2.0 KiB
2.0 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3a41000cf542c50feb7 | Problem 56: Powerful digit sum | 5 | 302167 | 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.
assert(typeof powerfulDigitSum(3) === 'number');
powerfulDigitSum(3)
should return 4
.
assert.strictEqual(powerfulDigitSum(3), 4);
powerfulDigitSum(10)
should return 45
.
assert.strictEqual(powerfulDigitSum(10), 45);
powerfulDigitSum(50)
should return 406
.
assert.strictEqual(powerfulDigitSum(50), 406);
powerfulDigitSum(75)
should return 684
.
assert.strictEqual(powerfulDigitSum(75), 684);
powerfulDigitSum(100)
should return 972
.
assert.strictEqual(powerfulDigitSum(100), 972);
--seed--
--seed-contents--
function powerfulDigitSum(n) {
return true;
}
powerfulDigitSum(3);
--solutions--
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;
}