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Find natural numbers>0 divisible by 37 using FindInstance?

Posted 5 years ago
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As part of a larger Problem I have to find natural numbers>0 that are divisible by 37, which is a trival Problem. I write:

b1[n_] := Divisible[n, 37]
FindInstance[{n > 0, b1[n]}, n, Integers]

I get:

FindInstance::naqs: n>0&&Divisible[n,37] is not a quantified system of equations and inequalities. >>

The same happens wit Solve. What is wrong with that Code?

Actually I have to find numbers which respect a lot of logical conditions, some of them quite complex. That is why I put the conditions into seperate functions.

But even if I write:

FindInstance[{n > 0, Divisible[n, 37]}, n, Integers]

I get the same error.

13 Replies

Divisible is somehow not 'recognized' but you can circumvent it by using a code like:

FindInstance[{n > 0, n == 37 m}, {n, m}, Integers]
In[1]:= b1[n_] := Mod[n, 37] == 0
FindInstance[{n > 0, b1[n]}, n, Integers]

Out[2]= {{n -> 37}}
Posted 5 years ago

Thanks Sander and Veleriu.

Of course I can test divisibility by other means than Divisible[n,m]. But what is special with the built-in function Divisible that it does not work?

Posted 5 years ago

Another example ist

In[151]:= FindInstance[Apply[Times, IntegerDigits[n]] == 23, n, Integers]
Out[151]= {{n -> 23}}

which looks for numbers with a product 23 of their digits. There are none of course since 23 ist prime. But FindInstance delivers 23, which is wrong.

I think FindInstance is designed for geometry, polynomials, and booleans. I don't think they had in mind applying those kind of operations for this function...

The one you mention seems to be a bug? Please report it.

I am not sure, but it seems that the problem

 FindInstance[Apply[Times, IntegerDigits[n]] == 23, n, Integers]

or in mathematical notation:

$x_1 * x_2 * \dots * x_n = a$

$x_i \in \{0, \dots, 9 \}, i =1, \dots, n$

may be related to knapsack problem, one of the very known NP-complete problems. The essential difference is operation * instead of +. If so, the problem is very difficult by itself.

Posted 5 years ago

It might be that the Problem is not solvable by FindInstance. But it should not deliver wrong results

You're absolutely right!

Posted 5 years ago

Hmm, Sander, very strange.

Two questions since I am new to Wolfram Language:

  1. Can I report bugs directly from the community threads? Or only via email out of the language environments?
  2. I cannot find any settings for the community language Settings. Since my location is Germany, it uses German spelling always. Which is annoying.

The spell check is (i guess) your browser. I'm in France, but I have everything in English since I'm Dutch my French is limited.

Bugs are submitted here: category: "Product feedback"

http://www.wolfram.com/support/contact/email/

Posted 5 years ago

Thanks, Sander. I was using Edge which seams to have no settings for languages. Now I tried IE and could install the language English (US). But I still have no spelling correction. I will find it some day :-)

I think the error message explains the problem. Divisible[n,37] is not an equation. All the methods that worked use equations.

Posted 5 years ago

Thanks, Frank for that hint. At first I thought this explains my mistake. I was misleaded by the documentation which tells:

FindInstance[expr,vars]
finds an instance of vars that makes the statement expr be True. 

But within the details-section they explain that expr has to be equations, etc. But within the examples there are some with pure logical expressions, not equations.

Anyway. If I write it as an equation, I get another error and ridiculous results:

In[206]:= FindInstance[{n > 0, Divisible[n, 37] = True}, n, Integers, 5]
During evaluation of In[206]:= Set::write: Tag Divisible in Divisible[n,37] is Protected. >>
Out[206]= {{n -> 31}, {n -> 33}, {n -> 267}, {n -> 335}, {n -> 400}}

If I write it with "==" instead of "=" I get:

In[205]:= FindInstance[{n > 0, Divisible[n, 37] == True}, n, Integers, 5]
During evaluation of In[205]:= FindInstance::exvar: The system contains a nonconstant expression True independent of variables {n}. >>
Out[205]= FindInstance[{n > 0, Divisible[n, 37] == True}, n, Integers,5]

I'm afraid, I still have some basic misunderstandings of the Wolfram Language.

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