# Project Euler 38 Solution: 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?

## Solution

This problem can be brute-forced quite easily, but lets do some analysis to shrink the search space. We need to find a 9-digit pandigital number greater than the given 918273645, which implies that the first digit of the concatenated number starts with 9 as well. Since $$n > 1$$, the trivial solution of $$987654321\cdot 1$$ is not allowed and with $$n$$ being at least two, the maximum digit count of the fixed number $$x$$ is forced to be $$\leq 4$$.

If the fixed number $$x$$ is a two-digit number, all resulting numbers will have $$<9$$ digits with $$n\leq 3$$ and $$>9$$ with $$n\geq 4$$, which allows us to exclude the range $$90\leq x < 100$$. If the fixed number $$x$$ is a three-digit number, all resulting numbers will be $$\neq 9$$ as well, which excludes the range $$900\leq x<1000$$. Finally, if $$x$$ is a four-digit number, we will get a 9 digit number, which leaves us with the first rough valid range of $$9000\leq x < 10000$$. This interval can be reduced to valid pandigital values, which results in an interval of $$9123\leq x \leq 9876$$. If we check again the length argument, we can lift the lower bound to 9213, since 9123 will not result in a 9 digit number, which gives the new interval $$9213\leq x \leq 9876$$.

When digging a little deeper, we see that if the second digit is $$>4$$, a carry over happens and two 9s are produced in the solution. A similar observation shows that non of the digits can be 1 since we will otherwise end up with two 1s in the solution. This allows us to further reduce the search space to $$9234\leq x\leq 9487$$.

Multiplied by two, the interval is $$18468 \leq x\leq 18974$$, which says that concatenating $$x$$ and $$2x$$ can be done with $$100000x + 2x$$ or better $$100002x$$. When we use the isPandigital function of Problem 32, the solution is just

function solution() {
for (let x = 9487; x >= 9234; x--) {
let res = 100002 * x;
if (isPandigital(res)) {
return res;
}
}
return null;
}

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