10-03-2002, 09:04 AM
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#1 (permalink) |
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Within my grasp!
 Join Date: Sep 2002
Location: India
Posts: 164
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Hi All
Can you guys work on this.... There are 20 packs of cigarettes. Each pack has 20 cigarettes. Each pack has cigarettes weighing 2gms each except one pack. Cigarettes in this only pack weigh 1gm each. All 20 packs are randomly mixed. You have a balance and you CAN balance ONLY ONE TIME. How will you find which pack has 1gm cigarretes? DKP |
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10-04-2002, 06:04 AM
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#4 (permalink) | |
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Within my grasp!
Join Date: Sep 2002
Location: India
Posts: 164
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Quote:
DKP |
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10-04-2002, 05:00 PM
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#5 (permalink) | |
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TestMagic Guru-in-Training
Join Date: May 2002
Location: India
Posts: 902
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Quote:
THE PROBLEM: There are 20 packs of cigarettes. Each pack has 20 cigarettes. Each pack has cigarettes weighing 2gms each except one pack. Cigarettes in this only pack weigh 1gm each. All 20 packs are randomly mixed. You have a balance and you CAN balance ONLY ONE TIME. How will you find which pack has 1gm cigarretes? I am not sure what exactly can be balanced ... - whether we should only balance cigarettes/packs against each other - or we can weigh a set of cigarettes/packs with standard weights to determine their weight In the second case, i.e., if we can find out the weight of a set of cigarettes we choose, by balancing them against standard weights, here's my solution... THE SOLUTION: Let's call our cigarette packs : P1, P2, P3, ... P20. We take as a set... 1 cigarette from P1 + 2 cigarettes from P2 + 3 cigarettes from P3 + ... so on ... 20 cigarettes from P20. We then find out the weight (Say, W) of this set of cigarettes. If all cigarettes were 2gm, the weight of the above chosen set would be ... W = (1x2gm)+(2x2gm)+(3x2gm)+...+(20x2gm) = (1+2+3+....+20)x2gm = (20x21/2)x2gm = (210)x2gm = 420gm However, one of the 20 packs contains only 1gm cigarettes. If P1 contained 1gm cigarettes instead of 2gm cigarettes, we have W = (420-1)gm If it's P2, we have W = (420-2)gm If P3, W = (420-3)gm .. so on .. If P20, W = (420-20)gm Thus if we weigh our set and find that it weighs 419gm, we can say that P1 is the pack that contained 1gm cigarettes. If it's 418gm, P2 is the one. .. so on ... If it's 400gm, P20 is the one. Hope that works! |
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10-05-2002, 05:44 AM
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#8 (permalink) |
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Within my grasp!
Join Date: Jul 2002
Location: Vietnam
Posts: 196
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Hi all,
I think Raghuveer's solution is the best one. I heard about this problem when I was in the 5th grade. Exactly most Vietnamese students who learn in Special Mathematics Classes know this problem. It is a beautiful one. In the 5th grade, we learn about arithmetic. There are many other "beautiful" problems in arithmetic, those anyone can understand (even the small children in primary schools) but their ideas attract many people from different ages. I want to introduce 2 more problems which I like very much, besides this great one. 1. Given a matrix n x n. Write the integer numbers from 1 to n x n into the matrix, using each one only once. The matrix you get in the end must have an equal sum in each column, each row and each criss-cross. (Tell me if you don't understand the problem, it is difficult for me to tell it in English) 2. Find the 6-digit number abcdef which: 2 x abcdef, 3 x abcdef, 4 x abcdef, 5 x abcdef, 6 x abcdef all have the product which are 6-digit numbers formed from a, b, c, d, e, f (for example: fabcde...) Hope you enjoy them. ************************* My 100th post ************************* |
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10-05-2002, 02:41 PM
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#9 (permalink) | |
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So many things to do...
Forum Admin
Join Date: Dec 2000
Location: USA
Posts: 8,303
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