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# Divisibility

Authors: Darren Yao, Michael Cao, Andi Qu, Kevin Sheng, Mihnea Brebenel

Contributor: Juheon Rhee

Using the information that one integer evenly divides another.

If you've never encountered any number theory before, AoPS is a good place to start.

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## Resources

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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:

$n = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}$

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.

C++

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;}

Java

List<Integer> factor(int n) {	List<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;}

Python

from typing import List

def factor(n: int) -> List[int]:	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 $n = 252$.

$i$$n$$\texttt{ret}$
$2$$252$$\{\}$
$2$$126$$\{2\}$
$2$$63$$\{2, 2\}$
$3$$21$$\{2, 2, 3\}$
$3$$7$$\{2, 2, 3, 3\}$

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 $\{2, 2, 3, 3, 7\}$.

Focus Problem – try your best to solve this problem before continuing!

### 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.

C++

#include <iostream>
using namespace std;
int main() {	int n;	cin >> n;	for (int q = 0; q < n; q++) {		int x;		int div_num = 0;

Java

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();

Python

### Warning!

Due to Python's big constant factor, the following code TLEs on quite a few test cases.

ans = []for _ in range(int(input())):	div_num = 0	x = int(input())	i = 1	while i * i <= x:		if x % i == 0:			div_num += 1 if i**2 == x else 2		i += 1	ans.append(div_num)
print("\n".join(str(i) for i in ans))

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:

$x = {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}$

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 use it with the above property to 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:

1. For each $k \leq 10^6$, find any prime number that divides $k$. To find this, we can use the Sieve of Eratosthenes which runs in $\mathcal{O}(n \log n)$, where $n$ is the larger numbers we consider. There's also a version of the sieve that runs in linear time, but we won't be needing it.
2. We can find the prime factorization of $x$ by repeatedly dividing it by the prime numbers we calculated earlier until $x = 1$.

Using this method gives us the following code:

C++

#include <iostream>
using namespace std;
const int MAX_N = 1e6;// max_div[i] contains the largest prime that goes into iint max_div[MAX_N + 1];
int main() {	for (int i = 2; i <= MAX_N; i++) {

Java

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++) {

Python

MAX_N = 10**6
# max_div[i] contains the largest prime that can go into imax_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:

$\gcd(a, b) = \begin{cases} a & b = 0 \\ \gcd(b, a \bmod b) & b \neq 0 \end{cases}$

This algorithm can be implemented with a recursive function as follows:

Java

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

C++

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

For C++14, you can use the built-in __gcd(a,b).

In C++17, there exists std::gcd and std::lcm in the <numeric> header, so there's no need to code your own GCD and LCM if you're using that version.

Python

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

You won't have to actually implement this in-contest, as the built-in math library has a gcd and lcm function.

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 $F_{n + 1}$. In this case, the algorithm will calculate that $\gcd(F_n, F_{n + 1}) = \gcd(F_{n - 1}, F_n) = \dots = \gcd(0, F_1)$. This takes a total of $n+1$ calls, 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 be calculated with the GCD using this property:

$\operatorname{lcm}(a, b) = \frac{a \cdot b}{\gcd(a, b)}$

### Warning!

Coding $\text{lcm}$ as a * b / gcd(a, b) might cause integer overflow if the value of a * b is greater than the max size of the data type of a * b (e.g. the max size of int in C++ and Java is around 2 billion). Dividng a by gcd(a, b) first, then multiplying it by b will prevent integer overflow if the result fits in an int.

Also, these two functions are associative, meaning that 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,

$\gcd(a_1, a_2, a_3, a_4) = \gcd(a_1, \gcd(a_2, \gcd(a_3, a_4))).$

## Euler's Totient Function

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Theory and Exercises

codeforces

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### Properties

Euler's totient function - written using phi $\phi(n)$ - counts the number of positive integers in the interval $[1,n]$ which are coprime to $n$. Two numbers $a$ and $b$ are coprime if their greatest common divisor is equal to 1 i.e. $gcd(a,b)=1$.

Here are the values of $\phi(n)$ for the first 20 numbers:

n1234567891011121314151617181920
$\phi(n)$112242646410412688166188

The totient function is multiplicative meaning that $\phi(nm)=\phi(n) \cdot \phi(m)$, where $n$ and $m$ are coprime - $gcd(n, m)=1$. For example $\phi(15)=\phi(3 \cdot 5)=\phi(3) \cdot \phi(5) = 2 \cdot 4 = 8$.

