Aug 20 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Aug 20 09:00 PM PDT  10:00 PM PDT Take 20% off the plan of your choice, now through midnight on Tuesday, 8/20 Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 09 Feb 2010
Posts: 40

A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
27 Jul 2010, 12:25
Question Stats:
65% (01:34) correct 35% (01:40) wrong based on 679 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 57155

Re: Permutation question
[#permalink]
Show Tags
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.
_________________



Retired Moderator
Joined: 02 Sep 2010
Posts: 755
Location: London

Re: Bank combination
[#permalink]
Show Tags
19 Sep 2010, 16:02
with k integers, the number of possible codes is k*k=k^2 We need k^2>15 Minimum k is 4. Ans : (b)
_________________



Senior Manager
Status: Current Student
Joined: 14 Oct 2009
Posts: 359
Schools: Chicago Booth 2013, Ross, Duke , Kellogg , Stanford, Haas

Re: Permutation question
[#permalink]
Show Tags
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?
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 57155

Re: Permutation question
[#permalink]
Show Tags
22 Sep 2010, 00:00
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? 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 2digit 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.
_________________



Manager
Joined: 20 Jul 2010
Posts: 203

Re: Permutation question
[#permalink]
Show Tags
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
_________________
If you like my post, consider giving me some KUDOS !!!!! Like you I need them



Manager
Joined: 30 Nov 2010
Posts: 208
Schools: UC Berkley, UCLA

Re: Permutation question
[#permalink]
Show Tags
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?
_________________
Thank you for your kudoses Everyone!!!
"It always seems impossible until its done." Nelson Mandela



Math Expert
Joined: 02 Sep 2009
Posts: 57155

Re: Permutation question
[#permalink]
Show Tags
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.
_________________



Intern
Joined: 17 Jul 2010
Posts: 14
Location: United States (AL)

Re: Permutation question
[#permalink]
Show Tags
31 Jan 2011, 19:29
mariyea, You used four digits: 0, 1, 2, 3



Manager
Joined: 30 Nov 2010
Posts: 208
Schools: UC Berkley, UCLA

Re: Permutation question
[#permalink]
Show Tags
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!
_________________
Thank you for your kudoses Everyone!!!
"It always seems impossible until its done." Nelson Mandela



Manager
Joined: 27 Jul 2010
Posts: 144
Location: Prague
Schools: University of Economics Prague

Re: Permutation question
[#permalink]
Show Tags
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...
_________________
You want somethin', go get it. Period!



Manager
Joined: 30 Nov 2010
Posts: 208
Schools: UC Berkley, UCLA

Re: Permutation question
[#permalink]
Show Tags
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
_________________
Thank you for your kudoses Everyone!!!
"It always seems impossible until its done." Nelson Mandela



Manager
Joined: 27 Oct 2010
Posts: 105

Re: Permutation question
[#permalink]
Show Tags
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.



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1034
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: A local bank that has 15 branches uses a twodigit code to r
[#permalink]
Show Tags
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
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Intern
Joined: 02 May 2012
Posts: 6
Location: Argentina
Concentration: General Management, Strategy
WE: Corporate Finance (Manufacturing)

Re: A local bank that has 15 branches uses a twodigit code to r
[#permalink]
Show Tags
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.



Math Expert
Joined: 02 Sep 2009
Posts: 57155

A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
27 Mar 2013, 04:48



Intern
Joined: 22 Jan 2010
Posts: 24
Location: India
Concentration: Finance, Technology
GPA: 3.5
WE: Programming (Telecommunications)

Re: A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
28 Mar 2013, 00:05
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.
_________________
Please press +1 KUDOS if you like my post.



Intern
Status: Learning
Joined: 07 Aug 2011
Posts: 36
Location: India
Schools: WBUT  Class of 2011
GMAT Date: 01062014
GPA: 2.6
WE: Research (Education)

Re: A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
19 Aug 2014, 08:24
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
_________________
If you like my post give me kudos.
Arindam Sur Researcher, Academian



Manager
Joined: 20 Jan 2014
Posts: 139
Location: India
Concentration: Technology, Marketing

Re: A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
02 Oct 2014, 08:35
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 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
_________________
Consider +1 Kudos Please



Director
Joined: 23 Jan 2013
Posts: 537

A local bank that has 15 branches uses a twodigit code to
[#permalink]
Show Tags
02 Dec 2015, 23:51
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!/(n2)!, 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




A local bank that has 15 branches uses a twodigit code to
[#permalink]
02 Dec 2015, 23:51



Go to page
1 2
Next
[ 21 posts ]



