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The product of two numbers is 2028 and their HCF is 13. The number of such pairs is
- a)2
- b)1
- c)4
- d)3
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The product of two numbers is 2028 and their HCF is 13. The number of ...
Here, HCF = 13
Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028
The possible pairs are : (1, 12), (3, 4), (2, 6)
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2
Let the numbers be 13x and 13y where x and y are Prime to each other.
Now, 13x × 13y = 2028
The possible pairs are : (1, 12), (3, 4), (2, 6)
But the 2 and 6 are not co-prime.
∴ The required no. of pairs = 2
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Community Answer
The product of two numbers is 2028 and their HCF is 13. The number of ...
Given:
The product of two numbers is 2028 and their HCF is 13.
To find:
The number of such pairs
Solution:
Prime Factorization:
Let the two numbers be a and b. Since the HCF of a and b is 13, we can express a and b as follows:
a = 13 * x
b = 13 * y
where x and y are co-prime numbers.
Product of the numbers:
Given that the product of a and b is 2028, we have:
13 * x * 13 * y = 2028
169xy = 2028
xy = 12
Number of pairs:
The possible pairs of x and y that satisfy the above equation are:
(1, 12) and (12, 1)
Therefore, there are 2 such pairs of numbers that have a product of 2028 and an HCF of 13.
Conclusion:
The number of such pairs is 2.
The product of two numbers is 2028 and their HCF is 13.
To find:
The number of such pairs
Solution:
Prime Factorization:
Let the two numbers be a and b. Since the HCF of a and b is 13, we can express a and b as follows:
a = 13 * x
b = 13 * y
where x and y are co-prime numbers.
Product of the numbers:
Given that the product of a and b is 2028, we have:
13 * x * 13 * y = 2028
169xy = 2028
xy = 12
Number of pairs:
The possible pairs of x and y that satisfy the above equation are:
(1, 12) and (12, 1)
Therefore, there are 2 such pairs of numbers that have a product of 2028 and an HCF of 13.
Conclusion:
The number of such pairs is 2.
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The product of two numbers is 2028 and their HCF is 13. The number of such pairs isa)2b)1c)4d)3Correct answer is option 'A'. Can you explain this answer?
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