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27 Jul 2010, 12:25
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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64% (01:34) correct
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A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?
A. 3
B. 4
C. 5
D. 6
E. 7
A. 3
B. 4
C. 5
D. 6
E. 7
27 Jul 2010, 12:44
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zest4mba
A local bank that has 15 branches uses a two-digit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of two-digit numbers that are the reverse of each other (such as 17 and 71) are considered as two separate codes. What is the fewest number of different integers required for the 15 codes?
Choices
A 3
B 4
C 5
D 6
E 7
Choices
A 3
B 4
C 5
D 6
E 7
Consider the code XY. If there are \(n\) digits available then X can take \(n\) values and Y can also take \(n\) values, thus from \(n\) digits we can form \(n^2\) different 2-digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2-digit numbers (00, 01, 02, ..., 99).
We want # of codes possible from \(n\) digit to be at least 15 --> \(n^2\geq{15}\) --> \(n\geq4\), hence min 4 digits are required.
Answer: B.
Hope it's clear.
General Discussion
19 Sep 2010, 16:02
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with k integers, the number of possible codes is k*k=k^2
We need k^2>15
Minimum k is 4.
Ans : (b)
We need k^2>15
Minimum k is 4.
Ans : (b)
