Again?!

Created by James

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{"parts":[{"partUri":"/matlab/document.xml","contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?><w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>I am thinking of a positive number X. To determine what number I am thinking of, I will give you two 1xN vectors. The first vector (V1) is several numbers, none of which will share a factor. The second vector (V2) is the remainder of X when divided by each of the numbers in V1. Calculate what the lowest possible value of X can be given these criteria. For example:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>V1 = [2 3] ; V2 = [1 2]</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>would give an X value of 5. There are an infinite number of other values of X that would satisfy V1 and V2, but I want the lowest one.</w:t></w:r></w:p></w:body></w:document>","relationship":null}],"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","target":"/matlab/document.xml","relationshipId":"rId1"}]}

<div style = "text-align: start; line-height: 20px; min-height: 0px; white-space: normal; color: rgb(0, 0, 0); font-family: Menlo, Monaco, Consolas, monospace; font-style: normal; font-size: 14px; font-weight: normal; text-decoration: none; white-space: normal; "><div style="display: block; min-width: 0px; padding-top: 0px; transform-origin: 332px 82.5px; vertical-align: baseline; perspective-origin: 332px 82.5px; "><div style="font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-bottom: 9px; margin-left: 4px; margin-right: 10px; margin-top: 2px; text-align: left; transform-origin: 309px 42px; white-space: pre-wrap; perspective-origin: 309px 42px; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; "><span style="display: inline; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; transform-origin: 0px 0px; perspective-origin: 0px 0px; "><span style="">I am thinking of a positive number X. To determine what number I am thinking of, I will give you two 1xN vectors. The first vector (V1) is several numbers, none of which will share a factor. The second vector (V2) is the remainder of X when divided by each of the numbers in V1. Calculate what the lowest possible value of X can be given these criteria. For example:</span></span></div><div style="font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-bottom: 9px; margin-left: 4px; margin-right: 10px; margin-top: 2px; text-align: left; transform-origin: 309px 10.5px; white-space: pre-wrap; perspective-origin: 309px 10.5px; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; "><span style="display: inline; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; transform-origin: 0px 0px; perspective-origin: 0px 0px; "><span style="">V1 = [2 3] ; V2 = [1 2]</span></span></div><div style="font-family: Helvetica, Arial, sans-serif; line-height: 21px; margin-bottom: 9px; margin-left: 4px; margin-right: 10px; margin-top: 2px; text-align: left; transform-origin: 309px 21px; white-space: pre-wrap; perspective-origin: 309px 21px; margin-left: 4px; margin-top: 2px; margin-bottom: 9px; margin-right: 10px; "><span style="display: inline; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; transform-origin: 0px 0px; perspective-origin: 0px 0px; "><span style="">would give an X value of 5. There are an infinite number of other values of X that would satisfy V1 and V2, but I want the lowest one.</span></span></div></div></div>

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Solution Stats

149 Solutions
14 Solvers

Last Solution submitted on Nov 26, 2020

Last 200 Solutions

Problem Comments

8 Comments

8 Comments

Show 5 older comments

Paul Berglund on 5 Jul 2013

I think that there's a bug in the test cases; putting
clear V1;
before test case 7 should fix it.

James on 8 Jul 2013

clear V1 has now been added before test case 7, and the solutions are being rescored. Hopefully, this will fix the problems you were having.

rifat on 14 Dec 2014

Chinese remainder theorem

http://en.wikipedia.org/wiki/Chinese_remainder_theorem

William on 10 May 2019

I have a solution that solves all of the test cases, but when I run it against the test suite it produces "Error using builtin
Undefined function or variable 'solutionTest'." for each case.

Karl Ezra Pilario on 12 May 2020

I have the same problem as William. Can someone help?

Rafael S.T. Vieira on 18 Aug 2020

Apparently James is deleting essential files for Cody by using rm at his script. I've informed him of the issue.

goc3 on 29 Sep 2020

I have commented out the initial lines so that this problem is solvable again.

Rafael S.T. Vieira on 29 Sep 2020

Thanks, goc3.

Solution Comments

Solution 290803

1 Comment

1 Comment

Claudio Gelmi on 27 Jul 2013

Sorry about that...it was only for learning purposes... :-)

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leftovers modulo

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Problem 1672. Leftovers? Again?!

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Problem 1672. Leftovers? Again?!

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Last 200 Solutions

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