It is currently 17 Oct 2017, 03:11

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Suppose a “Secret Pair” number is a four-digit number in which two adj

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
1 KUDOS received
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4417

Kudos [?]: 8419 [1], given: 102

Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 01 Mar 2017, 11:56
1
This post received
KUDOS
Expert's post
12
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

33% (01:58) correct 67% (02:24) wrong based on 114 sessions

HideShow timer Statistics

Suppose a “Secret Pair” number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other. For example, 2209 and 1600 are “Secret Pair” numbers, but 1333 or 2552 are not. How many “Secret Pair” numbers are there?

(A) 720
(B) 1440
(C) 1800
(D) 1944
(E) 2160


This is a somewhat easier problem from a list of 15 challenging problems for the GMAT Quant. For the whole collection, as well as the OE for this question, see:
Challenging GMAT Math Practice Questions

Mike :-)
[Reveal] Spoiler: OA

_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Kudos [?]: 8419 [1], given: 102

Magoosh Discount CodesMath Revolution Discount CodesOptimus Prep Discount Codes
Intern
Intern
avatar
B
Joined: 28 Sep 2016
Posts: 19

Kudos [?]: 7 [0], given: 34

Premium Member
Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 01 Mar 2017, 14:58
Thanks Mike for this Question, been a Fan of your questions many times.

If I am not wrong, this question has a flaw, The words like Number and digits makes confusing

If it is a Number Type it can't have 00 in the beginning
If it is a Digits Type it can have 00 in the beginning(like a Secret digit like ATM pin, Etc..)

If it is digit type the answer is E 2160

Approach :-

Given
There are _ _ _ _ 4 digits or 4 Positions named 1,2,3,4 respectively
And the total set of Consecutive digits are (00,11,22,33,44,55,66,77,88,99) there are 10 such pairs
The consecutive positions are 12, 23, and 34 -- there are 3 consecutive positions

Ans :-
Choosing one consecutive digit which is 10 ways
And these digits can go into any positions namely 12, 23, and 34 which is 3 ways
And the remaining digit places can be filled with 8*9 = 72 ways

10*3*72 = 2160

Hope it is correct, Fingers crossed :roll:

Kudos [?]: 7 [0], given: 34

Expert Post
1 KUDOS received
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4417

Kudos [?]: 8419 [1], given: 102

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 01 Mar 2017, 15:13
1
This post received
KUDOS
Expert's post
joepc wrote:
Thanks Mike for this Question, been a Fan of your questions many times.

If I am not wrong, this question has a flaw, The words like Number and digits makes confusing

If it is a Number Type it can't have 00 in the beginning
If it is a Digits Type it can have 00 in the beginning(like a Secret digit like ATM pin, Etc..)

If it is digit type the answer is E 2160

Approach :-

Given
There are _ _ _ _ 4 digits or 4 Positions named 1,2,3,4 respectively
And the total set of Consecutive digits are (00,11,22,33,44,55,66,77,88,99) there are 10 such pairs
The consecutive positions are 12, 23, and 34 -- there are 3 consecutive positions

Ans :-
Choosing one consecutive digit which is 10 ways
And these digits can go into any positions namely 12, 23, and 34 which is 3 ways
And the remaining digit places can be filled with 8*9 = 72 ways

10*3*72 = 2160

Hope it is correct, Fingers crossed :roll:

Dear joepc,

I'm happy to respond. :-)

My friend, to the best of my understanding, there is no flaw in this question. The question makes very clear that we are looking for a particularly category of numbers, four-digit numbers. The digits are mentioned to describe the restraint, but clearly we want to find numbers. You interpreted it as if we were just looking for four-digit expressions, such as 0027. This would be a possibility if the question were looking for something such as "sets of four digits," but for an actual number, this doesn't work.

Does all this make sense?
Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Kudos [?]: 8419 [1], given: 102

1 KUDOS received
Intern
Intern
avatar
B
Joined: 29 Dec 2016
Posts: 14

Kudos [?]: 9 [1], given: 8

Location: United States (CA)
GMAT 1: 620 Q49 V27
GMAT 2: 670 Q48 V35
GPA: 3.93
Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 01 Mar 2017, 21:09
1
This post received
KUDOS
one of the answer choice, which by the way is correct answer, is typed wrong.. I was breaking my head and wondering what is going on :)

Kudos [?]: 9 [1], given: 8

Intern
Intern
avatar
B
Joined: 28 Sep 2016
Posts: 19

Kudos [?]: 7 [0], given: 34

Premium Member
Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 02 Mar 2017, 05:19
Thanks Mike for the clarification

This is very good question, But it took some time to solve ; and a very Good Mind GYM question.

