Problem 42409. Divisible by 7

Created by goc3

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>Pursuant to the</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2\"><w:r><w:t>first problem</w:t></w:r></w:hyperlink><w:r><w:t> in this series, this one involves checking for divisibility by 7.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Write a function to determine if a number is divisible by 7. This can be done by a variety of methods. Some are:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Multiply each digit in the candidate number (from right to left) by the digit in the corresponding position in this pattern: 1, 3, 2, -1, -3, -2 (1 applies to the ones digit, 3 to the tens digit, etc.). This pattern should be repeated beyond the hundred-thousands digit. The resulting sum will be a smaller number. This method can be applied recursively to the absolute values of the resulting sum until a single digit results. Then, check that number for divisibility by seven.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>The previous method can also be used applying a slightly different pattern: 1, 3, 2, 6, 4, 5 (1 applies to the ones digit again, etc.). The absolute value of the resulting sum is not necessary as none of the factors are negative.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Multiply the last (ones) digit by two and subtract the result from the remaining number. Apply recursively, as noted in methods above.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Similar to the previous method, multiply the last digit by five and add to the remaining number. Apply recursively, as noted in methods above.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Double the last digit and subtract it from the remaining number (original number except for the ones digit). Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Similar to the previous method, multiply the ones digit by five and add it to the remaining number. Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Along similar lines, take the last three digits of the number and subtract that number from the remaining number. Once you reach a number less than 1000, another method can be applied to further reduce the number to check for divisibility by seven.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"2\"/></w:numPr></w:pPr><w:r><w:t>Etc. (there are others)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The restriction for multiplication has been lifted for this specific problem.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Previous problem:</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://www.mathworks.com/matlabcentral/cody/problems/42408-divisible-by-6\"><w:r><w:t>divisible by 6</w:t></w:r></w:hyperlink><w:r><w:t>. Next problem:</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://www.mathworks.com/matlabcentral/cody/problems/42410-divisible-by-8\"><w:r><w:t>divisible by 8</w:t></w:r></w:hyperlink><w:r><w:t>.</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>"}]}

Pursuant to the <a href = "http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2">first problem</a> in this series, this one involves checking for divisibility by 7.Write a function to determine if a number is divisible by 7. This can be done by a variety of methods. Some are:<ol><li>Multiply each digit in the candidate number (from right to left) by the digit in the corresponding position in this pattern: 1, 3, 2, -1, -3, -2 (1 applies to the ones digit, 3 to the tens digit, etc.). This pattern should be repeated beyond the hundred-thousands digit. The resulting sum will be a smaller number. This method can be applied recursively to the absolute values of the resulting sum until a single digit results. Then, check that number for divisibility by seven.</li><li>The previous method can also be used applying a slightly different pattern: 1, 3, 2, 6, 4, 5 (1 applies to the ones digit again, etc.). The absolute value of the resulting sum is not necessary as none of the factors are negative.</li><li>Multiply the last (ones) digit by two and subtract the result from the remaining number. Apply recursively, as noted in methods above.</li><li>Similar to the previous method, multiply the last digit by five and add to the remaining number. Apply recursively, as noted in methods above.</li><li>Double the last digit and subtract it from the remaining number (original number except for the ones digit). Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.</li><li>Similar to the previous method, multiply the ones digit by five and add it to the remaining number. Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.</li><li>Along similar lines, take the last three digits of the number and subtract that number from the remaining number. Once you reach a number less than 1000, another method can be applied to further reduce the number to check for divisibility by seven.</li><li>Etc. (there are others)</li></ol>The restriction for multiplication has been lifted for this specific problem.Previous problem: <a href = "http://www.mathworks.com/matlabcentral/cody/problems/42408-divisible-by-6">divisible by 6</a>. Next problem: <a href = "http://www.mathworks.com/matlabcentral/cody/problems/42410-divisible-by-8">divisible by 8</a>.

Pursuant to the <http://www.mathworks.com/matlabcentral/cody/problems/42404-divisible-by-2 first problem> in this series, this one involves checking for divisibility by 7.

Write a function to determine if a number is divisible by 7. This can be done by a variety of methods. Some are:

# Multiply each digit in the candidate number (from right to left) by the digit in the corresponding position in this pattern: 1, 3, 2, -1, -3, -2 (1 applies to the ones digit, 3 to the tens digit, etc.). This pattern should be repeated beyond the hundred-thousands digit. The resulting sum will be a smaller number. This method can be applied recursively to the absolute values of the resulting sum until a single digit results. Then, check that number for divisibility by seven.
# The previous method can also be used applying a slightly different pattern: 1, 3, 2, 6, 4, 5 (1 applies to the ones digit again, etc.). The absolute value of the resulting sum is not necessary as none of the factors are negative.
# Multiply the last (ones) digit by two and subtract the result from the remaining number. Apply recursively, as noted in methods above.
# Similar to the previous method, multiply the last digit by five and add to the remaining number. Apply recursively, as noted in methods above.
# Double the last digit and subtract it from the remaining number (original number except for the ones digit). Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.
# Similar to the previous method, multiply the ones digit by five and add it to the remaining number. Apply this recursively until a single digit results. If that number is divisible by seven, then so is the original number.
# Along similar lines, take the last three digits of the number and subtract that number from the remaining number. Once you reach a number less than 1000, another method can be applied to further reduce the number to check for divisibility by seven.
# Etc. (there are others)

The restriction for multiplication has been lifted for this specific problem.

Previous problem: <http://www.mathworks.com/matlabcentral/cody/problems/42408-divisible-by-6 divisible by 6>. Next problem: <http://www.mathworks.com/matlabcentral/cody/problems/42410-divisible-by-8 divisible by 8>.

Solve

Solution Stats

518 Solutions
85 Solvers

Last Solution submitted on Jan 13, 2022

Last 200 Solutions

Problem Comments

Solution Comments

Solution 691188

4 Comments

4 Comments

Show 1 older comment

James on 25 Jun 2015

I didn't see that the check for the multiplication symbol was commented out for this test suite until I checked your solution...

goc3 on 25 Jun 2015

Sorry about that. I added a statement that that function restriction is lifted in the problem description. Still, I'm impressed that you were able to solve it without using that function.

James on 25 Jun 2015

No worries. Multiplication without the * symbol is easy enough. There's a reason the final variable in my first solution is named "cheat" though; I had no idea how to accomplish the final check otherwise.

goc3 on 26 Jun 2015

Well, I'll take that as a compliment, then, because your problems are clever and often stump me.

Problem Recent Solvers85

Problem Tags

Community Treasure Hunt

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

Start Hunting!