Question

What number do you get when you multiply the distinct (different) prime factors of $28$?
A) $9$
B) $4$
C) $7$
D) $14$
E) $28$

Answer 1

Hint: First we have to know what prime factors of a number are. So, in simple words, the prime factor is finding which prime numbers multiply together to make the original number. We write the prime factors of a number in the form $n_1^{{l_1}} \times n_2^{{l_2}} \times .... \times n_m^{{l_k}}$, where, ${n_1},{n_2},....,{n_m}$ are prime numbers and ${l_1},{l_2},...,{l_k}$ are positive integers. So, first we have to find the prime factors of $28$. Then we have to find out whose, which are the distinct prime factors among them. Then we are to multiply the distinct prime factors among them to get the required result.

Complete step-by-step solution:
First we have to find the prime factors of the number $28$.
We know, we write the prime factors of a number in the form $n_1^{{l_1}} \times n_2^{{l_2}} \times .... \times n_m^{{l_k}}$.
Therefore, writing $28$ in this form, we get,
$28 = {2^2} \times 7$.
Therefore, the distinct prime factors of $28$ are $2$ and $7$.
So, the number that we get on multiplying the distinct prime factors of $28$$ = 2 \times 7$
$ = 14$
Therefore, the number that we get on multiplying the distinct prime factors of $28$ is $14$.

Note: Prime factors of a number helps us to know all the numbers that are factors of the given number. From the distinct prime numbers of the number, we can use permutation to find all the numbers that divide that given number. Prime factors let us easily convert factors into simpler factors. Like the factor $\dfrac{{24}}{{32}}$, if we write the prime factors of the numerator and denominator, we get, $\dfrac{{{2^3} \times 3}}{{{2^5}}}$.
Now, cancelling the common terms, we will get, $\dfrac{3}{4}$.