Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Oct 2014, 17:59

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M22#16

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Senior Manager
Senior Manager
avatar
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
Followers: 4

Kudos [?]: 28 [0], given: 0

M22#16 [#permalink] New post 26 Nov 2008, 19:17
7
This post was
BOOKMARKED
How many three-digit integers bigger than 710 are there such that all their digits are different?

(A) 98
(B) 202
(C) 207
(D) 209
(E) 212

Source: GMAT Club Tests - hardest GMAT questions

Not sure here, but I think the OA is wrong.
Thanks.
Kaplan Promo CodeKnewton GMAT Discount CodesVeritas Prep GMAT Discount Codes
6 KUDOS received
CEO
CEO
User avatar
Joined: 29 Aug 2007
Posts: 2500
Followers: 54

Kudos [?]: 512 [6] , given: 19

Re: M22#16 [#permalink] New post 26 Nov 2008, 21:15
6
This post received
KUDOS
ventivish wrote:
How many three-digit integers bigger than 710 are there such that all their digits are different?

# 98
# 202
# 207
# 209
# 212

Not sure here, but I think the OA is wrong.
Thanks.


3-digit integers bigger than 710 such that all their digits are different = (3 x 9 x 8) = 216
deduct all those 3-digit integers that are above > 699 but < 711 (do not count 700 and 707 becaue we already ecluded them in 216) = 11-2 = 9
so it is 216 - 9 = 207


intresting
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Manager
Manager
User avatar
Joined: 30 Apr 2009
Posts: 136
Followers: 1

Kudos [?]: 33 [0], given: 9

Re: M22#16 [#permalink] New post 30 Sep 2009, 11:22
Can you please let me know how you came up with 3*9*8..?

Got the right answer by using diff. approach, however do not understand the explanation that is given, that is the same as yours.
_________________

Trying to make CR and RC my strong points

"If you want my advice, Peter," he said at last, "you've made a mistake already. By asking me. By asking anyone. Never ask people. Not about your work. Don't you know what you want? How can you stand it, not to know?" Ayn Rand

31 KUDOS received
Manager
Manager
User avatar
Joined: 22 Jul 2009
Posts: 192
Followers: 4

Kudos [?]: 195 [31] , given: 18

Re: M22#16 [#permalink] New post 14 Oct 2009, 10:39
31
This post received
KUDOS
kt00381n wrote:
Can you please let me know how you came up with 3*9*8..?


hundreds: 3 possible digits (7,8,9)
tens: 9 possible digits (from 0 to 9 minus the one digit already selected for the hundreds)
units: 8 possible digits (from 0 to 9 minus the two digits already selected for the hundred and tens)

3*8*9=216. But that's for the 700-999 range. Question asks about the 711-999 range. So you have to substract 701, 702, 703, 704, 705, 706, 708, 709, 710 -> 216-9=207.
_________________

Please kudos if my post helps.

Intern
Intern
avatar
Joined: 20 Jul 2009
Posts: 2
Followers: 0

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

Re: M22#16 [#permalink] New post 09 Dec 2009, 08:49
The question asks for all different digits. In that case dont we have to subtract occurrences of #s like 777, 722, 855, etc. ?

Thanks!
4 KUDOS received
Manager
Manager
avatar
Joined: 05 Dec 2009
Posts: 127
Followers: 2

Kudos [?]: 78 [4] , given: 0

Re: M22#16 [#permalink] New post 13 Mar 2010, 16:38
4
This post received
KUDOS
another way in 7 series - first digit = 7, second = any digit but 7 and 0 = 8, 3 digit = any digit but 7 and second digit = 8
so total numbers = 8 X 8 = 64 but subtract 1 for 710 so = 64 - 1 = 63.

Now total numbers in 8 and 9 series = 2 X 9X 8 = 144
Total = 144 + 63 = 207
1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 13 Dec 2009
Posts: 264
Followers: 10

Kudos [?]: 111 [1] , given: 13

Reviews Badge
Re: M22#16 [#permalink] New post 19 Mar 2010, 07:32
1
This post received
KUDOS
ventivish wrote:
How many three-digit integers bigger than 710 are there such that all their digits are different?

# 98
# 202
# 207
# 209
# 212

Not sure here, but I think the OA is wrong.
Thanks.


100th digit 10 digit one's digit
7,8,9 1 to 9 1 to 8
3 * 9 * 9 = 216 - 9 occurences below 710 = 207 is the answer
_________________

My debrief: done-and-dusted-730-q49-v40

1 KUDOS received
Director
Director
avatar
Joined: 21 Dec 2009
Posts: 589
Concentration: Entrepreneurship, Finance
Followers: 15

Kudos [?]: 291 [1] , given: 20

Re: M22#16 [#permalink] New post 05 May 2010, 12:17
1
This post received
KUDOS
Neels wrote:
The question asks for all different digits. In that case dont we have to subtract occurrences of #s like 777, 722, 855, etc. ?

Thanks!