Answer is C 1800

Approach:-

Pairs -{ 00,11,22,33,44,55,66,77,88,99}- 10 Pairs

Treat the 00 Pair separate i.e Case I and Other pairs Together separate i.e Case II

Case I :- 00 Pair

Since 4 digits each digit place is named as position 1,2,3,4 respectively

A) 00 pair in position 2 and 3
First Position has 9 Options to fill
Fourth Position has 8 Options
= 72 Ways

B) 00 in position 3 and 4
First Position has 9 Options to fill
Second Position has 8 Options
= 72 Ways

C) 00 in Position 1 and 2 cannot be considered as they are not Numbers

Total for Case I = 72+72 = 144 pairs

Case II Other Pairs{11,22,33,44,55,66,77,88,99}
Take Any one pair let say 11

A)Put 11 in First and second position
Third position can be filled in 9 ways
Fourth Position can be filled in 8 ways
= 72 ways

B) Put 11 in second and third position
First position can be filled in 8 ways
Fourth Position can be filled in 7 ways
= 56 Ways

C) Put 11 in third and fourth position
First position can be filled in 8 ways
Second Position can be filled in 7 ways
= 56 Ways

Total = 72+56+56 = 184 ways
This for 11 and for remaining 22,33,44,55,66,77,88,99
Total for case II 184*9 = 1656 Ways


Total ways = Case I + case II
= 144+1656
= 1800 ways



Hope it is correct :-D

Kudos [?]: 7 [0], given: 34

Expert Post
1 KUDOS received
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4417

Kudos [?]: 8419 [1], given: 102

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 02 Mar 2017, 11:28
1
This post received
KUDOS
Expert's post
Dear joepc,

I'm happy to respond. :-)

My friend, I'm sorry to say that there are flaws in your method of counting. Look at this part
Quote:
Total for Case I = 72+72 = 144 pairs

Case II Other Pairs{11,22,33,44,55,66,77,88,99}
Take Any one pair let say 11

A )Put 11 in First and second position, Third position can be filled in 9 ways, Fourth Position can be filled in 8 ways = 72 ways

B) Put 11 in second and third position, First position can be filled in 8 ways, Fourth Position can be filled in 7 ways = 56 Ways

C) Put 11 in third and fourth position, First position can be filled in 8 ways, Second Position can be filled in 7 ways, = 56 Ways

Total = 72+56+56 = 184 ways
This for 11 and for remaining 22,33,44,55,66,77,88,99
Total for case II 184*9 = 1656 Ways

Here, part (A) is correct.

For part (B), suppose we have X11X. It's true, we can put only 8 choices in the first slot (not 0 or 1)--suppose we put 5 there. Now we have 511X. At this point, we have eight digits left for the fourth place (not 1 or 5, but any of the others). The mistake in part (C) is similar.

Your method undercounted the possibilities.

Does this make sense?
Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Kudos [?]: 8419 [1], given: 102

Intern
Intern
avatar
B
Joined: 28 Sep 2016
Posts: 19

Kudos [?]: 7 [0], given: 34

Premium Member
Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 02 Mar 2017, 13:18
Thanks Mike,

It was an eye opener, and It was a wonderful question, Keep Posting.

I got the answer, But not telling this time, waiting for your revelation

Long Live My friend Mike

Kudos [?]: 7 [0], given: 34

Manager
Manager
avatar
S
Joined: 02 Aug 2015
Posts: 70

Kudos [?]: 19 [0], given: 22

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 03 Mar 2017, 09:41
Three types of Secret pairs can be formed. Let X, Y, and S be different integers.

1. SSXY - 648 nos

2. XYSS - 648 nos

3. XSSY - 648 nos

Hence the total numbers - 1944. Hopefully this is correct.

Cheers!

Kudos [?]: 19 [0], given: 22

Intern
Intern
avatar
Joined: 16 Jan 2014
Posts: 3

Kudos [?]: [0], given: 107

Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 05 Mar 2017, 20:41
There are 10 pairs {00,11,22,33,44,55,66,77,88,99}

and 3 locations where these pairs can be fitted

Case 1 : when 00 is the pair

there are just two locations where his pair can go 23 and 34

no of possible options for other two numbers = 9*8
so total no of possible numbers= 1*2*8*9= 144

Case 2

When pair other than 00 is chosen. Now this pair has three locations where it can be fitted 12,23,34

But
Here again there can be two cases

A ) when the position of such a pair is 23 and 34

the number of options for other two digits is 8*8 because 0 can not come at 1st position

so total in such case is 9*2*8*8

B) When position of such pair is 12

The option for other two digits is 9*8 because 0 is also a valid number in this case

total = 9*1*9*8


Total = 1*2*9*8+ 9*1*9*8+9*2*8*8 = 1944

So I think 1944 is the answer

Kudos [?]: [0], given: 107

Senior Manager
Senior Manager
avatar
G
Joined: 24 Apr 2016
Posts: 335

Kudos [?]: 184 [0], given: 48

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 05 Mar 2017, 21:28
The answer is C (1800).
Attachments

