Question

The product of two numbers is 2028 and their H.C.F is 13. How many such pairs are there?
A). 1
B). 2
C). 3
D). 4

Answer 1

Hint: Consider the two number as a and b. So, we have ab= 2028 and the greatest divisor of them is 13. Write a and b in multiples of 13 and solve for the answer.

Complete step-by-step solution
Let the two numbers are a and b. Now their H.C.F is given as 13. So this means 13 is the greatest common divisor of both of them.
So we can write \[a=13{{k}_{1}}\] and \[b=13{{k}_{2}}\] where \[{{k}_{1}},{{k}_{2}}\] are positive integers and H.C.F of \[{{k}_{1}},{{k}_{2}}\] is 1.
Now it is given that, ab = 2028
So putting the values of a and b we get,\[(13{{k}_{1}})\cdot (13{{k}_{2}})=2028\]
\[\Rightarrow 169{{k}_{1}}{{k}_{2}}=2028\]
Alternatively, we can write \[{{k}_{1}}{{k}_{2}}=12\]
Now 12 can be written as a product of two positive integers where they are 12’s divisors.
We can write 12 as a product of 1 and 12 or 2 and 6 or 3 and 4 where order does not matter.
Now among the above three pairs, H.C.F of 2 and 6 is 2 which is not 1. Therefore, we can discard it.
The remaining two pairs are (1, 12) and (3, 4) respectively.
This can be our required values of \[{{k}_{1}},{{k}_{2}}\]
If we take \[{{k}_{1}}=1\] and \[{{k}_{2}}=12\] then we get \[a=13\] and \[b=156\]
Again if we take \[{{k}_{1}}=3\] and \[{{k}_{2}}=4\] then we get \[a=39\] and \[b=52\]
Hence, we get two pairs of numbers, which satisfies the given conditions. These are (13,156) and (39,52) respectively.
Hence, the correct option to this question is option (b) 2.

Note: As H.C.F of a and b is 13 and already taken out then the H.C.F of \[{{k}_{1}},{{k}_{2}}\] must be 1 in order to remain the H.C.F of a and b, 13. As the product is commutative so the pair (a,b) and (b, a) are equivalent and hence have not counted twice.