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id	title	challengeType	forumTopicId	dashedName
5900f3931000cf542c50fea5	Problem 38: Pandigital multiples	5	302042	problem-38-pandigital-multiples

--description--

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?

--hints--

pandigitalMultiples() should return a number.

assert(typeof pandigitalMultiples() === 'number');

pandigitalMultiples() should return 932718654.

assert.strictEqual(pandigitalMultiples(), 932718654);

--seed--

--seed-contents--

function pandigitalMultiples() {

  return true;
}

pandigitalMultiples();

--solutions--

function pandigitalMultiples() {

  function get9DigitConcatenatedProduct(num) {
    // returns false if concatenated product is not 9 digits
    let concatenatedProduct = num.toString();
    for (let i = 2; concatenatedProduct.length < 9; i++) {
      concatenatedProduct += num * i;
    }
    return concatenatedProduct.length === 9 ? concatenatedProduct : false;
  }

  function is1to9Pandigital(num) {
    const numStr = num.toString();

    // check if length is not 9
    if (numStr.length !== 9) {
      return false;
    }

    // check if pandigital
    for (let i = 9; i > 0; i--) {
      if (numStr.indexOf(i.toString()) === -1) {
        return false;
      }
    }
    return true;
  }

  let largestNum = 0;
  for (let i = 9999; i >= 9000; i--) {
    const concatenatedProduct =  get9DigitConcatenatedProduct(i);
    if (is1to9Pandigital(concatenatedProduct) && concatenatedProduct > largestNum) {
      largestNum = parseInt(concatenatedProduct);
      break;
    }
  }
  return largestNum;
}

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