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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;
}

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