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 10 of 10.
Parmanand Singh said:
1 year ago
It should be 3 pairs because the co-prime of product 12 is (1, 12) (2, 6) (3, 4).
So we get, 3 pairs.
So we get, 3 pairs.
(22)
Vina said:
8 months ago
It should be 3 pairs. So, the answer will be option C.
(5)
Gayathri said:
4 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:
4 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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers