Question

Write the number of factors of 60 and verify by listing the factors.

Answer 1

Hint:
Here, we will be using the concept of prime factorisation of numbers to express the number 60 as \[60 = {x^a} \times {y^b} \times {z^c} \times \]…, that is a product of its prime factors \[x,y,z,...\]. Prime factorisation is a method of finding the prime factors of the given number. We will use the formula \[\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)\]… to find the number of factors of 60 and list them to verify.

Complete step by step solution:
We know that in the number system, prime numbers are the numbers which have only two factors, 1 and the number itself. Hence, it is obvious that they will be divisible by 1 and the number itself.
Now, the given number is 60. Let us find its divisibility by the prime divisors, starting with the smallest prime number, which is 2.
We observe that the number 60 is an even number. Hence, it is divisible by 2.
\[\dfrac{{60}}{2} = 30\]

Next, we observe that the number 30 is also an even number. Hence, it is divisible by 2.
\[\dfrac{{30}}{2} = 15\]
As 15 is an odd number and it is not divisible by 2. Now, we sum up the digits of the result 15.
Hence, we get \[1 + 5 = 6\].
We see that the sum of the digits of the given number is divisible by 3.
So, the number 15 is divisible by 3.
Thus, dividing 15 by 3, we get
\[\dfrac{{15}}{3} = 5\]
Hence, resolving into factors, we get that 60 can be expressed as
\[60 = 2 \times 2 \times 3 \times 5\]
We will write this product as a product of numbers in exponential form to express 60 as \[60 = {x^a} \times {y^b} \times {z^c} \times \]….
Hence, writing the product in exponential form, we get
\[60 = {2^2} \times {3^1} \times {5^1}\]

Now, the number of factors of a number \[A = {x^a} \times {y^b} \times {z^c} \times \] … can be found using the formula \[\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)\]….
Thus, substituting 2 for \[a\], 1 for \[b\] and 1 for \[c\], we get the number of factors of 60, that is
\[\begin{array}{c}\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right) = \left( {2 + 1} \right)\left( {1 + 1} \right)\left( {1 + 1} \right)\\ = 3 \times 2 \times 2\\ = 12\end{array}\]
The number 60 has 12 factors.
Now, we will list all the factors of 60 to verify the number of factors.
We know that
\[\begin{array}{l}60 = 1 \times 60\\60 = 2 \times 30\\60 = 3 \times 20\\60 = 4 \times 15\\60 = 5 \times 12\\60 = 6 \times 10\\60 = 10 \times 6\\60 = 12 \times 5\\60 = 15 \times 4\\60 = 20 \times 3\\60 = 30 \times 2\\60 = 60 \times 1\end{array}\]
Therefore, the factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60 itself.

Hence, we have verified that the numbers of factors of 60 are 12.

Note:
In this problem, we can easily calculate the number of factors with the formula \[\left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right)\]…. Here, we have also used the divisibility rule to find the factors. Divisibility rule is a way of finding whether a number is completely divisible by another number or not without actually dividing the number.