Question

Do Prime factorisation of the number \[1729\].

Answer 1

Hint: To find the prime factors we divide the number by all of the prime numbers. And if it is divisible, then that number is a factor. Then repeat the process to find next.

Useful formula:
Prime numbers are those numbers with only factors one and the number itself.
Prime factorisation means write the number as a product of powers of prime factors.

Complete step by step solution:
Here the number given is $1729$.
Prime factorisation means write this number as a product of powers of prime factors.
We need to find all the prime factors of it.
Prime numbers are those numbers with only factors one and the number itself.
The list of the prime numbers taken in ascending order is $2,3,5,7,11,13,17,19,23,29,...$
So, to find the prime factors of $1729$, let’s divide it by each of these.
By observation, we can see that $1729$ is not divisible by $2,3,5$.
But it is divisible by $7$.
$ \Rightarrow 1729 = 7 \times 247 - - - (i)$
So $7$ is the least prime factor of $1729$.
Continuing, we have to factorise $247$ further.
Here also $247$ is not divisible by $2,3,5,7,11$.
But it is divisible by $13$, meaning the least factor is $13$.
We have, $247 = 13 \times 19$
Therefore, from $(i)$ we can write
$ \Rightarrow 1729 = 7 \times 13 \times 19 - - - (ii)$
All these numbers $7,13,19$ are primes. So no further factorisation is possible.

Therefore prime factorisation of $1729$ is $1729 = 7 \times 13 \times 19$.

Additional information:Numbers which are not prime are called composite. One is neither prime nor composite.

Note: While doing prime factorisation if a factor is repeated more than once, then it is written in powers. That is if we have $p \times p \times q \times q \times q \times r$, we write it as ${p^2} \times {q^3} \times r$. If the number itself is a prime, it is not possible to factorise.