Problem 42405. Divisible by 3

Created by goc3

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Pursuant to the first problem in this series, this one involves checking for divisibility by 3.

Write a function to determine if a number is divisible by 3. This can be checked by at least two different methods:

Sum the digits in the number. If the resulting sum is divisible by 3, then so is the original number. This routine can be applied recursively until the resulting sum is a single digit.
From the wikipedia reference page: "Subtract the quantity of the digits 2, 5, and 8 in the number from the quantity of the digits 1, 4, and 7 in the number." If the resulting sum (absolute value) is a multiple of 3, then so is the original number.

Previous problem: divisible by 2. Next problem: divisible by 4.

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Last Solution submitted on Nov 25, 2019

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Jean-Marie Sainthillier on 25 Jun 2015

I don't understand your restrictions. OK for Java functions, but here you need mod/rem no ?

goc3 on 25 Jun 2015

No, mod() or rem() isn't necessary because you can recursively apply the method, down to a single-digit number. The small number of single-digit numbers (i.e., 10, assuming absolute value) then makes checking against known divisible numbers trivial.

Alfonso Nieto-Castanon on 25 Jun 2015

I understand the java restriction, but the rest seem a bit heavy handed. Also the large number of similar problems might encourage solvers to build a general "divisible by n" routine instead

goc3 on 26 Jun 2015

I'm not trying to be authoritarian; just providing awareness to interesting mathematical trivia. There are some numbers, such as 7 and 11, for which a range of methods are possible. Without blocking a lot of those basic functions, very easy workarounds can be used, defeating the purpose of trying to optimize your score (akin to regexp, eval, etc. being banned due to defeating the scoring system).

goc3 on 26 Jun 2015

Also, building a general "divisible by n" routine is a good idea and I had thought about posting that at the end of the group of problems. I just think that it's too big for a standalone problem. There are many problems on Cody that are extremely simple and require no thought (like the recent string of beginner problems). There are also quite a few that require significant time, thought, and/or programming to solve, but they are all worth the same number of points. I prefer to break more difficult problems into simpler modules (just like is done in programming) to provide an encouraging approach, especially to those not already expert in Matlab. There are similarities between many of the problems in this set, but some numbers require quite different methods, such as 7 and 11.

Jan Orwat on 2 Jul 2015

Solved problems before reading comments and must agree with Alfonso. "divisible by N" approach worked nicely through all of them.

goc3 on 6 Jul 2015

Thanks for the comment. I'll be adding a few more problems to this set in the near future to round out and finish up the "divisible by N" approach.

Solution Comments

Solution 1890020

1 Comment

1 Comment

Nikolaos Nikolaou on 2 Aug 2019

Why is the test suite not visible ? Is it me (firefox display issue) or it is supposed to be that way?

Solution 1705227

2 Comments

2 Comments

Chris Breshike on 9 Jan 2019

How am I only failing test 12?
I get tf=0 when i run this in Matlab...

goc3 on 10 Jan 2019

That test case has the longest string input, by far. Your solution is coded to only sum digits twice. After those two steps, the resulting sum in this specific case is still greater than 9, and, therefore, fails your test of equal to a single-digit multiple of three (3, 6, or 9).

Solution 695335

1 Comment

1 Comment

Jean-Marie Sainthillier on 21 Jul 2015

First time I see ismembertol.Nice !

Solution 691474

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1 Comment

Andrew Newell on 18 Apr 2017

Is this machine dependent?

Solution 691434

1 Comment

1 Comment

Andrew Newell on 18 Apr 2017

I love the simplicity of this one!

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Problem 42405. Divisible by 3

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Problem 42405. Divisible by 3

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