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.

• Blue: would it be ok to print out each number as you calculate it up towards n=infinity?
• Fatalize: @BlueEyedBeast No, you have to take an input and return the corresponding number.
• george: When checking 10, is the product of its digits 0 or 1?
• Fatalize: @george its product is 0.
• cat: Can I arbitrarily limit the range of the input if the upper limit of the range wouldn't be computed before the heat death of the universe anyways?

Answer 1

05AB1E, (削除) 13 (削除ここまで) 12 bytes

Thanks to Emigna for saving a byte!

μNNSONSP‚ÖP1⁄2

Explanation:

μ 1⁄2 # Get the nth number for which the following holds:
 NSO # The sum of digits of the current number
 NSP # And the products of digits of the current number
 N ‚ÖP # Divides the current number
 # If the nth number has been reached, quit and implicitly print N

Uses the CP-1252 encoding. Try it online!

• Emigna: µNNSONSP‚ÖP½ works as well doesn't it?
• Adnan: @Emigna Nice one! Thanks :)

Answer 2

Pyke, 14 bytes (non-competitive) (1-indexed)

~1IY'sB]im%X)@

Try it here!

My god what a lot of new features.

~1 - infinite list of natural numbers
 IY'sB]im%X) - filter(^, V) - remove if any truthiness
 Y - digits(i)
 'sB] - [sum(^), product(^)]
 im% - map(^, %i)
 X - splat(^)
 @ - ^[input]

Of which are non-competitive

• a bugfix in I where it would only check if the first item on the stack was truthy
• digits - return a list of digits in the number
• @ used to get the nth item of an infinite list

Of which were being used for the first time:

• all of the above
• infinite lists

Remove the last 2 bytes to get all of these numbers.

Answer 3

Jelly, 13 bytes

DμP;SðḍȦ
1Ç#Ṫ

1-based.
TryItOnline!

How?

DμP;SðḍȦ - Link 1, test a number
D - convert to a decimal list
 μ - monadic chain separation
 ; - concatenate the
 P - product, and the
 S - sum
 ð - dyadic chain separation
 ḍ - divides n?
 Ȧ - all (i.e. both)
1Ç#Ṫ - Main link, get nth entry, 1-based: n
1 # - find the first n matches starting at 1 of
 Ç - the last link (1) as a monad
 Ṫ - tail (the ultimate result)

Answer 4

C#, 118 bytes

n=>{int x=0,c=0;for(;;){int s=0,p=1,i=++x;while(i>0){s+=i%10;p*=i%10;i/=10;}if((c+=p>0&&x%s+x%p<1?1:0)==n)return x;}};

Full program with ungolfed function and test cases:

using System;
public class Program
{
 public static void Main()
 {
 // x - output number
 // c - counter
 // s - sum
 // p - product
 // i - iterator
 Func<int,int>f= n=>
 {
 int x=0, c=0;
 for ( ; ; )
 {
 int s=0, p=1, i=++x;
 while (i > 0)
 {
 s += i%10;
 p *= i%10;
 i /= 10;
 }
 if ( (c += p> 0&& x%s+x%p<1 ? 1 : 0) == n)
 return x;
 }
 };
 
 // tests:
 Console.WriteLine(f(1)); //1
 Console.WriteLine(f(5)); //5
 Console.WriteLine(f(10)); //12
 Console.WriteLine(f(20)); //312
 Console.WriteLine(f(42)); //6912
 Console.WriteLine(f(50)); //11313
 }
}

• raznagul: for(int x=0,c=0;;) saves you 1 byte.

Answer 5

Jellyfish, 45 bytes

p
\Ai
\&
>(&]&|0
 <*&d
 &~bN
 10
 ( )/+
 /*

Try it online!

Explanation

This is by far the most elaborate (and also the longest) program I've written in Jellyfish so far. I have no idea whether I'll be able to break this down in an understandable manner, but I guess I'll have to try.

