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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,Mathematics

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# Prime Factors Decomposition

## Fraction A/B Prime Numbers Factorization

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

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

The decomposition into prime factors of a number $$N$$ requires attempting to divide the number $$N$$ by the set of prime factors $$p$$ which are less than $$N$$. If $$p$$ is a divisor of $$N$$ then we start again by taking $$N = N/p$$ as long as there are any possible divisors.

Consider the number $$N = 147$$, the prime numbers less than $$N = 147$$ are $$2, 3, 5, 7, 11, 13, ...$$. To find the decomposition into a product of prime factors of $$147$$, you begin by attempting the division by $$2$$, $$147$$ is not divisible by $$2$$. You then divide by $$3$$, $$147/3 = 49$$ so $$147$$ is divisible by $$3$$ and $$3$$ is a prime factor of $$147$$. Then, you no longer consider $$147$$ but $$147/3 = 49$$. The prime numbers less than $$49$$ are $$2, 3, 5, 7, 11, 13, ...$$ You try to divide \ (49 \) by \ (2 \) and so on.

Finally, you'll get the factors $$3, 7, 7$$ and you can check that $$3 * 7 * 7 = 149$$, an you can also write $$147 = 3 * 7 ^ 2$$.

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

$$123 = 3 * 41$$, \ (1234 = 2 * 617 \), \ (12345 = 3 * 5 * 823 \) or \ (123456 = 2 ^ 6 * 3 * 643 \

The problem with this method (or algorithm) is that it is very long when the numbers are very large.

dCode allows numbers up to 100 digits, but will stop calculation if it requires too many resources or takes too long.

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

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

### How to know if a number is a prime?

To know if a number is prime, it need to check if it has any divisor except 1 or itself, this test is called a primality test.

### What are primality tests algorithms?

It exists several test to know if a number is a prime number : Miller–Rabin or Lucas-Lehmer are the one used by dCode.

### 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 :

Consider $$P$$ a prime number and $$P\#$$ the Primorial of $$P$$ : the product of $$2*3*5*......*P$$ of all the prime numbers between $$2$$ and $$P$$. Consider $$Q = P\#+1$$, 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;};