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 5 of 10.
Subbu said:
10 years ago
I am not getting the concept of ab=12.
Abhishek s g said:
7 years ago
We should not take 2 and 6 because if we take 2 and 6, the product will be 2028 but HCF is 26 not 13.
Zoha said:
7 years ago
The product of two numbers is 2028 and their HCF is 13 find their LCM.
Ans: so, there LCM is 2 is that correct.
Ans: so, there LCM is 2 is that correct.
Umesh said:
5 years ago
The answer is incorrect. The answer could also be 2 and 6.
Subham said:
5 years ago
The factors that is said to be co prime only when their GSF OR HCF is 1.
Gayathri said:
3 months ago
Why can't we take pairs as four because (12, 1) and (4, 3) count as well right?
Anyone please explain.
Anyone please explain.
Amit Patel said:
3 months ago
Given:
Product of two numbers = 2028
H.C.F. (Highest Common Factor) = 13
Let the two numbers be:
a= 13x, b = 13y.
a= 13x, b = 13y.
Since their HCF is 13, the remaining parts xx and yy must be co-prime (i.e., gcd(x,y)=1gcd(x,y)=1).
Now,
ab = 13x⋅13y = 169xy = 2028⇒xy = 2028169.
ab = 13x⋅13y = 169xy = 2028⇒xy = 1692028.
Calculate:
2028169 = 12⇒xy = 12,
1692028 = 12⇒xy = 12,
So, we now need to find the number of co-prime pairs (x,y)(x,y) such that xy=12xy=12 and, gcd(x,y)=1gcd(x,y)=1.
Step: List all co-prime pairs with product 12
First, find all positive integer pairs (x,y)(x,y) such that xy=12xy = 12:
(1,12)(1,12)
(12,1)(12,1)
(2,6)(2,6)
(6,2)(6,2)
(3,4)(3,4)
(4,3)(4,3)
Now eliminate pairs where gcd(x,y)≠1gcd(x,y)=1:
(2,6)(2,6) → gcd=2gcd=2 ,wrong.
(6,2)(6,2) → gcd=2gcd=2, wrong.
(1,12)(1,12) → gcd=1gcd=1, correct.
(12,1)(12,1) → gcd=1gcd=1, correct.
(3,4)(3,4) → gcd=1gcd=1, correct.
(4,3)(4,3) → gcd=1gcd=1, correct.
So, valid co-prime pairs:
(1,12),(12,1),(3,4),(4,3)(1,12),(12,1),(3,4),(4,3).
Final Step: Multiply each x and y by 13 to get the original numbers.
So, there are 4 such pairs.
Product of two numbers = 2028
H.C.F. (Highest Common Factor) = 13
Let the two numbers be:
a= 13x, b = 13y.
a= 13x, b = 13y.
Since their HCF is 13, the remaining parts xx and yy must be co-prime (i.e., gcd(x,y)=1gcd(x,y)=1).
Now,
ab = 13x⋅13y = 169xy = 2028⇒xy = 2028169.
ab = 13x⋅13y = 169xy = 2028⇒xy = 1692028.
Calculate:
2028169 = 12⇒xy = 12,
1692028 = 12⇒xy = 12,
So, we now need to find the number of co-prime pairs (x,y)(x,y) such that xy=12xy=12 and, gcd(x,y)=1gcd(x,y)=1.
Step: List all co-prime pairs with product 12
First, find all positive integer pairs (x,y)(x,y) such that xy=12xy = 12:
(1,12)(1,12)
(12,1)(12,1)
(2,6)(2,6)
(6,2)(6,2)
(3,4)(3,4)
(4,3)(4,3)
Now eliminate pairs where gcd(x,y)≠1gcd(x,y)=1:
(2,6)(2,6) → gcd=2gcd=2 ,wrong.
(6,2)(6,2) → gcd=2gcd=2, wrong.
(1,12)(1,12) → gcd=1gcd=1, correct.
(12,1)(12,1) → gcd=1gcd=1, correct.
(3,4)(3,4) → gcd=1gcd=1, correct.
(4,3)(4,3) → gcd=1gcd=1, correct.
So, valid co-prime pairs:
(1,12),(12,1),(3,4),(4,3)(1,12),(12,1),(3,4),(4,3).
Final Step: Multiply each x and y by 13 to get the original numbers.
So, there are 4 such pairs.
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....
Swati said:
1 decade ago
Hey 13a*13b=2028 then a*b=156 if we take 13 common and divide it by 2028.
Then how it is written that a*b=12?
Then how it is written that a*b=12?
Priya said:
1 decade ago
Factors of:
6 = 1, 2, 3.
2 = 1, 2.
They have highest common factor 2. coprime are those pair that have 1 as hcf.
So only (1,12) and (4,3) are such pairs.
6 = 1, 2, 3.
2 = 1, 2.
They have highest common factor 2. coprime are those pair that have 1 as hcf.
So only (1,12) and (4,3) are such pairs.
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