FINDING THE LEAST COMMON MULTIPLE OF MONOMIALS

Find LCM and GCF :

Example 1  :

12x, 36y

Solution  :

Finding the LCM :

12x  =  2 × 2 × 3 × x

12x  =  2² × 3¹ × x

36y  =  2 × 2 × 3 × 3 × y

36y  =  2² × 3² × y

By taking the highest power of each common factor, we get

=  2² × 3² × x × y

=  36xy

So, the L.C.M of 12x, 36y is 36xy

Finding the GCF :

12x  =  2 × 2 × 3 × x

36y  =  2 × 2 × 3 × 3 × y

=  2 × 2 × 3

=  12

So, the G.C.F of 12x, 36y is 12

Example 2  :

24y³x, 40x³y

Solution  :

Finding the LCM :

24y³x  =  2 × 2 × 2 × 3 × y³ × x

24y³x  =  2³ × 3¹ × y³ × x

40x³y  =  2 × 2 × 2 × 5 × x³ × y

40x³y  =  2³ × 5¹ × x³ × y

By taking the highest power of each common factor, we get

=  2³ × 3 × 5 × y³ × x³

=  120y³x³

So, the L.C.M of 24y³x, 40x³y is 120y³x³

Finding the GCF :

24y³x  =  2 × 2 × 2 × 3 × y × y × y × x

40x³y  =  2 × 2 × 2 × 5 × x × x × x × y

=  2 × 2 × 2 × x × y

=  8xy

So, the G.C.F of 24y³x, 40x³y is 8xy

Example 3  :

16y², 24y²

Solution  :

Finding the LCM :

16y²  =  2 × 2 × 2 × 2 × y²

16y²  =  2⁴ × y²

24y²  =  2 × 2 × 2 × 3 × y²

24y²  =  2³ × 3¹ × y²

By taking the highest power of each common factor, we get

=  2⁴ × 3 × y²

=  48y²

So, the L.C.M of 16y², 24y² is 48y²

Finding the GCF :

16y²  =  2 × 2 × 2 × 2 × y × y

24y²  =  2 × 2 × 2 × 3 × y × y

=  2 × 2 × 2 × y × y

=  8y²

So, the G.C.F of 16y², 24y² is 8y²

Example 4  :

30a³b, 40a³

Solution  :

Finding the LCM :

30a³b  =  2 × 3 × 5 × a³ × b

30a³b  =  2¹ × 3¹ × 5¹ × a³ × b¹

40a³  =  2 × 2 × 2 × 5 × a³

40a³  =  2³ × 5¹ × a³

By taking the highest power of each common factor, we get

=  2³ × 3¹ × 5¹ × a³ × b¹

=  120a³b

So, the L.C.M of 30a³b, 40a³ is 120a³b

Finding the GCF :

30a³b  =  2 × 3 × 5 × a × a × a × b

40a³  =  2 × 2 × 2 × 5 × a × a × a

=  2 × 5 × a × a × a

=  10a³

So, the G.C.F of 30a³b, 40a³ is 10a³

Example 5  :

10b²a², 14a²

Solution  :

Finding the LCM :

10b²a²  =  2 × 5 × b² × a²

10b²a²  =  2¹ × 5¹ × b² × a²

14a²  =  2 × 7 × a²

14a²  =  2¹ × 7¹ × a²

By taking the highest power of each common factor, we get

=  2 × 5 × 7 × a² × b²

=  70a²b²

So, the L.C.M of 10b²a², 14a² is 70a²b²

Finding the GCF :

10b²a²  =  2 × 5 × b × b × a × a

14a²  =  2 × 7 × a × a

=  2 × a × a

=  2a²

So, the G.C.F of 10b²a², 14a² is 2a²

Example 6  :

27ab², 18b

Solution  :

Finding the LCM :

27ab²  =  3 × 3 × 3 × a × b²

27ab²  =  3³ × a¹ × b²

18b  =  3 × 3 × 2 × b

18b  =  3² × 2¹ × b¹

By taking the highest power of each common factor, we get

=  3³ × 2 × a × b²

=  54ab²

So, the L.C.M of 27ab², 18b is 54ab²

Finding the GCF :

27ab²  =  3 × 3 × 3 × a × b × b

18b  =  3 × 3 × 2 × b

=  3 × 3 × b

=  9b

So, the G.C.F of 27ab², 18b is 9b

Example 7  :