Jellyfish provides a fairly general iteration operator, \, which helps a lot with "finding the Nth something". One of its semantics is "iterate a function on a value until a separate test function gives something truthy" (in fact, the test function receives both the current and the last element, but we'll only make it look at the current element). We can use this to implement a "next valid number" function. Another overload of \ is "iterate a function on a starting value N times". We can use our previous function and iterate it on 0 N times, where N is the input. All of that is set up fairly concisely with this part of the code:

p
\Ai
\&
> 0

(The reasons why 0, the actual input to the resulting function, is over there are a bit complicated and I won't go into them here.)

The issue with all of this is, that we won't be passing the current value to the test function manually. The \ operator will do this for us. So we now have construct a single unary function (via compositions, hooks, forks and currying) which takes a number and tells us whether it's a valid number (i.e. one which is divided by its digit sum and digit product). This is fairly non-trivial when you can't refer to the argument. Ever. It's this beauty:

 (&]&|
 <*&d
 &~bN
 10
 ( )/+
 /*

The ( is a unary hook, which means that it calls the function below (f) on its input (the current value x), and then passes both of them to the test function to the right (g), that is it computes g(f(x), x).

In our case, f(x) is another composite function which obtains a pair with the digit product and digit sum of x. That means g will be a function that has all three values to check if x is valid.

We'll start by looking at how f computes the digit sum and digit product. This is f:

 &~b
 10
 ( )/*
 /+

& is also composition (but the other way round). ~ is currying so 10~b gives function that computes the decimal digits of a number, and since we're passing that to & from the right, that's the first thing that will happen to the input x. The remainder uses this list of digits to compute their sum and product.

To compute a sum, we can fold addition over it, which is /+. Likewise, to compute the product we fold multiplication over it with /*. To combine both of these results into a pair, we use a pair of hooks, ( and ). The structure of this is:

()g
f

(Where f and g are product and sum, respectively.) Let's try to figure out why this gives us a pair of f(x) and g(x). Note that the right hook ) only has one argument. In this case, the other argument is implied to be ; which wraps its arguments in a pair. Furthermore, hooks can also be used as binary functions (which will be the case here) in which case they simply apply the inner function only to one argument. So really ) on a single function g gives a function that computes [x, g(y)]. Using this in a left hook, together with f, we obtain [f(x), g(y)]. This, in turn is used in a unary context, which mean that it's actually called with x == y and so we end up with [f(x), g(x)] as required. Phew.

That leaves only one thing, which was our earlier test function g. Recall that it will be called as g([p, s], x) where x is still the current input value, p is its digit product and s is its digit sum. This is g:

 &]&|
 <*&d
 N

To test divisibility, we'll obviously use modulo, which is | in Jellyfish. Somewhat unusually, it take its right-hand operand modulo its left-hand operand, which means that the arguments to g are already in the right order (arithmetic functions like this automatically thread over lists, so this will compute the two separate moduli for free). Our number is divisible by both the product and sum, if the result is a pair of zeros. To check whether that's the case, we treat the pair as a list of base-2 digits (d). The result of this is zero, only when both elements of the pair are zero, so we can negate the result of this (N) to obtain a truthy value for whether both values divide the input. Note that |, d and N are simply all composed together with a pair of &s.

Unfortunately, that's not the full story. What if the digit product is zero? Division and modulo by zero both return zero in Jellyfish. While this may seem like a somewhat odd convention, it actually turns out to be somewhat useful (because we don't need to check for zero before doing the modulo). However it also means we can get a false positive, if the digit sum does divide the input, but the digit product is zero (e.g. input 10).

We can fix this by multiplying our divisibility result by the digit product (so if the digit product is zero it will turn our truthy value into a zero as well). It turns out to be simpler to multiply the divisibility result with the pair of product and sum, and extract the result from the product afterwards.

To multiply the result with the pair, we kinda need to get back at an earlier value (the pair). This is done with a fork (]). Forks are kinda like hooks on steroids. If you give them two functions f and g, they represent a binary function which computes f(a, g(a, b)). In our case, a is the product/sum pair, b is the current input value, g is our divisibility test, and f is the multiplication. So all of this computes [p, s] * ([p, s] % x == [0, 0]).

