The product of two numbers is 6760 and their | vedclass

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(A) Let the two numbers be $13x$ and $13y$,where $x$ and $y$ are co-prime numbers.Given that the product of the numbers is $6760$.$13x 	imes 13y = 67

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$2$

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(A) Let the two numbers be $13x$ and $13y$,where $x$ and $y$ are co-prime numbers.Given that the product of the numbers is $6760$.$13x 	imes 13y = 6760$$169xy = 6760$$xy = \frac{6760}{169} = 40$Now,we need to find pairs of co-prime numbers $(x, y)$ such that their product is $40$.The factors of $40$ are $(1, 40), (2, 20), (4, 10), (5, 8)$.For the numbers to have an $H.C.F.$ of $13$,$x$ and $y$ must be co-prime (i.e.,their $H.C.F.$ must be $1$).Checking the pairs:$(1, 40)$: $H.C.F.(1, 40) = 1$ (Valid)$(2, 20)$: $H.C.F.(2, 20) = 2$ (Invalid)$(4, 10)$: $H.C.F.(4, 10) = 2$ (Invalid)$(5, 8)$: $H.C.F.(5, 8) = 1$ (Valid)Thus,there are $2$ such pairs: $(1, 40)$ and $(5, 8)$.

The product of two numbers is $6760$ and their $H.C.F.$ is $13.$ How many such pairs can be formed?

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