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:
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 9 of 10.
Kanika said:
1 decade ago
When we assume 13a AND 13b as numbers please explain.
Dinesh said:
1 decade ago
Why to take the pairs of 12?
Dhanendra thakre said:
1 decade ago
Because in this problem contain highest common factor 13 if you take 2 and 6 then problem condition is not satisfied. Because 2and 6 contain 2 is become highest factor.
13*6 13*2 have HCF 2 not 13.
And 13*1 13*12 have HCF 13.
13*6 13*2 have HCF 2 not 13.
And 13*1 13*12 have HCF 13.
Anotny mathew said:
1 decade ago
Why not 2, 6?
SRI said:
1 decade ago
@Veena.
I too had the doubt of 2,6.... Nice clarification given. Never though to check the HCF formed by the numbers (2, 6).
I too had the doubt of 2,6.... Nice clarification given. Never though to check the HCF formed by the numbers (2, 6).
Dhivya said:
1 decade ago
13a*13b = 2028.
ab = 2028/(13*13).
ab = 2028/169.
ab = 12.
ab = 2028/(13*13).
ab = 2028/169.
ab = 12.
Naveen said:
1 decade ago
Why ab = 12?
Hemanth said:
1 decade ago
Guys you don't need to know what a coprimes for this sum. Just understand that the HCF given is 13.
Now if you take (2, 6) as another pair then the two numbers become 13*2 and 13*6.
Now the above two numbers have an HCF of 26 (because in 2 and 6 the common factor is 2) which does not satisfy the condition of the two numbers having HCF of 13 given in the question.
Hope you understand. :).
Now if you take (2, 6) as another pair then the two numbers become 13*2 and 13*6.
Now the above two numbers have an HCF of 26 (because in 2 and 6 the common factor is 2) which does not satisfy the condition of the two numbers having HCF of 13 given in the question.
Hope you understand. :).
Santhosh said:
1 decade ago
1 is neither prime nor composite. In the solution it was given a co prime pair. So can it be co-prime pair ?
Sriram said:
1 decade ago
We have to consider the co primes with 12.
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