All that's left now is to extract the first value of this, which is the final value of the test function used in the iterator. This is as simple as composing (&) the fork with the head function <, which returns the first value of a list.

• Zgarb: As the creator of Jellyfish, I approve this message. (Really, I would lose patience halfway through solving this challenge in Jellyfish.)

Answer 6

Perl 6, 44 bytes (0-indexed)

{grep({$_%%(.comb.sum&[*] .comb)},1..*)[$_]}

Explanation:

{ } # A function, with an argument n (`$_`)
 grep( ,1..*) # Filter the infinite list
 {$_ } # Check that the function's argument
 %%( ) # is divisible by
 & # both:
 .comb.sum # - the sum of the digits
 [*] .comb # - the product of the digits
 [$_] # Get the n-th value

Infinite lists ftw!

• Ven: @joshua thanks, but those parens are necessary for the precedence. Also, using a weird symbol instead of * would mean more bytes.
• Brad Gilbert b2gills: Dangit I forgot to check if it had a Perl 6 answer before posting. Also I would (did) handle Failures by using //0 in the grep block.
• Ven: @BradGilbertb2gills Don't hesitate to post a better version! I didn't use //0 because it's usually accepted in codegolf to print to stderr.
• Brad Gilbert b2gills: It was literally exactly the same except for //0

Answer 7

Actually, 20 bytes

Naive implementation of the sequence definition. Golfing suggestions welcome! Try it online!

u`;;$♂≈;Σ(%@π(%|Y`╓N

Ungolfing

 Implicit input n.
u Increment n, so that we don't accidentally include 0 in the sequence.
`...`╓ Starting with x=0, return the first n+1 values of x where f(x) is truthy.
 ;; Duplicate x twice.
 $♂≈ str(x) and convert each char (each digit) into an int. Call this digit_list.
 ; Duplicate digit_list.
 Σ Get sum(digit_list).
 (% Get x % sum(digit_list), which returns 0 if sum is a divisor of x.
 @ Swap the other duplicate of digit_list to TOS.
 π Get prod(digit_list).
 (% Get x % prod(digit_list), which returns 0 if prod is a divisor of x.
 | Get x % sum(digit_list) OR x % prod(digit_list).
 Y Logical negate, which only returns 1 if both are divisors, else 0.
N Return the last value in the list of x where f(x) is truthy,
 that is, the nth value of the sequence.

Answer 8

JavaScript (ES6), 72 bytes

k=>eval("for(n=0;(E=o=>n%eval([...n+''].join(o))!=0)`+`|E`*`||--k;)++n")

Demo

It tends to be slow for higher values, so I'm limiting it to 20 here.

let f =
k=>eval("for(n=0;(E=o=>n%eval([...n+''].join(o))!=0)`+`|E`*`||--k;)++n")
console.log(f(1)); // 1
console.log(f(5)); // 5
console.log(f(10)); // 12
console.log(f(20)); // 312

Answer 9

R, (削除) 132 (削除ここまで) 115 bytes

New version thanks to @Billywob nice comments !

n=scan();b=i=0;while(i<n){b=b+1;d=strtoi(el(strsplit(c(b,""),"")));if(na.omit(c(!b%%sum(d)&!b%%prod(d),0)))i=i+1};b

Ungolfed :

n=scan()
b=i=0
while(i<n)
 b=b+1;
 d=strtoi(el(strsplit(c(b,""),""))) #Splitting the number into its digits
 if(na.omit(c(!b%%sum(d)&!b%%prod(d),0)))
 i=i+1
b

(削除) Since R behave strangley with NAs, I had to add the whole ifelse(is.na(...)) part! (削除ここまで)
Or use na.omit(...)

