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Researcher -Coding

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Researcher -Coding

by vinay1983 » Mon Aug 26, 2013 4:53 am
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants,and each participant is to receive a different code?

(A) 4

(B) 5

(c) 6

(D) 7

(E) 8

OA: 5
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by ganeshrkamath » Mon Aug 26, 2013 5:43 am
vinay1983 wrote:A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants,and each participant is to receive a different code?

(A) 4

(B) 5

(c) 6

(D) 7

(E) 8

OA: 5
The answer is the smallest number n that satisfies the inequality:
12 <= n + nC2
12 <= n + n(n-1)/2
12 <= (2n + n^2 - n)/2
12 <= n(n+1)/2

Substituting n from the options gives the answer as 5.

Choose B

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by Brent@GMATPrepNow » Mon Aug 26, 2013 6:39 am
vinay1983 wrote:A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants,and each participant is to receive a different code?

(A) 4

(B) 5

(c) 6

(D) 7

(E) 8

OA: 5
Since the answer choices are so small, we might consider testing cases.
So, at this point, we need to figure out where to begin. This is where we need a bit of number sense.
I chose to start with A, because I could think of several possible codes using 4 letters.

So, let's manually list all of the codes that can be made with 4 letters
We get: A, B, C, D, AB, AC, AD, BC, BD, CD
10 possible codes in total (we need at least 12 codes)

So, we need to use MORE LETTERS.
At this point, it should be obvious that adding 1 letter will give us at least 2 extra codes, so the answer is [spoiler]5 letters (B)[/spoiler]

Cheers,
Brent
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