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stolyar
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evensflow
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satgates
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stolyar
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satgates
Please submit the solution too (Explaination) which book do I get such problems.

:shock: :x :cry:


The question is mine; probably you will not find it in books.
Meanwhile, the solution is not that difficult.

Z=3n+1
Z=5k+3
Z=7p+5 the trick is to see that each time a quotinent is 2 more than a remainder. Therefore, add 2 to both parts and get what you need.

Z+2=3n+1+2=3n+3, divisible by 3
Z+2=5k+3+2=5k+5, divisible by 5
Z+2=7p+5+2=7k+7, divisible by 7

so, Z+2 is divisible by 3, 5, and 7.

To find the smallest positive Z, we have to subtract 2 from the LCM of 3, 5, 7.

LCM = 3*5*7, since they all are primes = 105
105-2=103 is the answer.

To find the second smallest number: 2LCM-2=210-2=208 and so on.

Do you like the question? It is one of my best.
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evensflow
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Excellent!!! Can i say anything better than that... Well i logged to post my second answer but saw your post and was pleased to see it. The idea behind the question is simply brilliant. :cool
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anandnk
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This is one of the best solutions I have ever seen.
For these kinds of problems I simply used to mutiply the divisors and get a product. The desired number should lie in the visinity of this product. I used to search around this number using trial and error.
stolyar's solution has made it eaven easier.
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mdfrahim
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What a great logic stolyar.........i use to beat my head with such questions,but looks to me that it will not happen any more.........Kudos man for such a nice solution.



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