• Billywob: n=scan();b=i=0;while(i<n){b=b+1;d=strtoi(el(strsplit(c(b,""),"")));if(!b%%sum(d)&ifelse(is.na((p=!b%%prod(d))),F,p))i=i+1};b saves a few bytes by: el(...) instead of [[1]], using c(b,"") instead of paste(b),negating the logical expressions by ! instead of ==0 and skipping the curly brackets on the if statement. My guess is that there should be an easier way of handling the NA issue but couldn't figure out something clever.
• Billywob: Turns out we can circumvent it by appending a 0 to the expression evaluated in the if statement. However, this returns a warning when the product isn't equal to 0. n=scan();b=i=0;while(i<n){b=b+1;d=strtoi(el(strsplit(c(b,""),"")));if(na.omit(c(!b%%sum(d)&!b%%prod(d),0)))i=i+1};b
• Rudier: @Billywob Thanks a lot ! i didn't know about el(...) !

Answer 10

Brachylog, 22 bytes

:1yt
#>=.@e+:I*.@e*:J*

Try it online!

Explanation

:1y Evaluate the first N valid outputs to the predicate below given the
 main input as input
 t The output is the last one
#>=. Output is a strictly positive integer
 @e+ The sum of its digits...
 :I*. ...multiplied by an integer I results in the Output
 @e* The product of its digits...
 :J* ...multiplied by an integer J results in the Output

Answer 11

JavaScript (ES6), 78

n=>eval("for(i=0;n;!p|i%s|i%p||n--)[...++i+''].map(d=>(s-=d,p*=d),s=0,p=1);i")

Less golfed

n=>{
 for(i=0; n; !p|i%s|i%p || n--)
 s=0,
 p=1,
 [...++i+''].map(d=>(s-=d, p*=d));
 return i
} 

Answer 12

Pyth, 18 bytes

e.f!.xs%LZsM*FBsM`

Try it online: Demonstration

Explanation:

e.f!.xs%LZsM*FBsM`ZZQ1 implicit variables at the end
e print the last number of the 
 .f Q1 first Q (input) numbers Z >= 1, which satisfy:
 `Z convert Z to a string, e.g. "124"
 sM convert each digits back to a number, e.g. [1, 2, 4]
 *FB bifurcate with product, e.g. [[1, 2, 4], 8]
 sM take the sum of each, e.g. [7, 8]
 %LZ compute the modulo of Z with each number, e.g. [5, 4]
 s and add both numbers, e.g. 9
 .x Z if an exception occurs (mod 0), use number Z instead
 ! test, if this number is zero

Answer 13

Haskell, (削除) 94 (削除ここまで) (削除) 85 (削除ここまで) (削除) 72 (削除ここまで) 71 bytes

([n|n<-[0..],(==)=<<map(gcd n)$[product,sum]<*>[read.pure<$>show n]]!!)

1-indexed.

Thanks to @Zgarb for saving 13 bytes!

Thanks to @nimi for saving a byte!

• Zgarb: (==)=<<map(gcd n)$[sum k,product k] should save some bytes.
• Zgarb: And while we're doing that, [sum k,product k] can be map($read.pure<$>show n)[sum,product].
• nimi: One more byte: ([n|n<-[0..],(==)=<<map(gcd n)$[product,sum]<*>[read.pure<$>show n]]!!)

Answer 14

APL (Dyalog Unicode), 33 bytes

I followed an idea of @ngn when it comes to wrapping the function in some sort of a ⍣-based operation.

{1+⍣{0=+/((×ばつ/)⍎ ̈⍕⍺)|⍺ ⍺}⍣⍵⊢0}

Explanation:

{1+⍣{0=+/((×ばつ/)⍎ ̈⍕⍺)|⍺ ⍺}⍣⍵⊢0}
{1+⍣ ⍣⍵⊢0} ⍝ Execute the inner dfn until it returns one ⍵ times
 ⍝ and yield the last number it processed.
 {0=+/((×ばつ/)⍎ ̈⍕⍺)|⍺ ⍺} ⍝ 1 if ⍺ is divisible by sum and product of its digits.
 ⍎ ̈⍕⍺ ⍝ digits of ⍺
 , ⍝ a vector of...
 ×ばつ/ ⍝ product of digits
 1∘⊥ ⍝ sum of digits
 |⍺ ⍺ ⍝ each modulo ⍺
 0=+/ ⍝ both divisible?

