# Divisibility

Authors: Darren Yao, Michael Cao, Andi Qu

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.

Resources | |||
---|---|---|---|

AoPS | practice problems, set focus to number theory! | ||

AoPS | good book |

## Resources

Resources | |||
---|---|---|---|

IUSACO | module is based off this | ||

David Altizio | |||

CPH | |||

PAPS | |||

MONT |

## Prime Factorization

Focus Problem – read through this problem before continuing!

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.

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

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;}

Python

def factor(n):ret = []i = 2while i * i <= n:while n % i == 0:ret.append(i)n //= ii += 1if 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 $n = 252$.

$i$ | $n$ | $v$ |
---|---|---|

$2$ | $252$ | $\{\}$ |

$2$ | $126$ | $\{2\}$ |

$2$ | $63$ | $\{2, 2\}$ |

$3$ | $21$ | $\{2, 2, 3\}$ |

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

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

### 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 <bits/stdc++.h>using namespace std;int main() {ios_base::sync_with_stdio(0);cin.tie(0);int n;cin >> n;while (n--) {int x, ans = 0;

Java

import java.io.*;import java.util.*;public class Main {static int solve(int n) {int divisors = 0;for (int i = 1; i * i <= n; i++) {if (n % i == 0) {if (i * i == n) divisors++;else divisors += 2;

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.

C++

#include <bits/stdc++.h>using namespace std;int max_div[1000001];int main() {ios_base::sync_with_stdio(0);cin.tie(0);for (int i = 2; i <= 1000000; i++) {if (!max_div[i]) {

Java

### This section is not complete.

make java code consistent w/ C++ code ...

import java.io.*;import java.util.*;public class Main {static int solve(int n) {int divisors = 1;for (int i = 2; i * i <= n; i++) {int ct = 0;while (n % i == 0) {ct++;

### Optional

Apply the linear sieve.

## 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:

Java

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

C++

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

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

.

Python

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

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

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:

### 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`

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`

.

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,

C++

**Exercise:** What's wrong with the following code?

ll gcd(ll a, ll b){ return b == 0 ? a : gcd(b,a%b); }ll lcm(ll a, ll b) { return a/gcd(a,b)*b; }int main() { cout << lcm(1000000000,999999999); } // output: 1808348672

Solution

Java

Python

## Problems

Status | Source | Problem Name | Difficulty | Tags | |||||
---|---|---|---|---|---|---|---|---|---|

AC | Easy | ## Show TagsPrime Factorization | |||||||

CSES | Normal | ||||||||

CF | Normal | ## Show TagsPrime Factorization | |||||||

CSES | Hard | ||||||||

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