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Prime Factors Decomposition

Tool to decompose in prime factors. In Mathematics, the prime factors decomposition (also known as Prime Integer Factorization) consists in writing a positive integer with a product of prime factors

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Prime Factors Decomposition -

Tag(s) : Arithmetics

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# Prime Factors Decomposition

## Prime Numbers Decomposition

Very big numbers allowed - unlimited size (see FAQ)

## Fast Prime Decomposition

This function is limited to 100 integers each less than 1 billion

### How to decompose a number in a product of prime factors?

To find the prime factorization of a number $N$ there is no mathematical formula. To achieve this, there are algorithms including the most basic that attempt to divide the number $N$ by all prime factors $p$ which are less than $N$. If $p$ is a divisor of $N$ then start again by taking a new $N = N/p$ as long as there are any possible divisors.

Example: If $N = 147$, the prime numbers less than $N = 147$ are $2, 3, 5, 7, 11, 13, ...$. The prime factorisation algorithm for $147$, begins by attempting the division by $2$, $147$ is not divisible by $2$. Then divide by $3$, $147/3 = 49$ so $147$ is divisible by $3$ and $3$ is a prime factor of $147$. Then, no longer take $147$ but $147/3 = 49$. The prime numbers less than $49$ are $2, 3, 5, 7, 11, 13, ...$, try to divide $49$ by $2$ and so on.

Example: Finally, it remains the factors $3, 7, 7$ and check that $3 * 7 * 7 = 147$, or write $147 = 3 * 7 ^ 2$.

This decomposition is possible whatever the starting number, it is a fundamental theorem of arithmetic.

Example: $123 = 3 * 41$, $1234 = 2 * 617$, $12345 = 3 * 5 * 823$ or $123456 = 2 ^ 6 * 3 * 643$

### What decompositions algorithms are sometimes limited?

The problem with prime number decomposition methods (or algorithms) is that they are very long when the numbers are very large. As soon as the factors have several tens of digits and are not trivial, several minutes or even hours or even days of calculations can be necessary, even for the most powerful computers.

dCode performs the calculations on the server side as possible, but if the requested number has too many digits, the calculation will continue on your browser thanks to a WebAssembly applet (wasm) by Dario Alejandro Alpern (License GPL v3.0) the calculation time will therefore depend on the performance of your computer / phone.

### What are algorithms allowing to decompose in prime factors?

It exists several factorization algorithms : classical iterative division, Pollard rho algorithm, elliptic curves, and the quadratic sieve algorithm. dCode uses a combination of all them to fast factorize.

### Is there a list of prime numbers?

The whole list of prime numbers starts with : 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997... and there are an infinite number of primes.

### How to demonstrate that it exist an infinite number of primes?

The demonstration of the infinity of prime numbers is :

If $P$ is a prime number and $P\#$ the Primorial of $P$ : the product of $2*3*5*......*P$ of all the prime numbers between $2$ and $P$. If $Q = P\#+1$, then, the rest of the euclidean division of $Q$ by any prime number inferior to $P$ will be $1$. So, all prime factors of $Q$ ($Q$ can be prime) are prime numbers superior to $P$. It will always exists prime numbers superiors to $P$.

### How to code a prime factor decomposition?

// javascript function prime_factors(n) { if (!n || n < 2) return []; var f = []; for (var i = 2; i <= n; i++){ while (n % i === 0){ f.push(i); n /= i; } } return f;};

## Source code

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