Try it online!

• Adám: 27
• Adám: I'm fairly certain this 25er works.
• Jayant Choudhary: Could have saved 2 bites({}) by using ⎕ to get input instead of making it a dfn

Answer 15

Vyxal, 9 bytes

λ(∑ΠḊA;ȯt

Try it Online!

(削除) Apparently there's a 10 byte 0-indexed version. (削除ここまで)

-3 bytes from Lyxal.

• lyxal: 9 bytes. 10 bytes

Answer 16

MATL, 21 bytes

`@tFYAtswph\~?@]NG<]&

Long and inefficient...

Try it online!

How it works

` % Do...while
 @ % Push current iteration index (1-based)
 tFYA % Duplicate. Convert number to its digits
 ts % Duplicate. Sum of digits
 wp % Swap. Product of digits
 h\ % Concatenate. Modulo. This gives a length-2 array
 ~? % If the two values are zero: we found a number in the sequence
 @ % Push that number
 ] % End if
 NG< % True if number of elements in stack is less than input
] % End do...while. If top of the stack is true: next iteration. Else: exit
& % Specify only one input (top of stack) for implicit display

Answer 17

Python 2, (削除) 122 (削除ここまで) 110 Bytes

def a(m,i=1,k=1):n=map(int,`i`);p=reduce(lambda x,y:x*y,n);k+=p and 1>i%sum(n)+i%p;return(k>m)*i or a(m,i+1,k)

1 indexed, you need to use a Python interpreter with a quite high recursion limit.

Answer 18

Python3, (削除) 134 (削除ここまで) 80 Bytes

New version thanks to Flp.Tkc

(削除)

t=input();h=k=0;p=int
def g(x,q=0,w=1):
 for i in x:X=p(x);I=p(i);q+=I;w*=I
 return w!=0and X%q+X%w<1
while h<p(t):k+=1;h+=g(str(k))

(削除ここまで) New code, I remembered a golfing way to do the factorial

f,t=lambda x:0**x or x*f(x-1),0
for i in str(f(int(input()))):t+=int(i)
print(t)

(削除) The code itself isn't very golf-like, more like brute force golf

def g(x):
 q=0;w=1;X=int(x)
 for i in x:I=int(i);q+=I;w*=I
 return (w!=0+ X%q==0and X%w==0)
t=input();h=k=0
while h<int(t):
 k+=1
 if g(str(k))is True:h+=1

g(x) is a function that returns True if x fits the criteria.

(削除ここまで)

• FlipTack: In the future, use <1 instead of ==0. You don't need the is True, the point of an if statement is to check whether the condition is true anyway. You can use Python 2's backtick shortcut for str/repr to shave some bytes. There's also a lot of unneeded whitespace here.
• FlipTack: Also, you can use booleans as integer values: h+=g(str(k)) adds 1 if True, 0 if False.
• george: @Flp.Tkc can you explain the backtick trick. I've tried to use it and it threw a syntax error
• FlipTack: Doing (backtick)x(backtick) in Python 2 is the same is repr(x) or str(x) in Python 3 :)
• george: @Flp.Tkc that only works in pre Python 3. It was removed in 3.0

Answer 19

x86-16 machine code, 52 bytes

00000000: 33db 4351 8bc3 bf0a 0033 f6b1 0133 d2f7 3.CQ.....3...3..
00000010: f703 f291 f7e2 9142 e314 85c0 75ef 8bc3 .......B....u...
00000020: 33d2 f7f6 8bc3 85d2 7504 f7f1 85d2 5975 3.......u.....Yu
00000030: d1e2 cfc3 ....

