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Manager
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Find the smallest positive integer which will leave a [#permalink]
 24 Feb 2005, 18:31
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.
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Manager
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Just check your answer. 59 divided by 9 leaves a remainder of 6 and NOT 8.
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Manager
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Sorry 59 /9 leaves a remainder of 5 (not 6 or 8)
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anirban16 wrote: Just check your answer. 59 divided by 9 leaves a remainder of 6 and NOT 8.
Altho 59 divided by 9 leaves a remainder of 5, but u right my ans is wrong....I wud back solve it if there were ans choices. 
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Manager
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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
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Manager
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Yes 2519 it is.
The LCM of 1thru 10 is 2520 so 2520 will be divisible by by all the 10 numbers so 1 less than that will be the required number.
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Manager
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It shud be some multiple of 59. 59 satisfies the requirement till 6.. but 7 onwards it doesn't.. I think it shud be some multiple of 59.
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Re: Yes 2519 it is [#permalink]
25 Feb 2005, 07:23
anirban16 wrote: Yes 2519 it is.
The LCM of 1thru 10 is 2520 so 2520 will be divisible by by all the 10 numbers so 1 less than that will be the required number.
Will remember this, good one.
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Manager
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thats a very good example. thanks for posting
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Manager
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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.
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n=2k+1=3m+2=4s+3=...
n+1=2(k+1)=3(m+1)=4(s+1)=...
You can see n+1 is a product of 2, 3, 4, 5, 6 ...10
In other words, n+1=10!
n=10!-1
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HongHu wrote: n=2k+1=3m+2=4s+3=...
n+1=2(k+1)=3(m+1)=4(s+1)=...
You can see n+1 is a product of 2, 3, 4, 5, 6 ...10
In other words, n+1=10!
n=10!-1
Awesome ! u r the best. I always like algebraic approach.
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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. 
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Manager
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Although these problems are generally backsolve-able,a DS can trick you or waste lot of your time. So it is better to know how to solve it.
Thanks for the discussion.
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Manager
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Yup that's same as finding the LCM of first 10 natural numbers.
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what is the LCM rule ?
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what is the formular for finding the LCM of consecutive numbers?
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Manager
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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
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close
