I have a confusion in a problem of series .

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Raiyan Jamil
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I have a confusion in a problem of series .

Unread post by Raiyan Jamil » Mon Feb 03, 2014 10:32 pm

In this series , a group of (1,2,3,.............400) is made . In it , some numbers are to be removed such that , a new series is made and in that series , the sum of any two numbers won't be divisible by 7 . What is the maximum number of numbers in that series .

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asif e elahi
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Re: I have a confusion in a problem of series .

Unread post by asif e elahi » Fri Feb 07, 2014 9:59 pm

I think the answer is $217$.

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Fatin Farhan
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Re: I have a confusion in a problem of series .

Unread post by Fatin Farhan » Fri Feb 07, 2014 10:47 pm

I think the ans is $$170$$
Last edited by Fatin Farhan on Sat Feb 08, 2014 8:59 am, edited 2 times in total.
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Fatin Farhan
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Re: I have a confusion in a problem of series .

Unread post by Fatin Farhan » Fri Feb 07, 2014 10:51 pm

$$1+57+56+56= 170$$
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Labib
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Re: I have a confusion in a problem of series .

Unread post by Labib » Sat Feb 08, 2014 7:58 pm

Hi everyone. Please practise posting full solutions/hints. Otherwise it will be no help to the person posting.
Yes, the solution is $170$. Here's why-
We have to consider $7$ residue classes here, $0, \pm 1, \pm 2, \pm 3$.
We can only keep $1$ number from the $0$ residue class.
If we keep a number from any residue class $+a$ for $0<a\leq 3$, we cannot keep any number from $-a$ residue class and vice versa.
Since, $57$ elements of the set belong to $+1$ residue class, we'll keep it to maximize the number of elements. We can keep any two of the other four residue classes (who have $56$ elements each).
So the total number of elements $ = 1+57+56+56 = 170$
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