Listing:

33 DB XOR BX, BX ; X = 0 
 NUM_LOOP: 
43 INC BX ; increment X 
51 PUSH CX ; save loop counter (N) 
8B C3 MOV AX, BX ; AX = current X 
 START_N: 
BF 000A MOV DI, 10 ; DI = decimal divisor of 10 
33 F6 XOR SI, SI ; SI = sum of digits
B1 01 MOV CL, 1 ; CX = product of digits
 SPLIT_DIGITS: 
33 D2 XOR DX, DX ; clear high word of dividend 
F7 F7 DIV DI ; AX = quotient, DX = remainder 
03 F2 ADD SI, DX ; SI = SI + digit 
91 XCHG AX, CX ; AX = running product, CX = quotient 
F7 E2 MUL DX ; AX = AX * digit 
91 XCHG AX, CX ; CX = running product, AX = quotient 
42 INC DX ; ZF = NZ (hacky) 
E3 14 JCXZ END_N ; if digit is a 0, end with Falsey 
85 C0 TEST AX, AX ; is quotient 0? 
75 EF JNZ SPLIT_DIGITS ; loop until it is 
8B C3 MOV AX, BX ; AX = current X 
33 D2 XOR DX, DX ; clear high word of dividend 
F7 F6 DIV SI ; divide X by sum 
8B C3 MOV AX, BX ; AX = current X 
85 D2 TEST DX, DX ; if sum remainder is 0 then Truthy (ZF = ZR) 
75 04 JNZ END_N ; if not, end with Falsey (ZF = NZ) 
F7 F1 DIV CX ; divide X by product 
85 D2 TEST DX, DX ; if product remainder is 0 then Truthy (ZF = ZR) 
 END_N: 
59 POP CX ; restore loop counter 
75 D1 JNZ NUM_LOOP ; if falsey, keep looping and don't decrement counter 
E2 CF LOOP NUM_LOOP ; otherwise decrement counter and loop 
C3 RET ; return to caller

Callable function. Input N in CX, Output Nth positive integer in BX.

N of 50 is about 4 seconds on an IBM PC 5150:

enter image description here

Answer 20

JavaScript (ES6), 70 bytes

k=(b,n=1,f=c=>n%eval([...n+''].join(c))!=0)=>f`+`|f`*`||--b?k(b,n+1):n

This turned out quite a bit like @Arnauld's answer, but recursion is apparently 2 bytes shorter. Works in Chrome, though it's very slow on inputs greater than 30 or so (50 takes 6 seconds).

Answer 21

Wonder, 33 bytes

@:^#0(!>@!(| %#0sum#0)%#0prod#0)N

Zero-indexed. Usage:

(@:^#0(!>@!(| %#0sum#0)%#0prod#0)N)9

Explanation

More readable:

@
 iget #0 
 (fltr@
 not (or % #0 sum #0) % #0 prod #0
 ) N

Basically gets an infinite list of numbers divisible by its digital sum and product by filtering an infinite list of whole numbers through a predicate. Then the nth item is simply picked out of the list.

Answer 22

Julia, 81 bytes

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

This is an anonymous function that accepts an integer and returns an integer. To call it, give it a name. The approach is the obvious one: check every number until we've encountered n terms of the sequence. The all check is necessary to ensure we don't get a DivisionError from % when the product of the digits is 0.

Ungolfed:

function f(n)
 i = c = 1
 while c < n
 d = digits(i += 1)
 all(d .> 0) && i % sum(d) == i % prod(d) < 1 && (c += 1)
 end
 return i
end

Try it online! (includes all test cases)

• Martin Ender: You can save two bytes by assigning prod(d) to p or something and then replacing the all(d.>0) with p>0. And you can save another by moving the i%sum(d) onto the other side of the 1 i.e. p<1>i%sum(d).

Answer 23

C89, (削除) 381 (削除ここまで) (削除) 226 (削除ここまで) (削除) 195 (削除ここまで) (削除) 170 (削除ここまで) 169 bytes

1-indexed (same exact answers as in the challenge).

Assumes 4-byte (32 bit) int (most modern architectures).

I genuinely believe this can't go any shorter.

x,c,*b,m,t,i,a;g(n){for(b=malloc(0);c<n;b[c-1]=x++,t=1){char s[9];for(i=m=0;i<sprintf(s,"%d",x);m+=a,t*=a)a=s[i++]-48;b=m*t?x%m+x%t?b:realloc(b,4*++c):b;}return b[c-1];}

Function int g (int) leaks memory and accesses uninitialised memory once per call but doesn't segfault and returns the right number.

