Numbers divisible by the sum and product of their digits

Question

Take a positive integer X. This number is part of the sequence we are interested in if the sum of all digits of X is a divisor of X, and if the product of all digits of X is a divisor of X.

For example, 135 is such a number because 1 + 3 + 5 = 9 which divides 135 = 9 * 15 and 1 * 3 * 5 = 15 which also divides 135.

This is sequence A038186 in the OEIS.

Your task: given an integer N, output the Nth positive integer with such properties.

Inputs and outputs

• Numbers may be 0-indexed or 1-indexed; please indicate which one your answer use.

• The input may be taken through STDIN, as a function argument, or anything similar.

• The output may be printed to STDOUT, returned from a function, or anything similar.

Test cases

The test cases below are 1-indexed.

Input        Output

1            1
5            5
10           12
20           312
42           6912
50           11313

Scoring

This is code-golf, so the shortest answer in bytes wins.

Answer 1

BASH, 125 bytes

while ((n<$1));do
((i++))
p(){ fold -1<<<$i|paste -sd$1|bc;}
z=`p \*`
((z))&&[ $[i%`p +`]$[i%z] -eq 0 ]&&((n++))
done
echo $i

Answer 2

Perl 5 -MList::Util=sum,product -p, 63 bytes

($\%sum(@a=++$\=~/./g)+$\=~/0/||$\%product@a)&&redo for 1..$_}{

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1-indexed

Answer 3

Jelly, 11 bytes

DS,PƊḍ¹Ạµ#Ṫ

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0-indexed

          Ṫ  get the last element after
        µ#   counting the first (input) elements where
       Ạ     all (both) are true:
     ḍ¹      the element is divisible by
 S,P         the sum, and the products of
D            the digits

Answer 4

Julia, 72 bytes

n->(i=c=1;while c<n d=digits(i+=1);c+=i%sum(d)<all(d.>0)>i%prod(d)end;i)

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improvement over Alex A's answer (explanation there)