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 7 of 10.
SUDIP CHOUDHURY said:
8 years ago
LCM=2028/13=156 which means 13x12 or 12*13 so there are 2 ways.
(1)
Vishal said:
8 years ago
4 is not a co-prime because it has a factor 1*4 and 2*2 also.
Shefali said:
1 decade ago
If we divide 156 by 13 then it is 12. So there are 2 pairs.
Grateful dead said:
1 decade ago
What is co-prime and why would we use it in this question?
Umesh said:
6 years ago
The answer is incorrect. The answer could also be 2 and 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.
Vina said:
10 months ago
It should be 3 pairs. So, the answer will be option C.
(5)
Arvind said:
1 decade ago
There will be three pairs (26,78), (52,39), (156,13).
Kanika said:
1 decade ago
When we assume 13a AND 13b as numbers please explain.
Yashraj said:
9 years ago
I can't understand this. Please help me to get it.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers