Multiples of #8# are #{8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144......}#

Multiples of #12# are #{12,24,36,48,60,72,84,96,108,120,132,144......}#

Multiples of #18# are #{18,36,54,72,90,108,126,144......}#

Hence common multiples are #{72,144,.......}#

and Least Common Multiple is #72#

Another shorter way is to write numbers as multiplication of its prime factors

#8=2xx2xx2# #-># #2# comes three times

#12=2xx2xx3# #-># #2# comes twice and #3# once.

You are watching: What is the lcm of 8 and 18

#18=2xx3xx3# #-># #2# comes once and #3# twice

So maximum times is three times for #2# and two times for #3#.

Hence, Least Common Multiple is #2xx2xx2xx3xx3=72#.


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Tony B
Apr 9, 2018

Use Shwetyank"s method in the class and an exam. The marking schema will be set up for that approach.

#color(blue)("My alternative approach - helps with understanding.")#


Explanation:

Target: LCM of 8,12 and 18.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(magenta)("Once you are used to this it should look like")##color(magenta)("what follows (for these numbers).")#

#12=color(white)("d")8+4 color(white)("d") ->2xx4=color(white)(..)8" so "2xx12=24##24=18+6 color(white)("d")->3xx6=color(white)(.)18" so "color(white)(.)3xx24=72#

#"LCM "=72#

3 lines inclusive of the answer for 3 relationships

///////////////////////////////////////////////////////////////////////////////////////////////

#color(blue)("Preamble about the method")#

There are many ways of writing any particular value so we can make it look however we wish as long as final outcome is the correct value. Example: 6 can and may be written as #5+1# if we so chose.

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We can use this to our advantage.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(blue)("Answering the question")#

#color(brown)("Dealing with the 8 and 12 part")#

Write 12 as #8+4#

Now we start counting the 12"s but make sure it encompasses the 8 + 4"s. When we have accumulated enough 4"s to make another eight we have found the least common factor of 8 and 12

#1: -> color(green)(12)=color(red)(8)+4##2:-> ul(color(green)(12))=color(red)(8)+ul(4 larr" Add the 12"s and the 4"s")##color(white)("ddddd")color(green)(24) color(white)("ddd")ul(color(white)("ddd")color(red)(8))##color(white)("dddddddddddd")color(red)(24) larr" Adding all the 8"s"#

#2" of "12=24##3" of "color(white)("d")8=24#

So LCF (for 8 and 12) is 24

So we have a fixed ratio of 8 and 12 at every sum of 24. Consequently any other product of factors will need to include some multiple of 24~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(brown)("Dealing with the 18 part")#

As the value of 24 locks together the counts of 8 and 12 we carry that forward.

Write the 24 as #18+6#

We are now counting the 24"s

#1: ->color(green)( 24)=color(red)(18)+6##2: ->color(green)(24)=color(red)(18)+6##3:->ul(color(green)(24))=color(red)(18)+ul(6larr" Add the 24"s and 6"s")##color(white)("ddddd") color(green)(72)color(white)("dddddd")ul(color(red)(18))larr" Add all the 18"s"##color(white)("ddddddddddddd")color(red)(72)#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(magenta)(" LCM "= 72)#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~