I was almost going to ask the same question.
Take a look. HTU (3*9*8 )
step1: H =7xx, 8xx, and 9xx...7, 8, and 9 make 3 different hundred's digits
step2: T = 0 - 9 (excluding either 7, 8 or 9) => 9 digits left. 77x not possible. Got it?
step3: U = 0 - 9 (excluding any digit used as H and that for T) 8 digits left.... 855 not possible


Subtract numbers ( 9 in all) less than 710 excluding 700 and 707 -already taken care of
in steps 1 to 3.

Thus the calculation 3*9*8 - 9 = 207.

Do i make any sense?
_________________

KUDOS me if you feel my contribution has helped you.

1 KUDOS received
Manager
Manager
User avatar
Joined: 15 Apr 2010
Posts: 196
Followers: 3

Kudos [?]: 15 [1] , given: 29

Re: M22#16 [#permalink] New post 05 May 2010, 18:49
1
This post received
KUDOS
i did it in a slightly roundabout way....

total numbers in consideration: 999-710= 289

Each hundred set of numbers has 10 in-eligible series of numbers like 77*, 88*, 99* => subtract 30
Every set of 10 numbers (710-719 for eg, has 711, 717) has 2 numbers that are in-eligible (and exclude the set of 10 above):
=> 2 *3 (hundreds series) * 9 => 54 so we can subtract that as well

289-30-54 = 205, but we have to add back the 700, 707 numbers since the range in the question is from 710-999.

this is an alternative way (since i saw the pattern repeating and did it this way)- not elegant at all!
2 KUDOS received
Intern
Intern
avatar
Joined: 11 Jan 2010
Posts: 38
Followers: 1

Kudos [?]: 50 [2] , given: 6

Re: M22#16 [#permalink] New post 12 May 2010, 09:44
2
This post received
KUDOS
To find number of three digit integers that are bigger than 710:

(1) The numbers with digit 7 in hundreds place and digit 1 in tens place: 71 :?:
Excluding 7, 1 and 0, there can be seven different digits in the units place.
Therefore, count of different numbers = 7

(2) The number with digit 7 in hundreds place and greater than or equal to 2 in in tens place: 7 :?: :?:
Excluding 7, 1 and 0, there can be seven different digits in the tens place.
Excluding 7 and the digit at tens place, there can be eight different digits in the units place.
Therefore, count of different numbers = 7 * 8 = 56

(3) The number with digit 8 or 9 in hundreds place: 8 or 9 :?: :?:
There can be two different digits in the hundreds place.
Excluding the hundreds digit, there can be nine different digits in the tens place.
Excluding the hundreds digit and the tens digit, there can be eight different digits in the tens place.
Therefore, count of different numbers = 2 * 9 * 8 = 144

Thus, total count of different number bigger than 710 = 7 + 56 + 144 = 207

Answer is C.

Consider awarding me with Kudos if this helps!! Cheers!!! :)
1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 21 Dec 2009
Posts: 268
Location: India
Followers: 9

Kudos [?]: 72 [1] , given: 25

Re: M22#16 [#permalink] New post 20 May 2010, 07:00
1
This post received
KUDOS
[quote="ventivish"]How many three-digit integers bigger than 710 are there such that all their digits are different?

(A) 98
(B) 202
(C) 207
(D) 209
(E) 212

No of different 3 digit no greater than 700 = count of (7,8,9) X {count of (0,1,2,3....7,8,9) - 1} X {count of (0,1,2,3....7,8,9) - 2}
=3X9X8=216

Now between 700 to 710 three different digit numbers are 701, 702,703...710 total count 9.

So the number of three-digit integers bigger than 710 are there such that all their digits are different = 216 -9 =207

C is the answer.
_________________

Cheers,
SD

3 KUDOS received
SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1691
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 297 [3] , given: 36

Premium Member Reviews Badge
Re: M22#16 [#permalink] New post 09 May 2011, 05:07
3
This post received
KUDOS
0,1,2,3,4,5,6,7,8,9

7,8,9 - 3 ways

2nd digit - 9 ways

3rd digit - 8 ways

Total = 72 *3 = 216

700 - 710 < 710 = 11

Subtract 2 from this for 700 and 707

So total = 216 - 9 = 207

Answer - C
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 17 Dec 2010
Posts: 12
Followers: 0

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

Re: M22#16 [#permalink] New post 10 May 2011, 07:51
Correct Answer C

3-digit integers bigger than 710 such that all their digits are different = (3 x 9 x 8) = 216
deduct all those 3-digit integers that are above > 699 but < 711 (do not count 700 and 707 becaue we already ecluded them in 216) = 11-2 = 9
so it is 216 - 9 = 207
1 KUDOS received
Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 108
Followers: 1

Kudos [?]: 10 [1] , given: 28

Re: M22#16 [#permalink] New post 12 May 2012, 08:21
1
This post received
KUDOS
My approach was a littler different but should yield the same answer but its not. Can someone please help out.

So I broke the possibilities down.

