Reverse and add degeneracy

Question

Intro

Reverse and add is as simple as it sounds, take n and add it to its digits in reverse order. (e.g. 234 + 432 = 666).

If you apply this process repeatedly some numbers will eventually hit a prime number, and some will never reach a prime.

Example

I currently have

11431 rep.

11431 is not prime
11431 + 13411 = 24842 which is not prime
24842 + 24842 = 49684 which is not prime
49684 + 48694 = 98378 which is not prime
98378 + 87389 = 185767 which is prime!

This number hits a prime

In contrast any multiple of 3 will never hit a prime, this is because the all multiples of 3 have a digit sum that is a multiple of 3 and vice versa. Thus reverse and add on a multiple of 3 will always result in a new multiple of 3 and thus never a prime.

Task

Take a positive integer n and determine if repeatedly reversing and adding will ever result in a prime number. Output a truthy or falsy value. Either truthy for reaches a prime and falsy value for does not or the other way around both are acceptable.

Prime numbers will be considered to reach a prime number in zero iterations.

This is code-golf so try to make your code as short as possible.

Test Cases

True for reaches a prime false for never reaches a prime

11 -> True
11431 -> True
13201 -> True
13360 -> True
13450 -> True
1019410 -> True
1019510 -> True
22 -> False
1431 -> False
15621 -> False
14641 -> False

Hint

While I was writing this challenge I discovered a cool trick that makes this problem a good deal easier. It is not impossible without this trick and it is not trivial with it either but it does help. I had a lot of fun discovering this so I will leave it in a spoiler below.

Repeated reverse and add will always hit a multiple of 11 in 6 iterations or less. If it does not hit a prime before it hits a multiple of 11 it will never hit a prime.

Answer 1

Ruby, 84 79 77 74 bytes

->x{y=1;x+="#{x}".reverse.to_i while(2...x).any?{|z|0==y=x%z}&&x%11>0;y>0}

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If I got it right, when we reach a multiple of 11 we can stop (we will only get multiples of 11 after that)

Answer 2

Haskell, 65 bytes

f takes an Integer and returns a Bool. True means it reaches a prime.

f n=gcd(product[2..n-1])n<2||gcd 33n<2&&f(n+read(reverse$show n))

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Unfortunately the short but inefficient prime test means that the OP's True test cases other than 11 grow too big to finish. But for example 11432 is a True case that does finish.

You can also try this 3 bytes longer one, for which TIO can finish all the True test cases:

f n=and[mod n i>0|i<-[2..n-1]]||gcd 33n<2&&f(n+read(reverse$show n))

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Both versions' prime tests break on 1, but it so happens that it gets to a prime (2) anyway.

Otherwise, I noticed about the same thing as G.B. in the spoiler of the Ruby submission:

Once a number grows to even length, the next iteration will be divisible by 11. Once a number is divisible by 11, so will all following iterations.

Answer 3

Python 2, 123 110 bytes

Saved 13 bytes thanks to Ørjan Johansen and Wheat Wizard!

n=input()
while 1:
 if all(n%m for m in range(2,n)):print 1;break
 if n%11==0:print 0;break
 n+=int(`n`[::-1])

Returns 1 if it reaches a prime, 0 if it doesn't. Try it online!

Answer 4

Python 2, 78 70 69 bytes

f=lambda x:all(x%a for a in range(2,x))or x%11and f(x+int(`x`[::-1]))

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Explanation

This program relies on the fact that

Every number that loss forever will reach a multiple of 11 in less than 6 moves

This program is a recursive lambda with circuited logical comparatives. It first checks if n is prime.

all(x%a for a in range(2,x))

If this is true we return true.

If it is false we check if it is a multiple of 11.

x%11

If false we return false otherwise we return the result of f on the next iteration

f(x+int(`x`[::-1]))

Answer 5

Jelly, 11 bytes

ṚḌ$+$6СÆPS

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Answer 6

05AB1E, 14 13 bytes

EDIT: Saved one byte because input is reused if there aren't enough elements on the stack

[Dp#D11Ö#R+]p

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Uses the hint in the question

How it works

[              # begin infinite loop
               # implicit input
 D             # duplicate input
  p            # push primality of input
   #           # if prime, break
    D          # duplicate input
     11        # push 11
       Ö       # push input % 11 == 0
        #      # if multiple of 11, break
               # implicit push input
          R    # reverse input
           +   # add both numbers
            ]  # end infinite loop
             p # push primality of result; 1 if prime, 0 if multiple of 11
               # implicit print

Answer 7

MATLAB, 88 81 bytes

function r=f(n);r=0;for i=1:7 r=r+isprime(n);n=n+str2num(fliplr(num2str(n)));end;

Answer 8

JavaScript (ES6), 73 bytes

Returns 0 or true.

