My usual language of choice...
import std.stdio;
import std.string;
import std.conv;
import std.range;
import std.algorithm;
/*
1) LCM of the digits 1 through 9 is 2_520
2) Lower bound (smallest ten-digit multiple of 2_520) is 1_000_001_520
3) Generate all the ten-digit multiples of 2520
4) Filter out number Pandigital and Divisible by 1 through 9
*/
bool isPandigitalAndDivisibleBy1Through9(long number) {
foreach (i; 1 .. 10) {
if (number % i != 0) {
return false;
}
}
return text(number.text.array.sort) == "0123456789";
}
void main() {
auto seq10digits = iota(1_000_001_520, 10_000_000_000, 2_520)
.filter!isPandigitalAndDivisibleBy1Through9
.array;
writeln("Second-to-last Element of the Sequence: ", seq10digits[$ - 2]);
writeln("Last Element of the Sequence: ", seq10digits[$-1]);
}
Second-to-last Element of the Sequence: 9876134520
Last Element of the Sequence: 9876351240
Trying other paradigms is tempting... a Taste of Lisp feeling this time, Go for it! 🏃♂️🏃♂️🏃♂️
#lang racket
(require math)
; Function to split a number into its digits
(define (digits n)
(if (zero? n)
'(0)
(reverse (let loop ((n n) (acc '()))
(if (zero? n)
acc
(loop (quotient n 10) (cons (modulo n 10) acc)))))))
; Function to combine a list of digits into a number
(define (undigits lst)
(foldl (lambda (d acc) (+ (* 10 acc) d)) 0 lst))
; Generate the list of numbers and filter them
(define xs
(filter (lambda (n)
(= (undigits (sort (digits n) <)) 123456789))
(range 1000001520 10000000000 2520)))
; Check if xs is empty and handle it
(if (null? xs)
(displayln "No numbers satisfy the condition.")
(begin
(displayln (list (cadr (reverse xs))
(car (reverse xs))))
(void)))(9876134520 9876351240)