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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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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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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n * (n-1) is to say that 99 will not be chosen. To choose 99 once we are saying n*n should be the combo

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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mariyea,
You used four digits: 0, 1, 2, 3

Well this is kind of embarrassing Forgot about zero My bad!

Nice one zero is the everlasting problem, not only by you...

Thank you for trying to keep my confidence intact

Starting with choice A, 3 * 3 = 9 options are possible to code 15 branches. Not suff. Using 4 in choice B, 4 * 4 = 16 options are possible. We need the fewest. So B.

\(N\)= number of integers
We have \(N\)options for the first one, \(N-1\) options for the second. So a total of \(N(N-1)\) combinations, and we want that \(N(N-1)=15\). Now or you plug in the different options, or you solve \(N^2-N-15=0\); the first way seems faster, so lets try with A) 3 : 3*2=6 No B)4*3=12 No again C)5*4=20 YES

C

I started from an small matrix :

11 12 13 14 15 16 17
21 22 23 24 25 26 27
31 32 33 34 35 36 37
41 42 43 44 45 46 47
51 52 53 54 55 56 57
61 62 63 64 65 66 67
71 72 73 74 75 76 77

Then I highlight the possible combinations (not considering the numbers with repeated integers) For example with 1 integer there are nor any number possible, with 2 integers, 2 possible numbers, with 3 integers , 6 possible numbers, with 4 integers, 12 possible numbers, and with 5 ... 20 . Correct Answer C

I know it´s not the finest answer, I guess it should be explained with combinatory.

Similar questions to practice:
http://gmatclub.com/forum/a-researcher- ... 34584.html
http://gmatclub.com/forum/if-a-code-wor ... 26652.html
http://gmatclub.com/forum/all-of-the-st ... 26630.html
http://gmatclub.com/forum/the-simplasti ... 05845.html
http://gmatclub.com/forum/a-4-letter-co ... 59065.html
http://gmatclub.com/forum/a-certain-sto ... 86656.html
http://gmatclub.com/forum/a-5-digit-cod ... 32263.html

Hope it helps.
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Let the required number of digits be n.
Considering the given conditions,
i)The first digit can be filled up in n ways.
ii)The second digit can be filled up in n ways too.
So we will get \(n * n = n^2\) numbers.

\(n^2 \geq 15\)

=>\(n \geq 4\) [since n is an integer.]

So,option B will be the correct answer.
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Please press KUDOS if you like my post.

More easy solution. Think logically.

Pick any two integer.

Integers: 1 & 2

Code: 11, 12, 21, 22 = 4 Codes

Add one more integer: 3

13, 31, 33, 23, 32 = 5 Codes

Add one more integer: 4

44, 14, 41, 24, 42, 34, 43 = 7 Codes

Total = 16 Codes. Enough. Answer: B

2 integers create 4 codes. we need 15 codes.

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


We can write that
XC1 + XP2 = 15
Lets take X=3
3+ 6 = 9

Now lets take X=4
4 + 12 = 16 - This is close to our answer. Hence B is the answer

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