Bunuel
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
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.
Actually it could be A. B/c think of these arrangements for the 15 codes.
00, 01, 10, 02, 20, 03, 30, 11, 21, 12, 31, 13, 22, 23, 32 and 33. We have 16 arrangements so minimum # of different integers used can be 3.
What do you think?