Bdmo National2014: junior 5

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atiab jobayer
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Bdmo National2014: junior 5

Unread post by atiab jobayer » Sun Feb 23, 2014 11:15 pm

A new is to be formed by removing some terms from the series 1,2,3,4..............,30 such that no term of the new series is obtained if any term of the new series is doubled. Maximum how many terms can there in new series?
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Labib
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Re: Bdmo National2014: junior 5

Unread post by Labib » Mon Feb 24, 2014 8:42 pm

First of all, there are a few minor typos.
atiab jobayer wrote:A new is to be formed by removing some terms from the series 1,2,3,4..............,30 such that no term of the new series is obtained if any term of the new series is doubled. Maximum how many terms can there in new series?
The statement, I guess, should be "A new series is to be formed by removing some terms from the series $1,2,3,4..............,30$ such that no term of the new series is obtained if any term of the new series is doubled. Maximum how many terms can there be in the new series?"
Here's a possible approach.
Since we have to maximize the number of terms in the series, we must take any term in the given series doubling which gives us a value greater than $30$.
So we take all $15$ numbers from $16$ onwards of the given series. Now, it's easy to notice that we can no longer take any number greater than $7$. So we take the $5$ numbers that we can get from the first $7$ (How is that $5$ numbers? ;) ). Now, can you prove that this is the optimal solution and you cannot take more than $15+5=20$ numbers in the new sequence?
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atiab jobayer
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Re: Bdmo National2014: junior 5

Unread post by atiab jobayer » Thu Dec 04, 2014 10:46 am

Yes. I've made a solution. I agree with Labib.In the new series, there must be contain 20 term. The missing numbers are : 2,6,8,10,14,18,22,24,26,30.
And the new wanted series is : 1,3,4,5,7,9,11,12,13,15,16,17,19,20,21,23,25,27,28,29.
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