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 2 of 10.
Chakri said:
1 year ago
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.
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.
(4)
Shivang Dwivedi said:
2 years ago
@Veerbhadra.
Because (12*13, 13) and (13, 12*13) are considered as one unique pair, irrespective of the order of writing.
Because (12*13, 13) and (13, 12*13) are considered as one unique pair, irrespective of the order of writing.
(2)
Xing said:
10 years ago
Longer method, divide both side by 13.
=> 13a * 13b/13 = 2028/13.
=> 13a * b = 156.
=> ab = 156/13.
=> ab = 12.
=> 13a * 13b/13 = 2028/13.
=> 13a * b = 156.
=> ab = 156/13.
=> ab = 12.
(1)
SUDIP CHOUDHURY said:
8 years ago
LCM=2028/13=156 which means 13x12 or 12*13 so there are 2 ways.
(1)
Julie said:
6 years ago
Thank you for giving the clear answer @Snehesh.
(1)
Randy Ranjith said:
6 years ago
Then why we take (4, 3) 4 is not co-prime?
(1)
Chaitanya said:
5 years ago
Take 13x*13y( because HCF is 13)
Let 13x*13y=2028
xy=2028/13*13
xy becomes 12.
Next find out the coprimes which should be the form of x,y hence,
coprimes are (1,12) and (3,4)or (4,3),
so the real pair will be;
(13*1,13*12)and (13*3,13*4),
Hence the answer is 2.
Let 13x*13y=2028
xy=2028/13*13
xy becomes 12.
Next find out the coprimes which should be the form of x,y hence,
coprimes are (1,12) and (3,4)or (4,3),
so the real pair will be;
(13*1,13*12)and (13*3,13*4),
Hence the answer is 2.
(1)
Veerabhadra said:
5 years ago
Why not answer is 4?
(1*13, 12*13) (3*13, 4*13) (12*13, 1*13) (4*13, 3*13) these are four possible pairs.
(1*13, 12*13) (3*13, 4*13) (12*13, 1*13) (4*13, 3*13) these are four possible pairs.
(1)
Madhu said:
4 years ago
(2, 6) are not co prime numbers.
As.
2=1*2.
6=1*6, 2*3;.
So for 6. (2) is again a factor so they can't be.
As.
2=1*2.
6=1*6, 2*3;.
So for 6. (2) is again a factor so they can't be.
(1)
Pradeep said:
9 years ago
Why they take the two pair like (1, 12), (3, 4)?
Why not (12, 1), (4, 3)?
Why not (12, 1), (4, 3)?
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