Divisibility

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Prime Factorization

A positive integer $a$ is called a divisor or a factor of a non-negative integer $b$ if $b$ is divisible by $a$, which means that there exists some integer $k$ such that $b = ka$. An integer $n > 1$ is prime if its only divisors are $1$ and $n$. Integers greater than $1$ that are not prime are composite.

Every positive integer has a unique prime factorization: a way of decomposing it into a product of primes, as follows:

where the $p_i$ are distinct primes and the $a_i$ are positive integers.

Now, we will discuss how to find the prime factorization of any positive integer.

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vector<int> factor(int n) {
	vector<int> ret;
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret.push_back(i);
			n /= i;
		}
	}
	if (n > 1) ret.push_back(n);
	return ret;
}

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ArrayList<Integer> factor(int n) {
	ArrayList<Integer> factors = new ArrayList<>();
	for (int i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			factors.add(i);
			n /= i;
		}
	}
	if (n > 1) factors.add(n);
	return factors;
}

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def factor(n):
	ret = []
	i = 2
	while i * i <= n:
		while n % i == 0:
			ret.append(i)
			n //= i
		i += 1
	if n > 1:
		ret.append(n)
	return ret

This algorithm runs in $\mathcal{O}(\sqrt{n})$ time, because the for loop checks divisibility for at most $\sqrt{n}$ values. Even though there is a while loop inside the for loop, dividing $n$ by $i$ quickly reduces the value of $n$, which means that the outer for loop runs less iterations, which actually speeds up the code.

Let's look at an example of how this algorithm works, for $n = 252$.

At this point, the for loop terminates, because $i$ is already 3 which is greater than $\lfloor \sqrt{7} \rfloor$. In the last step, we add $7$ to the list of factors $v$, because it otherwise won't be added, for a final prime factorization of $\{2, 2, 3, 3, 7\}$.

Solution - Counting Divisors

The most straightforward solution is just to do what the problem asks us to do - for each $x$, find the number of divisors of $x$ in $\mathcal{O}(\sqrt x)$ time.

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#include <iostream>

using namespace std;

int main() {
	int n;
	cin >> n;
	for (int q = 0; q < n; q++) {
		int x;
		int div_num = 0;

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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Divisors {
	public static void main(String[] args) throws IOException {
		BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
		int queryNum = Integer.parseInt(read.readLine());
		StringBuilder ans = new StringBuilder();
		for (int q = 0; q < queryNum; q++) {

This solution runs in $\mathcal{O}(n \sqrt x)$ time, which is just fast enough to get AC. However, we can actually speed this up to get an $\mathcal{O}((x + n) \log x)$ solution!

First, let's discuss an important property of the prime factorization. Consider:

Then the number of divisors of $x$ is simply $(a_1 + 1) \cdot (a_2 + 1) \cdots (a_k + 1)$.

Why is this true? The exponent of $p_i$ in any divisor of $x$ must be in the range $[0, a_i]$ and each different exponent results in a different set of divisors, so each $p_i$ contributes $a_i + 1$ to the product.

$x$ can have $\mathcal{O}(\log x)$ distinct prime factors, so if we can find the prime factorization of $x$ efficiently, we can answer queries in $\mathcal{O}(\log x)$ time instead of the previous $\mathcal{O}(\sqrt x)$ time.

Here's how we find the prime factorization of $x$ in $\mathcal{O}(\log x)$ time with $\mathcal{O}(x \log x)$ preprocessing:

• For each $k \leq 10^6$, find any prime number that divides $k$.

  • We can use the Sieve of Eratosthenes to find this efficiently.
• For each $x$, we can then find the prime factorization by repeatedly dividing $x$ by a prime number that divides $x$ until $x = 1$.

Alternatively, we can slightly modify the the prime factorization code above.

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#include <iostream>

using namespace std;

const int MAX_N = 1e6;
// max_div[i] contains the largest prime that goes into i
int max_div[MAX_N + 1];

int main() {
	for (int i = 2; i <= MAX_N; i++) {

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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Divisors {
	private static final int MAX_N = (int) Math.pow(10, 6);
	public static void main(String[] args) throws IOException {
		// maxDiv[i] contains the largest prime that can divide i
		int[] maxDiv = new int[MAX_N + 1];
		for (int i = 2; i <= MAX_N; i++) {

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MAX_N = 10 ** 6

# max_div[i] contains the largest prime that can go into i
max_div = [0 for _ in range(MAX_N + 1)]
for i in range(2, MAX_N + 1):
	if max_div[i] == 0:
		for j in range(i, MAX_N + 1, i):
			max_div[j] = i

ans = []

GCD & LCM

GCD

The greatest common divisor (GCD) of two integers $a$ and $b$ is the largest integer that is a factor of both $a$ and $b$. In order to find the GCD of two non-negative integers, we use the Euclidean Algorithm, which is as follows:

This algorithm is very easy to implement using a recursive function, as follows:

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public int gcd(int a, int b) {
	return b == 0 ? a : gcd(b, a % b);
}

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int gcd(int a, int b) {
	return b == 0 ? a : gcd(b, a % b);
}

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def gcd(a, b):
	return a if b == 0 else gcd(b, a % b)

This function runs in $\mathcal{O}(\log ab)$ time because $a\le b \implies b\pmod a <\frac{b}{2}$.

The worst-case scenario for the Euclidean algorithm is when $a$ and $b$ are consecutive Fibonacci numbers $F_n$ and $F_{n + 1}$. for an explanation). In this case, the algorithm will calculate $\gcd(F_n, F_{n + 1}) = \gcd(F_{n - 1}, F_n) = \dots = \gcd(0, F_1)$. This means that finding $\gcd(F_n, F_{n + 1})$ takes $n + 1$ steps, which is proportional to $\log \left(F_n F_{n+1}\right)$.

LCM

The least common multiple (LCM) of two integers $a$ and $b$ is the smallest integer divisible by both $a$ and $b$. The LCM can easily be calculated from the following property with the GCD:

If we want to take the GCD or LCM of more than two elements, we can do so two at a time, in any order. For example,

Problems