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Find the smallest positive integer which will leave a [#permalink]
24 Feb 2005, 17:31

Find the smallest positive integer which will leave a remainder of 1 when divided by2, a remainder of 2 when divided by 3, a remainder of 3 when divided by 4 and so on ...a remainder of 9 when divided by 10.

You are right.
Since it is a fundamental problem I didn't give any choices. Since it'd be too easy to back solve.
But if you solve it logically you end up learning a very basic concept in number theory
I'd explain after some more answers are posted

My pleasure.
In between absurd problems I would put in reasonable ETS like questions.

I somehow feel that even if a question seems absurd and not possible be asked in GMAT but it can open some door in ur mind or clear some funda that might help you solving difficult ETS like questions.

Actually, that's not right.
As we want the smallest number. We can see that we have 2, 3, 4=2*2, 5, 6=2*3, 7, 8=2*2*2, 9=3*3, 10=2*5
We really only need 5*7*8*9=2520
Then n=2520-1.
Sorry about that.

Well there is no general formula for LCM. Just find the factors of all the consecutive numbers and multiply them. So LCM of 2, 4 and 8 will be 8 as 8 contains both the multiples 2 and 4.
LCM of 1,2,3,4,5,6,7,8,9,10 will be 9*8*7*5. (Since 1,2,3 hence also 6 and 10 are already contained in the product