Question

Which of the following pairs are a pair of co-prime numbers:-
$\left( a \right){\text{ 13 and 14}}$
$\left( b \right){\text{ 8 and 20}}$
$\left( c \right){\text{ 31 and 59}}$
$\left( d \right){\text{ 34 and 85}}$

Answer 1

Hint: We should be familiar with the concept of co-prime numbers in order to solve this question. Two numbers are said to be co-prime numbers if they do not have a common factor other than $1$ or in other words we can say that the highest common factor (H.C.F.) of two co-prime numbers is always $1$ . Let us see one example to understand this concept in a better manner. Example: $2{\text{ , 3}}$ ; check whether these two represent a pair of co-prime numbers or not. The first step is to find the prime factors of the two numbers; $\left( 1 \right){\text{ The prime factors of 2 = 2}} \times {\text{1}}$ . $\left( 2 \right){\text{ The prime factors of 3 are = 3}} \times {\text{1}}$ . We can clearly see that the H.C.F. of $\left( {2,3} \right){\text{ is 1}}$ . Therefore $\left( {2,3} \right)$ are co-prime.

Complete step by step answer:
$\left( a \right){\text{ 13 and 14}}$
Let us first find the prime factors of these two numbers,
The prime factors of $13 = 13 \times 1$
The prime factors of $14 = 2 \times 7 \times 1$ .
$ \Rightarrow {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\left( {13,14} \right) = 1$
Therefore, $13{\text{ and 14 }}$are co-prime .

$\left( b \right){\text{ 8 and 20}}$
Let us first find the prime factors of these two numbers,
The prime factors of $8 = 2 \times 2 \times 2 \times 1$ or $8 = {2^3} \times 1$
 The prime factors of $20 = 2 \times 2 \times 5 \times 1$ or $20 = {2^2} \times 5 \times 1$ .
$ \Rightarrow {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\left( {8,20} \right) = 4$
Therefore, ${\text{8 and 20 }}$are not co-prime .

$\left( c \right){\text{ 31 and 59}}$
Let us first find the prime factors of these two numbers,
The prime factors of $31 = 31 \times 1$
 The prime factors of $59 = 59 \times 1$
$ \Rightarrow {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\left( {31,59} \right) = 1$
Here, we can clearly notice that $31{\text{ and 59}}$ both are prime numbers and we know that the factors of prime numbers are only $1{\text{ and the number itself}}{\text{.}}$
Therefore, ${\text{31 and 59 }}$are co-prime .

$\left( d \right){\text{ 34 and 85}}$
Let us first find the prime factors of these two numbers,
The prime factors of $34 = 2 \times 17 \times 1$
 The prime factors of $85 = 5 \times 17 \times 1$
$ \Rightarrow {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\left( {34,85} \right) = 17$
Therefore, ${\text{34 and 85 }}$are not co-prime .
One thing to keep in mind here is to not get confused between the concept of prime and co-prime numbers. A prime number is a number which can either get divided by itself or by $1$ , while co-prime numbers are a pair of two numbers with a common factor as $1$ only means their H.C.F. is always $1$ .

So, the correct answer is “Option a and c”.

Note: Co-prime numbers are also referred to as ‘relatively prime numbers’. Some properties
co-prime numbers are: $\left( 1 \right)$ $1$ will always be co-prime with every number. $\left( 2
\right)$ Any two successive numbers are always co-prime ( like in the first question $13{\text{ and 14}}$) .
$\left( 3 \right)$The sum of any co-prime pair is always co-prime with their product. $\left( {{\text{eg: 6
and 7, Sum }} \Rightarrow 6 + 7{\text{ = 13 , Product}} \Rightarrow {\text{6}} \times {\text{7 = 42}}}
\right)$. Here
$13{\text{ and 42 are also co - prime}}{\text{.}}$ $\left( 4 \right)$ Any two prime numbers are always co-
prime but vice-versa is not always true. For example: $14,15$ are co-primes but none of the two
numbers are prime numbers instead they are composite numbers.