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A local bank that has 15 branches uses a twodigit code to [#permalink]
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27 Jul 2010, 12:25
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A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of twodigit 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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Re: A local bank that has 15 branches uses a twodigit code to [#permalink]
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28 Mar 2013, 00:05
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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.  Please press KUDOS if you like my post.
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Re: A local bank that has 15 branches uses a twodigit code to [#permalink]
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19 Aug 2014, 08:24
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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. zest4mba wrote: A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of twodigit 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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Re: Permutation question [#permalink]
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27 Jul 2010, 12:44
zest4mba wrote: A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of twodigit 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 2digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2digit 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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Re: Bank combination [#permalink]
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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]
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21 Sep 2010, 16:19
I'm confused how xy would have n*n possibilities...wouldn't it b n*(n1) 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]
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22 Sep 2010, 09:13
Michmax3 wrote: I'm confused how xy would have n*n possibilities...wouldn't it b n*(n1) 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? n * (n1) is to say that 99 will not be chosen. To choose 99 once we are saying n*n should be the combo
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Re: Permutation question [#permalink]
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31 Jan 2011, 17:58
Bunuel wrote: zest4mba wrote: A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of twodigit 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 2digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2digit 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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Re: Permutation question [#permalink]
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31 Jan 2011, 18:03
mariyea wrote: Bunuel wrote: zest4mba wrote: A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer can be used for both digits of a code, and a pair of twodigit 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 2digit codes: this is the same as from 10 digits (0, 1, 2, 3, ..., 9) we can form 10^2=100 different 2digit 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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Re: Permutation question [#permalink]
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31 Jan 2011, 19:29
mariyea, You used four digits: 0, 1, 2, 3



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Re: Permutation question [#permalink]
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01 Feb 2011, 06:53
Bunuel wrote: How many digits did you use?
Answer is B not A.
Well this is kind of embarrassing Forgot about zero My bad!
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Re: Permutation question [#permalink]
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01 Feb 2011, 18:07
mariyea wrote: Bunuel wrote: How many digits did you use?
Answer is B not A.
Well this is kind of embarrassing Forgot about zero My bad! Nice one zero is the everlasting problem, not only by you...
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Re: Permutation question [#permalink]
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02 Feb 2011, 09:56
craky wrote: mariyea wrote: Bunuel wrote: How many digits did you use?
Answer is B not A.
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
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Re: Permutation question [#permalink]
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02 Feb 2011, 10:22
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.



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A local bank that has 15 branches uses a twodigit code to r [#permalink]
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26 Mar 2013, 10:19
A local bank that has 15 branches uses a twodigit code to represent each of its branches. The same integer cannot be used for both digits of a code, 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
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Re: A local bank that has 15 branches uses a twodigit code to r [#permalink]
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26 Mar 2013, 10:35
\(N\)= number of integers We have \(N\)options for the first one, \(N1\) options for the second. So a total of \(N(N1)\) combinations, and we want that \(N(N1)=15\). Now or you plug in the different options, or you solve \(N^2N15=0\); 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 YESC
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Re: A local bank that has 15 branches uses a twodigit code to r [#permalink]
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26 Mar 2013, 10:43
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.



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