Full program that takes input in unary (./prog $(seq 1 10) for 10) with ungolfed (kinda):

x, c, * b, m, t, i, a;
g(n) {
 for (b = malloc(0); c < n; b[c - 1] = x++, t = 1) {
 char s[9];
 i = m = 0;
 for (; i < sprintf(s, "%d", x); m += a, t *= a) a = s[i++] - 48;
 b = m * t ? x % m + x % t ? b : realloc(b, 4 * ++c) : b;
 }
 return b[c - 1];
}
main (j) {
 printf("%d\n", g(--j));
}

Old answer:

C99, 381 bytes

#include <stdio.h>
#include <inttypes.h>
#include <string.h>
#include <stdlib.h>
#define U uint64_t
#define S size_t
S f(S n){U x=0;S c=1,l;U*b=malloc(sizeof(U));while(c<=n){char s[21];snprintf(s,20,"%"PRIu64,x);U m=0,t=1;l=strnlen(s,21);for(S i=0;i<l;i++){U a=(U)s[i]-48;m+=a,t*=a;}if(m*t?(!(x%m))&&(!(x%t)):0){b=realloc(b,sizeof(U)*++c);b[c-1]=x;}++x;}U o=b[n];free(b);return o;}

This can probably be golfed more.

Full program:

#include <stdio.h>
#include <limits.h>
#include <stdint.h>
#include <inttypes.h>
#include <stdbool.h>
#include <string.h>
#include <stdlib.h>
bool qualifies (const uint64_t);
size_t f (const size_t);
int main(const int argc, const char* const * const argv) {
 (void) argc;
 size_t arg = strtoull(argv[1], NULL, 10);
 uint64_t a = f(arg);
 printf("a: %" PRIu64 "\n", a);
 return 0;
}
bool qualifies (const uint64_t num) {
 char s[21];
 snprintf(s, 20, "%" PRIu64 "", num);
 uint64_t sum = 0,
 mult = 1;
 size_t len = strnlen(s, 400);
 for (size_t i = 0; i < len; i++) {
 uint64_t a = (uint64_t) s[i] - 48;
 sum += a, mult *= a;
 }
 //printf("sum: %" PRIu64 "\nmult: %" PRIu64 "\n", sum, mult);
 return sum * mult ? (! (num % sum)) && (! (num % mult)) : false;
}
size_t f (const size_t n) {
 uint64_t x = 0;
 size_t s_len = 1;
 uint64_t* nums = malloc(sizeof (uint64_t) * s_len);
 while (s_len <= n) {
 if (qualifies(x)) {
 ++s_len;
 //printf("len: %zu\n", s_len);
 nums = realloc(nums, sizeof (uint64_t) * s_len);
 nums[s_len - 1] = x;
 }
 ++x;
 }
 uint64_t o = nums[n];
 free(nums);
 return o;
}

• Dennis: tio.run/nexus/… generates a few warnings, but it's equivalent. However, I think it's fine to use int for everything, since it's the default integer type.
• cat: @Dennis I'm not used to omitting headers and ignoring warnings outside of C89, so I golf with all warnings enabled and as errors :P I'll rewrite in C89 though.
• cat: @Dennis Fixed :D

Answer 24

C, 110 bytes

p;s;i;j;f(n){j=0;while(n){i=++j;p=1;s=0;do p*=i%10,s+=i%10;while((i/=10)>0);if(p>0&&j%p+j%s==0)--n;}return j;}

Ungolfed and usage:

p;s;i;j;
f(n){
 j=0;
 while(n){
 i=++j;
 p=1;
 s=0;
 do
 p*=i%10, //product
 s+=i%10; //sum
 while((i/=10)>0);
 //check if product is not zero since floating point exception
 if(p>0 && j%p + j%s == 0)--n;
 }
 return j;
}
int main(){
 int n;
 scanf("%d",&n);
 printf("\n%d\n", f(n));
}

• Toby Speight: Some shavings: i is never negative, so while(i/=10) is sufficient. Instead of the if, you could use short-circuiting logical operators and similar knowledge that p is non-negative: p&&j%p+j%s==0&&--n;; further golfing gives !p||j%p+j%s||--n. The do/while can become a for: for(i=++j,p=1,s=0;p*=i%10,s+=i%10,i/=10;);...
• Toby Speight: Combined, these savings get us down to 92 bytes

Answer 25

Scala, (削除) 80 (削除ここまで) 68 bytes

Stream from 1 filter{n=>val x=n+""map(_-48.0);n%x.sum+n%x.product<1}

Try it online!