Solution.png
Solution.png [ 134.83 KiB | Viewed 1052 times ]

Kudos [?]: 184 [0], given: 48

Manager
Manager
avatar
B
Joined: 23 Nov 2016
Posts: 57

Kudos [?]: 7 [0], given: 19

Premium Member CAT Tests
Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 08 Sep 2017, 14:34
Mike,

I'm curious here. I got the right answer, but only because I was able to eliminate A, B, C, and E. This question revealed an issue in my logic and I'm hoping you can help me determine why one method works and the other does not.

This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third.
AABC: (9 options)(1 option)(9 options)(8 options) = 648
BAAC: (8 options)(10 options)(1 option)(7 options) = 720
BCAA: (8 options)(7 options)(10 options)(1 options) = 720

Wrong answer: 2088

Now I picked D because of intuition, but after looking over the explanation, I'm very interested because in math you can often take more than one approach, as we did, but mine of course was incorrect :).

I see that when picking numbers for the three cases, I picked the digits from A to C, like so: AABC, BAAC, and BCAA.
However you picked them going from left to right: AABC, ABBC, and ABCC.

For example, for a "Secrect Number" such as BAAC, I got 720 options (8*10*7), whereas you got 648 (9*9*8) for the same type of "Secret Number", just you picked the first digit first (e.g. ABBC).

So, what gives? What did my approach forget to account for? I was not under the impression that when choosing combinations, one had to go from left to right, but now I am wondering if that was the error in my logic. Or, perhaps I should start with the most restrictive conditions first, then, move to least restrictive?

Kudos [?]: 7 [0], given: 19

Intern
Intern
avatar
B
Joined: 03 Aug 2017
Posts: 10

Kudos [?]: 5 [0], given: 52

CAT Tests
Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 08 Sep 2017, 15:24
brooklyndude wrote:
Mike,

This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third.
AABC: (9 options)(1 option)(9 options)(8 options) = 648
BAAC: (8 options)(10 options)(1 option)(7 options) = 720
BCAA: (8 options)(7 options)(10 options)(1 options) = 720

Wrong answer: 2088


Your fatal flaw comes with BAAC and BCAA.

BAAC: (8 options)(10 options)(1 option)(7 options) is incorrect

BAAC should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(1 option)(8 options [any of the remaining #s]) =648
BCAA should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(8 options [any of the remaining #s])(1 option) =648

648 * 3 - 1944

Kudos [?]: 5 [0], given: 52

Manager
Manager
avatar
B
Joined: 23 Nov 2016
Posts: 57

Kudos [?]: 7 [0], given: 19

Premium Member CAT Tests
Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 08 Sep 2017, 16:18
subsauce wrote:
brooklyndude wrote:
Mike,

This was my (erroneous) approach. A is the first digit I chose, then B is the second, then C is the third.
AABC: (9 options)(1 option)(9 options)(8 options) = 648
BAAC: (8 options)(10 options)(1 option)(7 options) = 720
BCAA: (8 options)(7 options)(10 options)(1 options) = 720

Wrong answer: 2088


Your fatal flaw comes with BAAC and BCAA.

BAAC: (8 options)(10 options)(1 option)(7 options) is incorrect

BAAC should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(1 option)(8 options [any of the remaining #s]) =648
BCAA should be: (9 options [can use anything except 0])(9 options [the remainder of the first #s & 0])(8 options [any of the remaining #s])(1 option) =648

648 * 3 - 1944


Completely understand that. My question is - why? I want to understand the theory why one method works and the other does not.

Kudos [?]: 7 [0], given: 19

Intern
Intern
avatar
B
Joined: 21 Sep 2016
Posts: 28

Kudos [?]: 1 [0], given: 261

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj [#permalink]

Show Tags

New post 19 Sep 2017, 14:41
Good question.

Let's assume the pair of equal digits in in the first 2 cases, hence 0 is not usable.

We'll have 9*1*9*8 options, which equals 648 numbers.

Now, if the pair is something like ABBC -> 9*9*1*8 (since now we can use 0 in the middle)

If the pair is in the end, which will give a number like ABCC: 9*9*8*1

Kudos [?]: 1 [0], given: 261

Re: Suppose a “Secret Pair” number is a four-digit number in which two adj   [#permalink] 19 Sep 2017, 14:41
Display posts from previous: Sort by

Suppose a “Secret Pair” number is a four-digit number in which two adj

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.