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:
91 comments Page 2 of 10.
Vaani said:
1 decade ago
Why should we take co-primes?
Utsavchavda said:
1 decade ago
Meaning of co-prime ?
Veena said:
1 decade ago
@vaani
In the given problem the H.C.F is 13..so if take the pair other than co-prime(i.e., the pair 2,6)....the H.C.F will become 26...but when we take the pairs 1,12;3,4....there is no common factor other than 1.. so that we have to take the pair of co-primes....
In the given problem the H.C.F is 13..so if take the pair other than co-prime(i.e., the pair 2,6)....the H.C.F will become 26...but when we take the pairs 1,12;3,4....there is no common factor other than 1.. so that we have to take the pair of co-primes....
Aajush said:
1 decade ago
Its not compulsory to get from prime or co-prime.
The thing is to get it not exceeding 2 pair.
The thing is to get it not exceeding 2 pair.
Pallavi said:
1 decade ago
Why not (13*6, 13*2) be another pair?
Pallavi said:
1 decade ago
Got the point veena, sorry to post the earlier comment. Thank !
RAHUL KUMAR SARASWAT said:
1 decade ago
Sir we have a formula that the product of two numbers= LCM * HCM.
Why this formula is not used in this question ?
Why this formula is not used in this question ?
Samaria vincent said:
1 decade ago
First of all why not 5, 7 because even these numbers are co-prime numbers. Isn't it?
S ADITYA GAUTAM said:
1 decade ago
We take either (1,12) or (4,3) as coprimes and not (2,6) becouse of two reasons:
1)we want to get the product 12 as ab=12 only.
2)we choose coprimes (1,12) or (3,4) not non-coprimes(2,6) because if we choose the non-coprimes(2,6) then we get two nos as (26,78) whose H.C.F is 26 not 13 which against the question.
On the other hand if coprime numbers are taken like(1,12) then we get the two nos. as (13,156) and for coprime pairs(3,4) we get the numbers (39,52).In both of the cases u may observe that the H.C.F is 13 only.
1)we want to get the product 12 as ab=12 only.
2)we choose coprimes (1,12) or (3,4) not non-coprimes(2,6) because if we choose the non-coprimes(2,6) then we get two nos as (26,78) whose H.C.F is 26 not 13 which against the question.
On the other hand if coprime numbers are taken like(1,12) then we get the two nos. as (13,156) and for coprime pairs(3,4) we get the numbers (39,52).In both of the cases u may observe that the H.C.F is 13 only.
Anup said:
1 decade ago
Is there a simpler method to solve this?
This was my solution but I don't know why is it wrong.
13 is the H.C.F.
So the numbers should have 13 in them. Also as a*b = 12.
We can have these combinations.
(13,13*2*2*4) (13*2,13*2*3) (13*3,13*2*2)
Please correct me if my logic is not appropriate.
This was my solution but I don't know why is it wrong.
13 is the H.C.F.
So the numbers should have 13 in them. Also as a*b = 12.
We can have these combinations.
(13,13*2*2*4) (13*2,13*2*3) (13*3,13*2*2)
Please correct me if my logic is not appropriate.
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