Question 810357
1st due to the multiples of 2s its an even number
2nd due to the multiples of 3s its a number where the digits when added together are divisible by 3
3rd due to the multiples of 5s its a number that could end in a 5 or 0 but since it has to be even the last digit has to be 0.
4th due to the multiples of the prime number 7 and that it has to be a number that ends in 0 will significantly narrow your search: 7*2=14 no, *3=21 no, *4=28 no, (*5,*6,*7,*8,*9)= no so only multiples of 10 times 7 are possible.
5th due to multiples of 10 this confirms all the previous. 
6th the multiples of 8 will also include the multiples of 4 & 2 and the multiples of 9 will include the multiples of 3, and the multiples of 10 will include the multiples of 5. with this knowledge we can line up the numbers (2*3*4*5*6*7*8*9*10) we can eliminate 2, 3, 4, 5 from this group due to the above.

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so we are left with
(6*7*8*9*10)
now if u look at the prime numbers that make these 5 numbers up:
2*3=6 , 7 , 2*2*2=8 , 3*3=9, 2*5=10
from my experience in the past i know that since six is made up by a 2 and a 3 and both a 2 and a 3 are multiplied together due to the 8 and 9 [can be seen like this: (8*9)=72=(2*2*2*3*3)= (4*6*3)] the 6 can also be removed.
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[[update wanted to make this more clear with {{{(6*7*8*9*10)}}}, writing them out as just prime numbers so u can see why 6 is removed. {{{(2*3 * 7 * 2*2*2 * 3*3 * 2*5)}}} with using the multiplication property it can be shown like this: {{{(2 * 3 * 2*2 * 5 * 2*3 * 2*3 * 7 )}}} which also looks like this {{{(2*3*4*5*cross(6)*6*7)}}} now this shows you that there are two 6s in here when you only need one, that is why the 6 can be canceled out. also: {{{2*3*4*5*6*7=5040}}} ]]
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so now all we have left is {{{(7*8*9*10)=5040}}}
check to see if this number meets the above guidelines we know of from 1st-5th
1st its EVEN and divisible by 2
2nd add the digits of the number 5+0+4+0=9 9 is divisible by 3 so this works
3rd it ends in 0, so 5 is divisible
4th 7 was multiplied by 10 in the problem
5th you can now test the number with a calculator for the other numbers 4, and 6. but we already said 8 is a multiple of 4 so its guaranteed to work and 6 is the only other number that isn't obviously divisible. so 5040/6= 840...yep it works.
so the answer for lowest common multiple is {{{highlight_green(5040)}}}

update...believe it or not this looks like a lot of work but its all done in seconds in your head and on paper once you get the hang of how to do these its like 10-20secs and you get the answer.