Aptitude - Problems on H.C.F and L.C.M - Discussion

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 9)
9.
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
1
2
3
4
Answer: Option
Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Discussion:
94 comments Page 10 of 10.

Pragya said:   1 decade ago
Why are we considering coprimes?

Siva said:   1 decade ago
@Arvind, you must not consider (26, 78) because then the HCF will become 26; as given in the question HCF must be only 13.

Arvind said:   1 decade ago
There will be three pairs (26,78), (52,39), (156,13).

Chinnu said:   1 decade ago
(3, 4)4 = 1, 2, 4 why it is coprime number 2 number is also in the factor of 4.


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