Question

How many pairs of numbers are possible whose LCM is 280 and HCF is 7?

Answer 1

Dear student
$$\begin{gathered} Letthenumberbe7xand7y \\ wherexandyareco-prime...\left(1\right) \\ Productoftwonumbers=productoftheirHCFandJCM \\ \Rightarrow 7x\times 7y=7\times 280 \\ \Rightarrow xy=40...\left(2\right) \\ 40=2\times 2\times 2\times 5\times 1 \\ Therefore,valuesofxandywhichsatifies\left(1\right)and\left(2\right)are \\ 8,5 \\ 1,40 \\ Therfore,thenumbers(7x,7y)are \\ 56,35 \\ 7,280 \\ Thus,twopairofnumbersarepossible \end{gathered}$$
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