Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by BdMO » Fri Jan 28, 2011 9:44 pm

In a box there are $50$ gold rings of $10$ different sizes and $75$ silver rings of $12$ different sizes. The size of a gold ring might be the same as that of a silver ring. What is the minimum number of rings one need to pick up to be sure of having at least two rings different both in size and material?

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Tahmid Hasan
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by Tahmid Hasan » Sat Jan 29, 2011 7:32 pm

soln for num 1:
i've done it with worst luck method. there are $10$ different size of gold coins, at most $41$ coins can be of same size. similarly at most $64$ silver coins can be of same size. If we take $64+41+1$ coins, we will be sure that either we have picked at least two gold coins of different size or at least two silver coins of different size. In both cases, we are getting two coins which are different both in size and colour. so the ans is $106$
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by *Mahi* » Sat Jan 29, 2011 10:32 pm

The answer isn't the one Tahmid posted,it is as follows-
The answer is 76.
Here we must know that in the silver ring set,there are 12 kinds,and in the gold ring set,there are 10 kinds.So at least 2 kinds of silver ring is different from the gold kinds.Now if we pick up 75 rings , all of them can be of silver(in the worst case scenario),and then picking up one ring ensures us of the two rings we want.
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by Zzzz » Sun Jan 30, 2011 7:28 am

*Mahi* wrote:The answer isn't the one Tahmid posted,it is as follows-
The answer is 76.
Here we must know that in the silver ring set,there are 12 kinds,and in the gold ring set,there are 10 kinds.So at least 2 kinds of silver ring is different from the gold kinds.Now if we pick up 75 rings , all of them can be of silver(in the worst case scenario),and then picking up one ring ensures us of the two rings we want.
Worst scenario of picking 76 rings will be mmm... may be 50 silver rings and 26 gold rings which are of same size.
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by *Mahi* » Sun Jan 30, 2011 9:41 am

Zzzz wrote:
*Mahi* wrote:The answer isn't the one Tahmid posted,it is as follows-
The answer is 76.
Here we must know that in the silver ring set,there are 12 kinds,and in the gold ring set,there are 10 kinds.So at least 2 kinds of silver ring is different from the gold kinds.Now if we pick up 75 rings , all of them can be of silver(in the worst case scenario),and then picking up one ring ensures us of the two rings we want.
Worst scenario of picking 76 rings will be mmm... may be 50 silver rings and 26 gold rings which are of same size.
It was the official solution as far as I know...........but I did exactly what tahmid did :P
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Tahmid Hasan
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by Tahmid Hasan » Sun Jan 30, 2011 10:12 pm

if anyone shows the answer is less than 106 then i'll put them in the set of 41 g coins and 64 s coins of the same size.so it's not possible.if anyone tries to disprove my ans then plz try to make the ans greater than
106
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by Hasib » Sun Jan 30, 2011 10:31 pm

i think the ans is 106
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by Hasib » Sun Jan 30, 2011 10:36 pm

if 76 coins picked, then the following things may b happened!

Let, pick all 75 silver ring, and pick 1 gold ring! :O may unfortunately 1 gold will be the same size with the silver ring! Slap me!!!!
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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by leonardo shawon » Sun Jan 30, 2011 11:39 pm

i also agreed with Tahmid and Hasib. Size of gold rings might be the same size of silver rings.
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long way to go .....

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Re: Dhaka Secondary 2011/1 (Higher Secondary 2011/1)

Unread post by sakib.creza » Fri Dec 28, 2012 6:58 am

What if among the 12 sizes of the silver ring 10 of them are same as the 10 sizes of the gold ring.so shudnt the size of the gold rings be a subset of the sizes of the silver rings. so with worst case scenario I think the answer shud be (50+73+1)=124. Please let me know if there is any bug in my answer.

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