2.4 KiB
2.4 KiB
id | challengeType | title | forumTopicId |
---|---|---|---|
5900f3931000cf542c50fea5 | 5 | Problem 38: Pandigital multiples | 302042 |
Description
Take the number 192 and multiply it by each of 1, 2, and 3:
192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
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
tests:
- text: <code>pandigitalMultiples()</code> should return a number.
testString: assert(typeof pandigitalMultiples() === 'number');
- text: <code>pandigitalMultiples()</code> should return 932718654.
testString: assert.strictEqual(pandigitalMultiples(), 932718654);
Challenge Seed
function pandigitalMultiples() {
return true;
}
pandigitalMultiples();
Solution
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;
}