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A local bank that has 15 branches uses a two-digit code to

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zest4mba
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A local bank that has 15 branches uses a two-digit code to [#permalink] A local bank that has 15 branches uses a two-digit code to   27 Jul 2010, 12:25
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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
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B
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Bunuel
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Re: Permutation question [#permalink] A local bank that has 15 branches uses a two-digit code to   27 Jul 2010, 12:44
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zest4mba wrote:
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.
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shrouded1
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Re: Bank combination [#permalink] A local bank that has 15 branches uses a two-digit code to   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)
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Re: Permutation question [#permalink] A local bank that has 15 branches uses a two-digit code to   21 Sep 2010, 16:19
I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?
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Re: Permutation question [#permalink] A local bank that has 15 branches uses a two-digit code to   22 Sep 2010, 00:00
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Michmax3 wrote:
I'm confused how xy would have n*n possibilities...wouldn't it b n*(n-1) possibilities because you would have to have two different digits? For example 17 and 71 are two different codes but 99 and 99 are the same code. Can someone explain?


It's always good to test theoretical thoughts on practice:

How many codes can be formed using 2 digits (\(n=2\)), 0 and 1.:

00;
01;
10;
11.

4=2^2.

Or consider the following: how many 2-digit codes can be formed out of 10 digits (0, 1, 2, 3, ..., 9)?
00;
01;
02;
...
99.

Total 100=10^2.

Hope it's clear.
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Re: Permutation question [#permalink] A local bank that has 15 branches uses a two-digit code to   31 Jan 2011, 17:58
Bunuel wrote:
zest4mba wrote:
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?
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Bunuel
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Re: Permutation question [#permalink] A local bank that has 15 branches uses a two-digit code to   31 Jan 2011, 18:03
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mariyea wrote:
Bunuel wrote:
zest4mba wrote:
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?


How many digits did you use?

Answer is B not A.
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Bunuel
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A local bank that has 15 branches uses a two-digit code to [#permalink] A local bank that has 15 branches uses a two-digit code to   27 Mar 2013, 04:48
Expert Reply
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Hope it helps.
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A local bank that has 15 branches uses a two-digit code to [#permalink] A local bank that has 15 branches uses a two-digit code to   02 Dec 2015, 23:51
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Try to explain why Permutation formula does not fully work here

if we have 4 distinct numbers, e.g. 0,1,2,3

00,01,02,03
10,11,12,13
20,21,22,23
30,31,32,33
we have 16 options (of 2 numbers taken)

permutation formula is P=n!/(n-2)!, so if n=4 we get only 12 options.

But formula defines how many times the arrangements of 4 distinct objects is that of any 2 distinct objects.
So, it excludes 00,11,22,33 from the list, i.e. 4 options

So, 12+4=16 and 4 numbers is enough

B
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Re: A local bank that has 15 branches uses a two-digit code to [#permalink] A local bank that has 15 branches uses a two-digit code to   08 Jun 2024, 08:22
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