Multiples that #8# are #8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144......#
Multiples that #12# space #12,24,36,48,60,72,84,96,108,120,132,144......#
Multiples that #18# are #18,36,54,72,90,108,126,144......#
Hence typical multiples are #72,144,.......#
and Least typical Multiple is #72#
Another shorter way is to write numbers as multiplication that its prime factors
#8=2xx2xx2# #-># #2# comes 3 times
#12=2xx2xx3# #-># #2# come twice and #3# once.
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#18=2xx3xx3# #-># #2# come once and also #3# twice
So maximum time is 3 times because that #2# and also two times for #3#.
Hence, Least typical Multiple is #2xx2xx2xx3xx3=72#.
Answer attach

Tony B
Apr 9, 2018
Use Shwetyank"s method in the class and an exam. The noting schema will certainly be collection up for that approach.
#color(blue)("My different approach - helps v understanding.")#
Explanation:
Target: LCM that 8,12 and 18.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(magenta)("Once friend are supplied to this it need to look like")##color(magenta)("what adheres to (for these numbers).")#
#12=color(white)("d")8+4 color(white)("d") ->2xx4=color(white)(..)8" therefore "2xx12=24##24=18+6 color(white)("d")->3xx6=color(white)(.)18" so "color(white)(.)3xx24=72#
#"LCM "=72#
3 lines inclusive that the answer for 3 relationships
///////////////////////////////////////////////////////////////////////////////////////////////
#color(blue)("Preamble around the method")#
There are numerous ways of writing any details value for this reason we can make the look however we great as lengthy 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 have the right to use this come our advantage.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(blue)("Answering the question")#
#color(brown)("Dealing with the 8 and 12 part")#
Write 12 as #8+4#
Now we begin counting the 12"s yet make certain it includes the 8 + 4"s. Once we have built up enough 4"s to make an additional eight we have found the least usual factor of 8 and 12
#1: -> color(green)(12)=color(red)(8)+4##2:-> ul(color(green)(12))=color(red)(8)+ul(4 larr" include the 12"s and also 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" including 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 in ~ every sum of 24. Consequently any kind of other product of factors will need to encompass some lot of of 24~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#color(brown)("Dealing through the 18 part")#
As the worth of 24 locks together the counts that 8 and 12 we lug that forward.
Write the 24 together #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)#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~