Question

**Question no 1**
The product of two number is 2028 and H.E.C. is 13. The number of such pairs is?
a, 3
b, 4
c, 2

**Question No 2**
The difference between a two-digits number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that
number?
a, 4
b, 3
c, 9​

Answer 1

Answer

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=13k1

and b=13k2

where k1,k2

are positive integers and H.C.F of k1,k2

is 1.

Now it is given that, ab = 2028

So putting the values of a and b we get,(13k1)⋅(13k2)=2028

⇒169k1k2=2028

Alternatively, we can write k1k2=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 k1,k2

If we take k1=1

and k2=12

then we get a=13

and b=156

Again if we take k1=3

and k2=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 k1,k2

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.

Answer 2

Answer:

The product of two numbers is 2028 and their HCF is 13. The number of such pairs is :

[A]1

[B]2

[C]3

[D]4

2

Here, HCF = 13

Let the numbers be 13x and 13y, where x and y are prime to each other.

Now, 13x \times 13y = 2028

=> xy = \frac{2028}{13\times 13} = 12

The opposite pairs are : (1, 12), (3, 4), (2, 6)

But the 2 and 6 are not co-prime.

So, required no. o