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The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is
- A4
- B3
- C2
- D1
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CCommunity
Given, HCF of two numbers is 15.
Let, the numbers are 15a and 15b (where a and b are coprime)
According to the question,
Their product = 6300
∴ 15a × 15b = 6300
⇒ ab = 28
∴ Possible values of (a, b) are (1, 28), (28, 1), (4, 7), (7, 4), (14, 2) and (2, 14)
The number of the pairs (14, 2) and (2, 14) is not co prime, so they will change the HCF.
∴ Possible pairs are (1, 28), (28, 1), (4, 7) and (7, 4).
Again, (1, 28) and (28, 1) give same pair.
And (4, 7) and (7, 4) give same pair.
∴ Number of Possible pairs is 2
Let, the numbers are 15a and 15b (where a and b are coprime)
According to the question,
Their product = 6300
∴ 15a × 15b = 6300
⇒ ab = 28
∴ Possible values of (a, b) are (1, 28), (28, 1), (4, 7), (7, 4), (14, 2) and (2, 14)
The number of the pairs (14, 2) and (2, 14) is not co prime, so they will change the HCF.
∴ Possible pairs are (1, 28), (28, 1), (4, 7) and (7, 4).
Again, (1, 28) and (28, 1) give same pair.
And (4, 7) and (7, 4) give same pair.
∴ Number of Possible pairs is 2