
Four prime numbers are written in ascending order of their magnitude. The product of the first three is 715 and that of the last three is 2431. What is the largest given prime number?
$
{\text{A}}{\text{. 5}} \\
{\text{B}}{\text{. 19}} \\
{\text{C}}{\text{. 17}} \\
{\text{D}}{\text{. 23}} \\
$
Answer
601.8k+ views
Hint- Here, we will proceed by letting the given prime numbers arranged in ascending order and then obtain the relations between these prime numbers according to the problem statement. Then, we will solve for the largest prime number using the HCF concept and prime factorisation method.
Complete step-by-step solution -
Let us suppose we have four given prime numbers arranged in ascending order as a, b, c and d
i.e., aGiven, the product of the first three prime numbers is 715 i.e., \[a \times b \times c = 715{\text{ }} \to {\text{(1)}}\]
Also given that the product of the last three prime numbers is 2431 i.e., \[b \times c \times d = 2431{\text{ }} \to {\text{(2)}}\]
Here, HCF of the numbers \[a \times b \times c\] and \[b \times c \times d\] is \[b \times c\]
By prime factorisation method,
The numbers 715 and 2431 can be represented as the product of its prime factors as given below
$715 = 5 \times 11 \times 13$ and $2431 = 11 \times 13 \times 17$
So, the HCF of the numbers 715 and 2431 = $11 \times 13 = 143$
Using equations (1) and (2), we can write
HCF of the numbers \[a \times b \times c\] and \[b \times c \times d\] = HCF of the numbers 715 and 2431
$ \Rightarrow b \times c = 143{\text{ }} \to {\text{(3)}}$
By dividing equation (3) from equation (2), we get
\[
\dfrac{{b \times c \times d}}{{b \times c}} = \dfrac{{2431}}{{143}} \\
\Rightarrow d = 17 \\
\]
Since, d is the largest prime number out of all the given prime numbers
Therefore, the largest given prime number is 17.
Hence, option C is correct.
Note- In this particular problem, if instead of ascending order we would have given that the prime numbers a, b, c and d are arranged in descending order then we would be having a>b>c>d such that the prime number a is the largest of the given prime numbers and the prime number d is the smallest of the given prime numbers.
Complete step-by-step solution -
Let us suppose we have four given prime numbers arranged in ascending order as a, b, c and d
i.e., aGiven, the product of the first three prime numbers is 715 i.e., \[a \times b \times c = 715{\text{ }} \to {\text{(1)}}\]
Also given that the product of the last three prime numbers is 2431 i.e., \[b \times c \times d = 2431{\text{ }} \to {\text{(2)}}\]
Here, HCF of the numbers \[a \times b \times c\] and \[b \times c \times d\] is \[b \times c\]
By prime factorisation method,
The numbers 715 and 2431 can be represented as the product of its prime factors as given below
$715 = 5 \times 11 \times 13$ and $2431 = 11 \times 13 \times 17$
So, the HCF of the numbers 715 and 2431 = $11 \times 13 = 143$
Using equations (1) and (2), we can write
HCF of the numbers \[a \times b \times c\] and \[b \times c \times d\] = HCF of the numbers 715 and 2431
$ \Rightarrow b \times c = 143{\text{ }} \to {\text{(3)}}$
By dividing equation (3) from equation (2), we get
\[
\dfrac{{b \times c \times d}}{{b \times c}} = \dfrac{{2431}}{{143}} \\
\Rightarrow d = 17 \\
\]
Since, d is the largest prime number out of all the given prime numbers
Therefore, the largest given prime number is 17.
Hence, option C is correct.
Note- In this particular problem, if instead of ascending order we would have given that the prime numbers a, b, c and d are arranged in descending order then we would be having a>b>c>d such that the prime number a is the largest of the given prime numbers and the prime number d is the smallest of the given prime numbers.
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