Problem 38 "Pandigital multiples"

Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?

問 38 「パンデジタル倍数」

192 に 1, 2, 3 を掛けてみよう.

192 × 1 = 192 192 × 2 = 384 192 × 3 = 576

積を連結することで1から9の パンデジタル数 192384576 が得られる.

192384576 を 192 と (1,2,3) の連結積と呼ぶ.

同じようにして, 9 を 1,2,3,4,5 と掛け連結することでパンデジタル数 918273645 が得られる. これは 9 と (1,2,3,4,5) との連結積である.

整数と (1,2,..., n) (n; > 1) との連結積として得られる9桁のパンデジタル数の中で最大のものはいくつか?

fn is_pandigital(a: u32, b: u32) -> bool {
    let mut bits = 0u16;
    for n in [a, b].iter_mut() {
        while *n > 0 {
            let d = *n % 10;
            *n /= 10;
            bits |= 1 << d;
        }
    }
    bits == 0b1111111110u16
}

fn largest_pandigital_concatenated_product() -> u32 {
    for n in (9183..=(19000 / 2) as u32).rev() {
        if is_pandigital(n, n * 2) {
            return n * 100_000 + 2 * n;
        }
    }
    918_273_645
}

fn main() {
    let max = largest_pandigital_concatenated_product();

    println!("{}", max);
    assert_eq!(max, 932718654);
}