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# In a certain lab, chemicals are identified by a color-coding

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Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

In a certain lab, chemicals are identified by a color-coding [#permalink]  28 Jun 2005, 16:08
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In a certain lab, chemicals are identified by a color-coding system. There are 20 different chemicals. Each one is coded with either a single color or a unique two color pair. If the order of the colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

Is it possible to do this problem without knowing the answer choices (w/ the choices, it is simply a matter of putting them into a quick equation)?

I also had another quick question - I have come across the <> symbol quite often in these boards, such as n<>0 etc. What does <> mean?

Thanks.
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

<> means 'not equal to'

sure you can do it w/o answer choices

just need to solve

n + 0.5*n*(n-1) >= 20
or
n^2 + n - 40 >=0
for a minimum positive integer n
Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

Could you fully solve your equation.

I dont seem to be getting the right OA after solving the equations you have given.

Thanks.
Director
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

if n=5, the expression equals -10

if n=6, the expression equals 2 > 0 so 6 is the answer
Intern
Joined: 19 Aug 2004
Posts: 47
Followers: 1

Kudos [?]: 3 [0], given: 0

Got it

Thanks.
Senior Manager
Joined: 30 May 2005
Posts: 374
Followers: 1

Kudos [?]: 7 [0], given: 0

If n is the required number, then it is the smallest positive integer that fulfils the inequality:

C(n,1) + C (n,2) > 19

Senior Manager
Joined: 06 Apr 2005
Posts: 355
Location: USA
Followers: 1

Kudos [?]: 25 [0], given: 1

Thanks ABJ, thats how I arrived at the answer = 6.
6C2 = 15, 6C1 = 6, therefore 6C2 + 6C1 = 21 which is > 19
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