18y³, 32y

Solution  :

Finding the LCM :

18y³  =  3 × 3 × 2 × y³

18y³  =  3² × 2¹ × y³

32y  =  2 × 2 × 2 × 2 × 2 × y

32y  =  2⁵ × y

=  2⁵ × 3² × y³

=  288y³

So, the L.C.M of 18y³, 32y is 288y³

Finding the GCF :

18y³  =  3 × 3 × 2 × y × y × y

32y  =  2 × 2 × 2 × 2 × 2 × y

=  2 × y

=  2y

So, the G.C.F of 18y³, 32y is 2y

Example 8  :

20n², 30m²n²

Solution  :

Finding the LCM :

20n² =  2 × 2 × 5 × n²

20n² =  2² × 5¹ × n²

30m²n²  =  2 × 3 × 5 × m² × n²

30m²n²  =  2¹ × 3¹ × 5¹ × m² × n²

=  2² × 3 × 5 × m² × n²

=  60m²n²

So, the L.C.M of 20n², 30m²n² is 60m²n²

Finding the GCF :

20n² =  2 × 2 × 5 × n × n

30m²n²  =  2 × 3 × 5 × m × m × n × n

=  2 × 5 × n × n

=  10n²

So, the G.C.F of 20n², 30m²n² is 10n²

Example 9  :

24b³, 12ab²

Solution  :

Finding the LCM :

24b³  =  2 × 2 × 2 × 3 × b³

24b³  =  2³ × 3¹ × b³

12ab²  =  2 × 2 × 3 × a × b²

12ab²  =  2² × 3¹ × a¹ × b²

=  2³ × 3¹ × a¹ × b³

=  24ab³

So, the L.C.M of 24b³, 12ab² is 24ab³

Finding the GCF :

24b³  =  2 × 2 × 2 × 3 × b × b × b

12ab²  =  2 × 2 × 3 × a × b × b

=  2 × 2 × 3 × b × b

=  12b²

So, the G.C.F of 24b³, 12ab² is 12b²

Example 10  :

40xy, 32x²y

Solution  :

Finding the LCM :

40xy  =  2 × 2 × 2 × 5 × x × y

40xy  =  2³ × 5¹ × x¹ × y¹

32x²y  =  2 × 2 × 2 × 2 × 2 × x² × y

=  2⁵ × x² × y¹

=  2⁵ × 5 × x² × y¹

=  160x²y

So, the L.C.M of 40xy, 32x²y is 160x²y

Finding the GCF :

40xy  =  2 × 2 × 2 × 5 × x × y

32x²y  =  2 × 2 × 2 × 2 × 2 × x × x × y

=  2 × 2 × 2 × x × y

=  8xy

So, the G.C.F of 40xy, 32x²y is 8xy

Example 11  :

14xy, 21y², 28y

Solution  :

Finding the LCM :

14xy  =  2 × 7 × x × y

14xy  =  2¹ × 7¹ × x¹ × y¹

21y²  =  3 × 7 × y²

21y²  =  3¹ × 7¹ × y²

28y  =  2 × 2 × 7 × y

28y  =  2² × 7¹ × y¹

=  2² × 3¹ × 7¹ × x × y²

=  84xy²

So, the L.C.M of 14xy, 21y², 28y is 84xy²

Finding the GCF :

14xy  =  2 × 7 × x × y

21y²  =  3 × 7 × y × y

28y  =  2 × 2 × 7 × y

=  7y

So, the G.C.F of 14xy, 21y², 28y is 7y

Example 12  :

28a², 40a, 24a

Solution  :

Finding the LCM :

28a²  =  2 × 2 × 7 × a²

28a²  =  2² × 7¹ × a²

40a  =  2 × 2 × 2 × 5 × a

40a  =  2³ × 5¹ × a¹

24a  =  2 × 2 × 2 × 3 × a

24a  =  2³ × 3¹ × a¹

=  2³ × 3¹ × 5¹ × 7¹ × a²

=  840a²

So, the L.C.M of 28a², 40a, 24a is 840a²

Finding the GCF :

28a²  =  2 × 2 × 7 × a × a

40a  =  2 × 2 × 2 × 5 × a

24a  =  2 × 2 × 2 × 3 × a

=  2 × 2 × a

=  4a

So, the G.C.F of 28a², 40a, 24a is 4a

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