Question

What number do you get when you multiply distinct prime number factors of 56?
(A). 48
(B). 7
(C). 14
(D). 28
(E). 56

Answer 1

Hint: In this question it is given that we have to find the multiplication of the factors of the number 56. So to find the solution we need to factorise the given number, after that we have to take those factors which are prime numbers and multiplication of those prime factors will give us the required solution.

Complete step-by-step solution:
Given number is 56, now we are going to factorise this number,
So this number 56 can be divided by 2, after division we will get two factors one is 2 and another one is 28, so we can write, $$56=2\times 28$$
Again we can divide 28 by 2 then we again get two factors,one is 2 and another one is 14.
Therefore, 56 can be written as, $$56=2\times 2\times 14$$
And lastly 14 can be divided by 7 and we get 2 and 7 as a factor of 14.
So ultimately,
$$56=2\times 2\times 2\times 7$$
Since we have to take distinct prime numbers which means different or unique, so the prime numbers are 2 and 7.
Therefore the multiplication = $$2\times 7$$= 14.
Hence the correct option is option C.

Note: To solve this type of problem you need to know that ‘factors’ are the numbers you multiply to get another number. For example, factors of 15 are 3 and 5, i.e, $$15=3\times 5$$. Some numbers have more than one factorization (more than one way of being factored). For instance, 12 can be factored as $$\times 12$$, $$2\times 6$$, or $$3\times 4$$. A number that can only be factored as 1 times itself is called "prime". The first few primes are 2, 3, 5, 7, 11, and 13. The number 1 is not regarded as a prime, and is usually not included in factorizations, because 1 goes into everything.
You most often want to find the "prime factorization" of a number: the list of all the prime-number factors of a given number. The prime factorization does not include 1, but does include every copy of every prime factor. For instance, the prime factorization of 8 is 2×2×2, here 2 is the only factor, i.e, the number of identical factors is only one, but you need three copies of it to multiply back to 8, so the prime factorization includes all three copies.