7 1 __ (one slot) The blank space's numbers can be 1 through 9 (hence 9 numbers) but we must subtract out 7,1 because they have already been picked. Therefore a total of 7 numbers

7 _ _ (two slots) The second slot's number can range from 2 through 9 (hence 8 numbers) but we must subtract a 7, therefore 7 possibilities for the second slot. The third slot can be 6 possibilities because we have one less number that went into the second slot. Hence 7*6=42

_ _ _ (three slots) First slot can be 8 though 9 (2 numbers). The second slot can range from 0 though 9 (10 numbers) but we must subtract the number we placed in the first slot, therefore 9 possible numbers. The third slot can range from 0 though 9 (10 numbers) but we must subtract out the two numbers that we picked for the first and second slot, hence 8 total numbers. Therefore we have 2*9*8=144.

Now we add all the numbers together we get 7+42+144=193. What am I missing here?
1 KUDOS received
Intern
Intern
User avatar
Joined: 22 Dec 2011
Posts: 46
Concentration: Entrepreneurship
GMAT 1: 680 Q47 V36
GMAT 2: 700 Q49 V37
GPA: 3.5
Followers: 1

Kudos [?]: 11 [1] , given: 2

Re: M22#16 [#permalink] New post 13 May 2012, 19:55
1
This post received
KUDOS
alphabeta1234 wrote:
My approach was a littler different but should yield the same answer but its not. Can someone please help out.



7 _ _ (two slots) The second slot's number can range from 2 through 9 (hence 8 numbers) but we must subtract a 7, therefore 7 possibilities for the second slot. The third slot can be 6 possibilities because we have one less number that went into the second slot. Hence 7*6=42


Now we add all the numbers together we get 7+42+144=193. What am I missing here?


The thing you are missing for your two slot method is the 2 other possibilities for the ones digit, 1 and 0, which can not be duplicated in the tens or hundreds digit. So instead of 6 possibilities, there are actually 8. So 7*8 = 56, which is 14 more than the 42 you got, which is the difference between your result, (193) and the answer (207).
Intern
Intern
avatar
Joined: 11 Jan 2010
Posts: 38
Followers: 1

Kudos [?]: 50 [0], given: 6

Re: M22#16 [#permalink] New post 15 May 2013, 05:01
How many three-digit integers bigger than 710 are there such that all their digits are different?

Here is another way to look at the same solution:

When 100's digits is 7:
10's digits can be 1 through 9 excluding 7. So 10's digit can be selected in 8 ways.
1's digits can be 0 through 9 excluding 7 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 7 = 8 * 8 = 64
But exclude 710, so the resulting count will be 64 - 1 = 63

When 100's digits is 8:
10's digits can be 0 through 9 excluding 8. So 10's digit can be selected in 9 ways.
1's digits can be 0 through 9 excluding 8 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 8 = 9 * 8 = 72

When 100's digits is 9:
10's digits can be 0 through 9 excluding 9. So 10's digit can be selected in 9 ways.
1's digits can be 0 through 9 excluding 8 and the 10's digit. So 1's digit can be selected in 8 ways.
So, count of ways when 100's digit is 8 = 9 * 8 = 72

Number of count for three-digit numbers greater than 710 = 63 + 72 + 72 = 207 ways
1 KUDOS received
Intern
Intern
avatar
Joined: 09 Nov 2012
Posts: 13
Location: India
saurav: suman
Concentration: Finance, Technology
GMAT Date: 09-02-2014
GPA: 3.38
WE: Engineering (Energy and Utilities)
Followers: 1

Kudos [?]: 9 [1] , given: 28

GMAT ToolKit User
Re: M22#16 [#permalink] New post 04 May 2014, 11:50
1
This post received
KUDOS
Here we have to find all different digit number between 710 to 999.
first of all we find the no of digits between 800 to 809
800, 801, 802, 803, 804,805,806, 807, 808, 809 total 8 nos are different
therefore between 800 to 899 = 8*10= 80 is not right, because we 8 number twice.
i.e from 880 to 899 so we have to subtract this no from 80 therefore total 72 nos are different
now we can easily find total different digit no between 700 to 999 as 72*3 =216.
Now subtract 9 from 216 (why 9) because 700 to 709 contains 8 different numbers and in question it is written we have to find three digit no greater than 710. here 1 extra no is 701. This makes a total of 8+1=9)
therefore 216-9=207
hence C is the correct ans
Intern
Intern
avatar
Joined: 14 Mar 2014
Posts: 1
Followers: 0

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

Re: M22#16 [#permalink] New post 29 May 2014, 18:55
I just wanna ask, where did the 707?
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23357
Followers: 3604

Kudos [?]: 28727 [0], given: 2839

Re: M22#16 [#permalink] New post 30 May 2014, 00:40
Expert's post
Irawaty wrote:
I just wanna ask, where did the 707?


Do you mean why don't we count 707? If yes, then notice that we need the numbers which are greater than 710 and with distinct digits. 707 does not satisfy any of these conditions.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: M22#16   [#permalink] 30 May 2014, 00:40
Display posts from previous: Sort by

M22#16

  Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: WoundedTiger, Bunuel



cron

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

Powered by phpBB © phpBB Group and phpBB SEO

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®.