factors of 1357913579
Answers
Step-by-step explanation:
) We need the list of the first prime numbers, ordered from 2 up to, let's say, 100: 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. Prime numbers are the building blocks of the composite numbers' prime factorization process.
1) Start by dividing 220 by the first prime number in our list, 2 (start with the smallest prime number). We could have skipped the division by 2 if our number were an odd number, ie. 221.
220 ÷ 2 = 110; remainder = 0 => 220 is divisible by 2 (2 | 220) => 2 is a prime factor of 220. => 220 = 2 × 110.
2) Divide the result of the previous operation, 110, by 2, again:
110 ÷ 2 = 55; remainder = 0 => 110 is divisible by 2 (2 | 110) => 2 is once again a prime factor of 220 => 220 = 2 × 2 × 55.
3) Divide the result of the previous operation, 55, by 2, again:
55 ÷ 2 = 27 + 1; remainder = 1 => 55 is not divisible by 2.
4) Move on, divide 55 by the next prime number, 3 (notice that the sum of digits of 55 is 5 + 5 = 10 is not divisible by 3, so 55 is not divisible by 3):
55 ÷ 3 = 18 + 1; remainder = 1 => as we already said, 55 is not divisible by 3.
5) Divide 55 by the next prime number, 5:
55 ÷ 5 = 11; remainder = 0 => 55 is divisible by 5 (5 | 55) => 5 is yet another prime factor of 220 => 220 = 2 × 2 × 5 × 11.
6) Notice that the remaining number, 11, is also a prime number, so we've got already all the prime factors of 220.
7) Conclusion, 220 prime factorization: 220 = 2 × 2 × 5 × 11.
This can be written in a condensed form, in the exponent notation: 220 = 22 × 5 × 11.
Step-by-step explanation:
Final answer:
1,357,913,579 is not a prime, is a composite number.
Integer prime factorization, as a product of prime factors:
1,357,913,579 = 11 × 37 × 367 × 9,091;