--- id: 5900f3931000cf542c50fea5 challengeType: 5 title: '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?
## Instructions
## Tests
```yml tests: - text: pandigitalMultiples() should return 932718654. testString: 'assert.strictEqual(pandigitalMultiples(), 932718654, "pandigitalMultiples() should return 932718654.");' ```
## Challenge Seed
```js function pandigitalMultiples() { // Good luck! return true; } pandigitalMultiples(); ```
## Solution
```js 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; } ```