* fix: clean-up Project Euler 161-180 * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
1005 B
1005 B
id, title, challengeType, forumTopicId, dashedName
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f4181000cf542c50ff2a | Problem 171: Finding numbers for which the sum of the squares of the digits is a square | 5 | 301806 | problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square |
--description--
For a positive integer n
, let f(n)
be the sum of the squares of the digits (in base 10) of n
, e.g.
$$\begin{align} & f(3) = 3^2 = 9 \\ & f(25) = 2^2 + 5^2 = 4 + 25 = 29 \\ & f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 \\ \end{align}$$
Find the last nine digits of the sum of all n
, 0 < n < {10}^{20}
, such that f(n)
is a perfect square.
--hints--
lastDigitsSumOfPerfectSquare()
should return 142989277
.
assert.strictEqual(lastDigitsSumOfPerfectSquare(), 142989277);
--seed--
--seed-contents--
function lastDigitsSumOfPerfectSquare() {
return true;
}
lastDigitsSumOfPerfectSquare();
--solutions--
// solution required