IMO 1970

Discussion on International Mathematical Olympiad (IMO)
tanmoy
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IMO 1970

Unread post by tanmoy » Mon Sep 14, 2015 9:27 pm

Find all positive integers $n$ such that the set
${n,n+1,n+2,n+3,n+4,n+5}$
can be split into to disjoint subsets such that the products of elements in these subsets are the same.
[This probolem can be solved without using any theorem! ;) ]
"Questions we can't answer are far better than answers we can't question"

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Kazi_Zareer
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Location: Malibagh,Dhaka-1217

Re: IMO 1970

Unread post by Kazi_Zareer » Thu May 19, 2016 12:22 am

Try yourself ;)
The primes which can divide this numbers will be 2,3 or 5 because if the prime factor is larger than this three then it can divide only one number.
Three number will be odd. At least one number will be the multiple of 3 and also at least one number will be the multiple of 5.Other odd number's can't have any prime factor.
So, there will not exist such set. :?
We cannot solve our problems with the same thinking we used when we create them.

Golam Musabbir Joy
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Location: Barisal, Bangladesh

Re: IMO 1970

Unread post by Golam Musabbir Joy » Tue May 24, 2016 7:05 am

By contradiction we assume that there exist a n such that that satisfy this proposition. there is a number in those there must be a number which is divisible by 5. so that , we have another number which is divisible by 5. suppose those number are n , n+5 . now n+1, n+2, n+3, n+4 none of them will not be divisible by any p>5 because we can't have two such prime divisor into another's. two of them must be even. so that they must be power of 3 . two consecutive even number can't be power of 3 unless they are 1,3 . so we have such a answer n=0. but this is not positive integer. so that there does not exist such an integer n that satisfy the proposition

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