0-indexed. Streams are not only infinite collections but also functions taking an Int and giving the element at that index.

Stream from 1 //Infinite stream starting at 1
 filter { n => //Filter each n according to this predicate:
 val x = //List of digits
 n + "" //Convert to string
 map(_ - 48.0); //Turn each character into its corresponding number (as a double, so n%0 works later)
 n % x.sum + //n modulo the sum of its digits
 n%x.product //plus n modulo the product of its digits
 < 1 //Make sure that's less than 1, i.e. both are 0
 }

Answer 26

Husk, 13 bytes

!f§Λ`¦o§eΠΣdN

Try it online!

Combinators good but combinators long :(

-1 byte from Leo in SE Chat.

Answer 27

APL (Dyalog Unicode), 25(削除) 27 (削除ここまで) bytes

This even better answer curtesy of Adam

1+⍣{0=⍺|⍨(×ばつ/)⍎ ̈⍕⍺}⍣⎕⊢0

This is essentially the same just hcf is taken of the sum and product saving 2 bytes

1+⍣{0=∨/⍺|⍨(×ばつ/)⍎ ̈⍕⍺}⍣⎕⊢0

Try it online!

⎕ ⍝ takes user input
{...}⍣n ⍵ ⍝ repeats the function n times 
1∘+⍣{0=∨/⍺|⍨(×ばつ/)⍎ ̈⍕⍺} ⍝ given a number repeatetly adds 1(1∘+) untill the condition is satisfied 
{⍵|⍨(×ばつ/)⍎ ̈⍕⍵} ⍝ converts numbers to a list of numbers and then take there sum and product and then take there reminder after dividing with the given number
∨/ ⍝ takes HCF of the list escecially checking if both the numbers are not zero
0= ⍝returs true if the hcf is zero 

• Adám: Golfs down the the same as this.
• Adám: TIO link for you: Try it online!

Answer 28

Thunno 2 ct, 6 bytes

ƘçSpḊp

Try it online!

Explanation

ƘçSpḊp # Implicit input, n
Ƙ # nth positive integer, k, where:
 ç # Parallelly apply:
 S # Sum of digits of k
 p # Product of digits of k
 Ḋ # Is k divisible by them
 p # Both return true?
 # Implicit output

Answer 29

APL(NARS), 59 chars

r←f ×ばつ⍳∼{∧/0=⍵∣⍨(×ばつ/k),+/k←⍎ ×ばつ⍳0≥w-←1
r+←1⋄→2

//+/6 4 41 8 = 59. 1-indexed. Input integer w>0 output the number in sequence position w.

test:

 f 1
1
 f 0
1
 f 10
12
 f 50
11313

Answer 30

PHP, 96 bytes

Takes n as a command line argument.

Golfed

for(;$i<$argv[1];)!($p=array_product($d=str_split(++$j)))|$j%array_sum($d)||$j%$p?:$i++;echo $j;

Ungolfed

for (; $i < $argv[1];) // Loop until we've found the nth number as pass by command line
 !($p = array_product($d = str_split(++$j))) || // Split our current number into its digits and assign that to a variable, then assign the product of that array to another variable.
 // As well, we're checking if the product equals 0, to prevent an error from trying to mod by 0 later. The condition short circuits before it happens.
 $j % array_sum($d) || // Check if the sum of the array is a divisor
 $j % $p // Check if the product is a divisor
 ?: $i++; // Increment the number of found instances only if all conditions are met.
echo $j; // Output to screen.