Question 810357: Jackie had many stones in a bag. When she counted them by twos threes fours fives sixes sevens eights nines and tens there was always one left over, What was the smallest number of stones Jackie could have had in her bag?
Answer by JBarnum(2146)   (Show Source):
You can put this solution on YOUR website! 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  , writing them out as just prime numbers so u can see why 6 is removed.  with using the multiplication property it can be shown like this:  which also looks like this  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:  ]]
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so now all we have left is 
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 
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.
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