Question

The product of two numbers is 4107. If the H.C.F of these number is 37, then the greater number is,
(a)101
(b)107
(c)111
(d)185

Answer 1

Hint: HCF is the highest common factor. 37 is a common factor for the 2 numbers. So take 2 numbers as 37a and 37b. Find the product and get the value of ab. By using the condition of co-prime, find the greatest number.

Complete step-by-step answer:
It is given that the product of 2 numbers is 4107.
It is said that the HCF of these numbers = 37.
Now let us assume that the 2 numbers are 37a and 37b. Here, HCF is the highest common factor, which means the 37 is a common factor in both the numbers.
So, the product of 2 numbers = 4107.
\[\therefore 37a\times 37b=4107\]
Simplifying the expression,
\[\begin{align}
  & \therefore ab=\dfrac{4107}{37\times 37} \\
 & ab=3 \\
\end{align}\]
In number theory 2 integers a and b are said to be co – prime if the positive integer that divides both of them is 1.
Any prime number that divides one does not divide the other.
\[\therefore \]The co – prime with product 3 are (1, 3).
\[\therefore \]The 2 numbers are 37a and 37b.
 \[\begin{align}
  & 37a=37\times 1=37 \\
 & 37b=37\times 3=111 \\
\end{align}\]
\[\therefore \]The two numbers are (37, 111) and the greatest number is 111.
\[\therefore \]Option (c) is the correct answer.

Note: Another method is let us take LCM = x and HCF = 37.
HCF\[\times \]LCM = Product of 2 numbers.
\[\begin{align}
  & \therefore 37x=4107 \\
 & \therefore x=\dfrac{4107}{37}=111 \\
\end{align}\]
\[\therefore \]2 numbers are 37 and 111. Greatest number = 111