2.0 KiB
2.0 KiB
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3b41000cf542c50fec7 | Problem 72: Counting fractions | 5 | 302185 | problem-72-counting-fractions |
--description--
Consider the fraction, \frac{n}{d}
, where n
and d
are positive integers. If n
< d
and highest common factor, {HCF}(n, d) = 1
, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d
≤ 8 in ascending order of size, we get:
\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}
It can be seen that there are 21
elements in this set.
How many elements would be contained in the set of reduced proper fractions for d
≤ limit
?
--hints--
countingFractions(8)
should return a number.
assert(typeof countingFractions(8) === 'number');
countingFractions(8)
should return 21
.
assert.strictEqual(countingFractions(8), 21);
countingFractions(20000)
should return 121590395
.
assert.strictEqual(countingFractions(20000), 121590395);
countingFractions(500000)
should return 75991039675
.
assert.strictEqual(countingFractions(500000), 75991039675);
countingFractions(1000000)
should return 303963552391
.
assert.strictEqual(countingFractions(1000000), 303963552391);
--seed--
--seed-contents--
function countingFractions(limit) {
return true;
}
countingFractions(8);
--solutions--
function countingFractions(limit) {
const phi = {};
let count = 0;
for (let i = 2; i <= limit; i++) {
if (!phi[i]) {
phi[i] = i;
}
if (phi[i] === i) {
for (let j = i; j <= limit; j += i) {
if (!phi[j]) {
phi[j] = j;
}
phi[j] = (phi[j] / i) * (i - 1);
}
}
count += phi[i];
}
return count;
}