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Find n Niven Numbers

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Challenge

A Niven number is a positive integer which is divisible by the sum of its digits.

For example, 81 -> 8+1=9 -> 81%9=0.

Your task is to find the first n Niven numbers, given n.

Tests

Reference implementation(takes input)

first 16 values:

0 => 1
1 => 2
2 => 3
3 => 4
4 => 5
5 => 6
6 => 7
7 => 8
8 => 9
9 => 10
10 => 12
11 => 18
12 => 20
13 => 21
14 => 24
15 => 27

code-golf math sequence

posted about 2 years ago

CC BY-SA 4.0

2y ago by user‭

Razetime‭

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is a duplicate

This question has been asked before and has already been answered. It should be marked as a duplicate.

Please enter the URL of the proposed duplicate in the details field below.

not constructive

This question cannot be answered in a way that is helpful to anyone. It's not possible to learn something from possible answers, except for the solution for the specific problem of the asker.

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2 comment threads

tangent: Niven? (2 comments)

Alternate output formats (2 comments)

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+2

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Python, 98 bytes

from itertools import*
f=lambda n:[*islice(filter(lambda k:k%sum(map(int,str(k)))==0,count(1)),n)]

itertools.count() generates integer numbers starting from 1,
filter() filters out the Niven numbers,
itertools.islice() limits the result to n items,
and [*...] collects the numbers in a list.

posted about 2 years ago

CC BY-SA 4.0

2y ago

__blackjack__‭

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Without itertools (3 comments)

Peter Taylor‭ wrote about 2 years ago

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The sum(map(int trick is shorter than I'd managed, but itertools is expensive. As a one-liner, a recursive lambda gets 76: f=lambda n,i=1,a=[]:a if a[n-1:]else f(n,i+1,a+[i][i%sum(map(int,str(i))):]). But if you don't insist on a list, the best I've found is an explicit loop for 73: def f(n,i=0): while n: i+=1 if i%sum(map(int,str(i)))<1:yield i;n-=1 where the tabs also indicate newlines