Let's take a look at some edge cases for $\phi(n)$:

• If n is a prime number then $\phi(n)=n-1$ because $gcd(n, x)=1$ for all $1 \leq x \le n$
• If n is a power of a prime number, $n=p^q$ where p is a prime number and $1 \leq q$ then
there are exactly $p^{q-1}$ numbers divisible by $p$, so $\phi(p^q)=p^{q} - p^{q-1} = p^{q-1}(p - 1)$

Using the multiplicative property and the last edge case we can compute the value of $\phi(n)$ from the factorization of number $n$. Let the factorization be $n=p_1^{q_1} \cdot p_2^{q_2} \cdot \ldots \cdot p_k^{q_k}$ where $p_i$ is a prime factor of $n$, then:

$\phi(n)=\phi(p_1^{q_1}) \cdot \phi(p_2^{q_2}) \cdot \ldots \cdot \phi(p_k^{q_k}) = p_1^{q_1-1}(p_1 - 1) \cdot p_2^{q_2-1}(p_2 - 1) \cdot \ldots \cdot p_k^{q_k-1}(q_k - 1)$

Below is an implementation for factorization in $\mathcal{o}(\sqrt{n})$. It can be a little bit tricky to understand why we substract $and/p$ from $ans$. For example $ans=p_q \cdot x$ ,where $p$ is a prime factor and $x$ is the rest of the prime factorization. By substracting $\frac{ans}{p}=p_{q-1} \cdot x$ we end up with: $p_q \cdot x - p_{q-1} \cdot x = p_{q-1} \cdot x \cdot (p - 1)$ which is exactly the form of $\phi(n)$ described a few linew above.

C++

int phi(int n) {	int ans = n;	for (int p = 2; p * p <= n; p++) {		if (n % p == 0) {			while (n % p == 0) { n /= p; }			ans -= ans / p;		}	}	if (n > 1) { ans -= ans / n; }	return ans;}

Usually in problems we need to precompute the totient of all numbers between $1$ and $n$, then factorizing is not efficient. The idea is the same as the Sieve of Eratosthenes. Since it's almost the same with the Sieve of Eratosthenes the time complexity will be: $\mathcal{n\log{n}\log{n}}$.

C++

void precompute() {	for (int i = 1; i < MAX_N; i++) { phi[i] = i; }	for (int i = 2; i < MAX_N; i++) {		// If i is prime		if (phi[i] == i) {			for (int j = i; j < NMAX; j += i) { phi[j] -= phi[j] / i; }		}	}}

Focus Problem – try your best to solve this problem before continuing!

### Solution

The queries ask us for the sum of the greatest common divisor between all the numbers smaller than $n$.

Let's start by defining a function $f(n, d)$ as how many pairs of numbers less or equal than $n$ have the greatest common divisor equal to $d$. i.e. $gcd(a, b)=d$.

Defining the function this way gives us the following equation: $f(n,d)=f(\frac{n}{d}, 1)$. The reasong why is simple: By choosing two coprime numbers $a$ and $b$ such that $a,b \le \frac{n}{d}$, we can make sure $a \cdot d, b \cdot d \le n$ and $gcd(a \cdot d, b \cdot d) = d$, thus contributing to the global answer.

Shifting away to our new function $f(\frac{n}{d}, 1)$ allows us to compute $f(n, d)$ more easily using a slightly modified phi function preprocessing. To find the sum we multiply $f(n, d)$ with $d$.

This way the answer for a query is $\sum_{d=1}^{n}d \cdot f(n, d) = \sum_{d=1}^{n}d \cdot f(\frac{n}{d}, 1)$.

C++

#include <bits/stdc++.h>using namespace std;
const int MAX_N = 1e6;// phi[i] = how many pairs of numbers <=i are coprimelong long phi[MAX_N + 1];// Store the values found to speed up the recursion processunordered_map<int, long long> memo;
long long solve(long long n) {

## Problems

StatusSourceProblem NameDifficultyTags
ACEasy
Show TagsPrime Factorization
CFEasy
Show TagsDivisibility, Modular Arithmetic
CFEasy
Show TagsNT
CFEasy
Show TagsDivisibility
CSESNormal
CFNormal
Show TagsPrime Factorization
CCNormal
Show TagsDivisibility
CSESHard
CFHard
Show TagsDivisibility

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