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:
95 comments Page 2 of 10.
Chakri said:
2 years 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)
Abdul Raziq P said:
4 months ago
(12, 1) and (1, 12) is the same as coprimilarity doesn't affect when the order is changed, so it is (1, 12) and (4, 3).
(2)
Shivang Dwivedi said:
3 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)
SUDIP CHOUDHURY said:
9 years ago
LCM=2028/13=156 which means 13x12 or 12*13 so there are 2 ways.
(1)
Xing said:
1 decade 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)
Julie said:
7 years ago
Thank you for giving the clear answer @Snehesh.
(1)
Randy Ranjith said:
7 years ago
Then why we take (4, 3) 4 is not co-prime?
(1)
Chaitanya said:
6 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:
6 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:
5 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)
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