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The product of two numbers is 2008 and their HCF is 13. The number of such pairs is = ?
Answer & Solution
Correct Answer:
Option
B
Let the numbers be 13a and 13b
Then, 13a × 13b = 2028
ab = 12
Now, co - primes with product 12 are (1, 12) and (3, 4)
So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
Clearly, there are 2 such pairs
Then, 13a × 13b = 2028
ab = 12
Now, co - primes with product 12 are (1, 12) and (3, 4)
So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
Clearly, there are 2 such pairs
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