102 lines
2.0 KiB
Markdown
102 lines
2.0 KiB
Markdown
---
|
|
id: 5900f3a41000cf542c50feb7
|
|
title: 'Problem 56: Powerful digit sum'
|
|
challengeType: 5
|
|
forumTopicId: 302167
|
|
dashedName: problem-56-powerful-digit-sum
|
|
---
|
|
|
|
# --description--
|
|
|
|
A googol ($10^{100}$) is a massive number: one followed by one-hundred zeros; $100^{100}$ is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
|
|
|
|
Considering natural numbers of the form, $a^b$, where `a`, `b` < `n`, what is the maximum digital sum?
|
|
|
|
# --hints--
|
|
|
|
`powerfulDigitSum(3)` should return a number.
|
|
|
|
```js
|
|
assert(typeof powerfulDigitSum(3) === 'number');
|
|
```
|
|
|
|
`powerfulDigitSum(3)` should return `4`.
|
|
|
|
```js
|
|
assert.strictEqual(powerfulDigitSum(3), 4);
|
|
```
|
|
|
|
`powerfulDigitSum(10)` should return `45`.
|
|
|
|
```js
|
|
assert.strictEqual(powerfulDigitSum(10), 45);
|
|
```
|
|
|
|
`powerfulDigitSum(50)` should return `406`.
|
|
|
|
```js
|
|
assert.strictEqual(powerfulDigitSum(50), 406);
|
|
```
|
|
|
|
`powerfulDigitSum(75)` should return `684`.
|
|
|
|
```js
|
|
assert.strictEqual(powerfulDigitSum(75), 684);
|
|
```
|
|
|
|
`powerfulDigitSum(100)` should return `972`.
|
|
|
|
```js
|
|
assert.strictEqual(powerfulDigitSum(100), 972);
|
|
```
|
|
|
|
# --seed--
|
|
|
|
## --seed-contents--
|
|
|
|
```js
|
|
function powerfulDigitSum(n) {
|
|
|
|
return true;
|
|
}
|
|
|
|
powerfulDigitSum(3);
|
|
```
|
|
|
|
# --solutions--
|
|
|
|
```js
|
|
function powerfulDigitSum(n) {
|
|
function sumDigitsOfPower(numA, numB) {
|
|
let digitsSum = 0;
|
|
let number = power(numA, numB);
|
|
while (number > 0n) {
|
|
const digit = number % 10n;
|
|
digitsSum += parseInt(digit, 10);
|
|
number = number / 10n;
|
|
}
|
|
return digitsSum;
|
|
}
|
|
|
|
function power(numA, numB) {
|
|
let sum = 1n;
|
|
for (let b = 0; b < numB; b++) {
|
|
sum = sum * BigInt(numA);
|
|
}
|
|
return sum;
|
|
}
|
|
|
|
const limit = n - 1;
|
|
let maxDigitsSum = 0;
|
|
for (let a = limit; a > 0; a--) {
|
|
for (let b = limit; b > 0; b--) {
|
|
const curDigitSum = sumDigitsOfPower(a, b);
|
|
if (curDigitSum > maxDigitsSum) {
|
|
maxDigitsSum = curDigitSum;
|
|
}
|
|
}
|
|
}
|
|
return maxDigitsSum;
|
|
}
|
|
```
|