Question: The product of two numbers is 2028 and their HCF is 13. What are the number of such pairs?
- 4
- 3
- 2
- 1
- 5
“The product of two numbers is 2028 and their HCF is 13.”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation:
Approach Solution 1:
Let us consider that the two numbers are x and y respectively. In the problem, it is stated that the product of these two numbers is 2028,
So, xy=2028
It is given that the HCF of two numbers is 13, which makes the numbers divisible by 13. So, let x=13a and y=13b, then the equation becomes
13a×13b=2028
Implies that 169ab=2028
Implies that ab=2028/169
Implies that ab=12
Hence, the required possible pair of values of x and y which are prime to each other is (1,12) and (3,4).
Therefore, the required numbers are (12,156) and (39,52).
Thus, the number of possible pairs is 2.
Correct Answer: C
Approach Solution 2:
Let us assume the required numbers 13a and 13b.
Then, 13a x 13b = 2028; which implies ab = 12
Now, the co-primes with product 12 are (1, 12) and (3, 4)
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4)
Clearly, there are 2 such pairs.
Note: Two integers a and b are said to be co-prime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1
Correct Answer: C
Approach Solution 3:
Let the two numbers be x and y respectively.
It is given that the product of the two numbers is 2028, therefore,
xy=2028
Also 13 is their HCF, thus both numbers must be divisible by 13.
So, let x=13a and y=13b, then
13a×13b=2028
⇒169ab=2028
⇒ab=2028/69
⇒ab=12
Therefore, required possible pair of values of x and y which are prime to each other are (1,12) and (3,4).
Thus, the required numbers are (12,156) and (39,52).
Hence, the number of possible pairs is 2.
Correct Answer: C
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