f=n=>{for(d=n;n%--d;);return d<2||n%11&&f(+[...n+''].reverse().join``+n)}

Commented

This is based on the magic-spoiler formula described by Wheat Wizard.

f = n => {              // given n:
  for(d = n; n % --d;); // find the highest divisor d of n
  return                //
    d < 2 ||            // if this divisor is 1, return true (n is prime)
    n % 11 &&           // else: if 11 is a divisor of n, return 0
    f(                  // else: do a recursive call with
      +[...n + '']      // the digits of n
      .reverse().join`` // reversed, joined,
      + n               // coerced to a number and added to n
    )                   //
}                       //

Test cases

I've removed the two largest inputs from the snippet, as they take a few seconds to complete. (But they do work as well.)

f=n=>{for(d=n;n%--d;);return d<2||n%11&&f(+[...n+''].reverse().join``+n)}

console.log(f(11))      // -> True
console.log(f(11431))   // -> True
console.log(f(13201))   // -> True
console.log(f(13360))   // -> True
console.log(f(13450))   // -> True
console.log(f(22))      // -> False
console.log(f(1431))    // -> False
console.log(f(15621))   // -> False
console.log(f(14641))   // -> False

Answer 9

Mathematica, 45 bytes

Or@@PrimeQ@NestList[#+IntegerReverse@#&,#,6]&

Answer 10

Microsoft Sql Server, 826 786* bytes

* I've recalled about the IIF function that was introduced in Microsoft Sql Server 2012

set nocount on
use rextester
go
if object_id('dbo.n','IF')is not null drop function dbo.n
go
create function dbo.n(@ bigint,@f bigint)returns table as return
with a as(select 0 c union all select 0),b as(select 0 c from a,a t),c as(select 0 c from b,b t),
d as(select 0 c from c,c t),e as(select 0 c from d,d t),f as(select 0 c from e,e t),
v as(select top(@f-@+1)0 c from f)select row_number()over(order by(select 0))+@-1 n from v
go
with u as(select cast(a as bigint)a from(values(11),(11431),(13201),(13360),(13450),(1019410),(1019510),(22),(1431),
(15621),(14641))u(a)),v as(select a,a c from u union all select a,c+reverse(str(c,38))from v
where 0=any(select c%n from dbo.n(2,c/2))and c%11>0)select a,iif(0=any(select max(c)%n from dbo.n(2,max(c)/2)),0,1)
from v group by a option(maxrecursion 0)

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The more neat formatting

SET NOCOUNT ON;
USE rextester;
GO
IF OBJECT_ID('dbo.n', 'IF') IS NOT NULL DROP FUNCTION dbo.n;
GO
CREATE FUNCTION dbo.n(@ BIGINT,@f BIGINT)RETURNS TABLE AS RETURN
  WITH
    a AS(SELECT 0 c UNION ALL SELECT 0),
    b AS(SELECT 0 c FROM a,a t),
    c AS(SELECT 0 c FROM b,b t),
    d AS(SELECT 0 c FROM c,c t),
    e AS(SELECT 0 c FROM d,d t),
    f AS(SELECT 0 c FROM e,e t),
    v AS(SELECT TOP(@f-@+1)0 c FROM f)
    SELECT ROW_NUMBER()OVER(ORDER BY(SELECT 0))+@-1 n FROM v;
GO
WITH u AS(
  SELECT CAST(a AS BIGINT) a
  FROM(VALUES (11), (11431), (13201), (13360), (13450), (1019410), (1019510),
              (22), (1431), (15621), (14641)) u(a)
),
v AS(
  SELECT a, a c FROM u
    UNION ALL
  SELECT a, c + reverse(str(c, 38))
  FROM v
  WHERE 0 = ANY(SELECT c % n FROM dbo.n(2, c / 2)) AND c % 11 > 0
)
SELECT a, IIF(0 = ANY(SELECT MAX(c) % n FROM dbo.n(2, MAX(c) / 2)), 0, 1)
FROM v
GROUP BY a
OPTION (MAXRECURSION 0);

Answer 11

Jelly, 9 bytes

ṚḌ+Ɗ6СẒẸ

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How it works

ṚḌ+Ɗ6СẒẸ    Monadic main link.
ṚḌ+Ɗ         Monad: Reverse and add to original.
    6С      Repeatedly apply the above 6 times, collecting all iterations
       ẒẸ    Is any of them a prime?

Answer 12

PHP 114 bytes

<?php function f($n){$n1=strrev((string)$n);$r=$n+(int)$n1;for($i=2;$i<$r;$i++){if($r%$i==0){die('0');}}die('1');}

More readable version:

<?php function f($n)
{
    $n1 = strrev((string)$n);
    $r = $n + (int)$n1;
    for ($i = 2; $i < $r; $i++) {
        if ($r % $i == 0) {
            die('0');
        }
    }
    die('1');
}

f(11431 );

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I used this thing for counting bytes.