fix(curriculum): rework Project Euler 38 (#42321)

* fix: rework challenge to use argument in function

* fix: update solution to challenge changes

* fix: use MathJax to improve look of math notation
This commit is contained in:
gikf
2021-06-05 10:16:40 +02:00
committed by GitHub
parent 2ac98e39a2
commit 989347387f

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@ -10,30 +10,36 @@ dashedName: problem-38-pandigital-multiples
Take the number 192 and multiply it by each of 1, 2, and 3:
<div style='margin-left: 4em;'>
192 × 1 = 192<br>
192 × 2 = 384<br>
192 × 3 = 576<br>
</div>
$$\begin{align}
192 × 1 = 192\\\\
192 × 2 = 384\\\\
192 × 3 = 576\\\\
\end{align}$$
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?
What is the largest 1 to `k` pandigital `k`-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.
`pandigitalMultiples(8)` should return a number.
```js
assert(typeof pandigitalMultiples() === 'number');
assert(typeof pandigitalMultiples(8) === 'number');
```
`pandigitalMultiples()` should return 932718654.
`pandigitalMultiples(8)` should return `78156234`.
```js
assert.strictEqual(pandigitalMultiples(), 932718654);
assert.strictEqual(pandigitalMultiples(8), 78156234);
```
`pandigitalMultiples(9)` should return `932718654`.
```js
assert.strictEqual(pandigitalMultiples(9), 932718654);
```
# --seed--
@ -41,38 +47,37 @@ assert.strictEqual(pandigitalMultiples(), 932718654);
## --seed-contents--
```js
function pandigitalMultiples() {
function pandigitalMultiples(k) {
return true;
}
pandigitalMultiples();
pandigitalMultiples(8);
```
# --solutions--
```js
function pandigitalMultiples() {
function get9DigitConcatenatedProduct(num) {
// returns false if concatenated product is not 9 digits
function pandigitalMultiples(k) {
function getKDigitConcatenatedProduct(num, k) {
// returns false if concatenated product is not k digits
let concatenatedProduct = num.toString();
for (let i = 2; concatenatedProduct.length < 9; i++) {
for (let i = 2; concatenatedProduct.length < k; i++) {
concatenatedProduct += num * i;
}
return concatenatedProduct.length === 9 ? concatenatedProduct : false;
return concatenatedProduct.length === k ? concatenatedProduct : false;
}
function is1to9Pandigital(num) {
function is1toKPandigital(num, k) {
const numStr = num.toString();
// check if length is not 9
if (numStr.length !== 9) {
// check if length is not k
if (numStr.length !== k) {
return false;
}
// check if pandigital
for (let i = 9; i > 0; i--) {
for (let i = k; i > 0; i--) {
if (numStr.indexOf(i.toString()) === -1) {
return false;
}
@ -81,11 +86,13 @@ function pandigitalMultiples() {
}
let largestNum = 0;
for (let i = 9999; i >= 9000; i--) {
const concatenatedProduct = get9DigitConcatenatedProduct(i);
if (is1to9Pandigital(concatenatedProduct) && concatenatedProduct > largestNum) {
largestNum = parseInt(concatenatedProduct);
break;
for (let i = 10 ** Math.floor(k / 2) + 1; i >= 1; i--) {
const concatenatedProduct = getKDigitConcatenatedProduct(i, k);
if (is1toKPandigital(concatenatedProduct, k)) {
const number = parseInt(concatenatedProduct, 10);
if (number > largestNum) {
largestNum = number;
}
}
}
return largestNum;