Step-by-step explanation: Let the numbers be 12x and 12y, where x and y are co-primes. Product of the numbers = 144xy 144xy=6336 xy=44 44 can be written as the product of two factors in three ways, i.e., 1×44,2×22,4×11 As x and y are relatively prime, (x,y) can be (1,44) or (4,11) but not (2,22). Hence two possible pairs exist. Reply
[tex] \: \huge\colorbox{pink}{Question}[/tex] The H.C.F. Of two numbers is 12 and their product is 2160. Find their L.C.M [tex] \: \huge\colorbox{pink}{Recommended Answer}[/tex] Product of two numbers = H.C.F × L.C.M [tex] \: ⇒ \: \: \: \: \: \: 2160 = 12 \times L.C.M[/tex] [tex] \: ⇒ \: \: \: \: \: L.C.M \: = \frac{2160}{12} = 180[/tex] Thus, the L.C.M of Two Numbers is 180 Reply
Step-by-step explanation:
Let the numbers be 12x and 12y, where x and y are co-primes.
Product of the numbers = 144xy
144xy=6336
xy=44
44 can be written as the product of two factors in three ways, i.e., 1×44,2×22,4×11
As x and y are relatively prime, (x,y) can be (1,44) or (4,11) but not (2,22).
Hence two possible pairs exist.
[tex] \: \huge\colorbox{pink}{Question}[/tex]
The H.C.F. Of two numbers is 12 and their product is 2160. Find their L.C.M
[tex] \: \huge\colorbox{pink}{Recommended Answer}[/tex]
Product of two numbers = H.C.F × L.C.M
[tex] \: ⇒ \: \: \: \: \: \: 2160 = 12 \times L.C.M[/tex]
[tex] \: ⇒ \: \: \: \: \: L.C.M \: = \frac{2160}{12} = 180[/tex]
Thus, the L.C.M of Two Numbers is 180