Inconsummate numbers in base 10

Inconsummate numbers in base 10 is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A consummate number is a non-negative integer that can be formed by some integer N divided by the the digital sum of N.

For instance

47 is a consummate number.

   846 / (8 + 4 + 6) = 47

On the other hand, there are integers that can not be formed by a ratio of any integer over its digital sum. These numbers are known as inconsummate numbers.

62 is an inconsummate number. There is no integer ratio of an integer to its digital sum that will result in 62.

The base that a number is expressed in will affect whether it is inconsummate or not. This task will be restricted to base 10.

Task

• Write a routine to find inconsummate numbers in base 10;
• Use that routine to find and display the first fifty inconsummate numbers.

Stretch

• Use that routine to find and display the one thousandth inconsummate number.

See also

• Numbers Aplenty - Inconsummate numbers
• OEIS:A003635 - Inconsummate numbers in base 10

Raku

Not really pleased with this entry. It works, but seems inelegant.

my $upto = 1000;

my @ratios = unique (^∞).race.map({($_ / .comb.sum).narrow})[^($upto²)].grep: Int;
my @incons = (sort keys (1..$upto * 10) (-) @ratios)[^$upto];

put "First fifty inconsummate numbers (in base 10):\n" ~ @incons[^50]».fmt("%3d").batch(10).join: "\n";
put "\nOne thousandth: " ~ @incons[999]

Output:

First fifty inconsummate numbers (in base 10):
 62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth: 6996

Wren

Library: Wren-math

Library: Wren-fmt

It appears to be more than enough to calculate ratios for all numbers up to 999,999 (which only takes about 0.4 seconds on my machine) to be sure of finding the 1,000th inconsummate number.

import "./math" for Int
import "./fmt" for Fmt

// Maximum ratio for 6 digit numbers is 100,000
var cons = List.filled(100001, false)
for (i in 1..999999) {
    var ds = Int.digitSum(i)
    var ids = i/ds
    if (ids.isInteger) cons[ids] = true
}
var incons = []
for (i in 1...cons.count) {
    if (!cons[i]) incons.add(i)
}
System.print("First 50 inconsummate numbers in base 10:")
Fmt.tprint("$3d", incons[0..49], 10)
Fmt.print("\nOne thousandth: $,d", incons[999])

Output:

First 50 inconsummate numbers in base 10:
 62  63  65  75  84  95 161 173 195 216 
261 266 272 276 326 371 372 377 381 383 
386 387 395 411 416 422 426 431 432 438 
441 443 461 466 471 476 482 483 486 488 
491 492 493 494 497 498 516 521 522 527 

One thousandth: 6,996