Question

Q.17. Find the smallest number which when increased by 27 is exactly divisible by 660 and 594?

Answer 1

Dear student
$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{numbers}\mathrm{are}660\mathrm{and}594. \\ \mathrm{The}\mathrm{smallest}\mathrm{number}\mathrm{which}\mathrm{when}\mathrm{increased}\mathrm{by}27\mathrm{is}\mathrm{exactly}\mathrm{divisible}\mathrm{by}660 \\ \mathrm{and}594\mathrm{is}\mathrm{obtained}\mathrm{by}\mathrm{subtracting}27\mathrm{from}\mathrm{the}\mathrm{LCM}\mathrm{of}660\mathrm{and}594. \\ \mathrm{Prime}\mathrm{factorisation}\mathrm{of}660=2\times 2\times 3\times 5\times 11 \\ \mathrm{Prime}\mathrm{factorisation}\mathrm{of}594=2\times 3\times 3\times 3\times 11 \\ \mathrm{LCM}\mathrm{of}660\mathrm{and}594=2\times 2\times 3\times 3\times 3\times 5\times 11=5940 \\ \mathrm{Smallest}\mathrm{number}\mathrm{which}\mathrm{when}\mathrm{increased}\mathrm{by}27\mathrm{is}\mathrm{exactly}\mathrm{divisible}\mathrm{by}\mathrm{both}660\mathrm{and}594=5940-27=5913 \end{gathered}$$
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