It is currently 11 Dec 2017, 14:08

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

Of the three-digit positive integers whose three digits are all differ

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

Hide Tags

3 KUDOS received
Current Student
avatar
Joined: 09 Mar 2014
Posts: 19

Kudos [?]: 32 [3], given: 50

Reviews Badge
Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 25 Apr 2016, 11:28
3
This post received
KUDOS
28
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

49% (01:19) correct 51% (01:29) wrong based on 582 sessions

HideShow timer Statistics

Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help
[Reveal] Spoiler: OA

Last edited by ENGRTOMBA2018 on 25 Apr 2016, 11:36, edited 2 times in total.
Reformatted the question and the title.

Kudos [?]: 32 [3], given: 50

17 KUDOS received
Current Student
avatar
B
Joined: 20 Mar 2014
Posts: 2671

Kudos [?]: 1787 [17], given: 796

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 25 Apr 2016, 11:42
17
This post received
KUDOS
7
This post was
BOOKMARKED
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.

Kudos [?]: 1787 [17], given: 796

Current Student
avatar
Joined: 09 Mar 2014
Posts: 19

Kudos [?]: 32 [0], given: 50

Reviews Badge
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 25 Apr 2016, 12:22
Thanks Engr2012 for helping with this one. Will keep in mind the posting rules.

Thank you

Kudos [?]: 32 [0], given: 50

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42544

Kudos [?]: 135259 [2], given: 12679

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 26 Apr 2016, 01:10
2
This post received
KUDOS
Expert's post
40
This post was
BOOKMARKED
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


Similar questions to practice:
of-the-three-digit-integers-greater-than-700-how-many-have-135188.html
how-many-integers-between-0-and-1570-have-a-prime-tens-digit-10443.html
how-many-integers-between-324-700-and-458-600-have-tens-110744.html
how-many-numbers-between-0-and-1670-have-a-prime-tens-digit-127319.html
of-the-three-digit-positive-integers-that-have-no-digits-41663.html
how-many-positive-integers-less-than-10-000-are-there-in-85291.html
a-credit-card-number-has-6-digits-between-1-to-9-the-firs-65745.html
how-many-natural-numbers-that-are-less-than-10-000-can-be-134571.html
how-many-four-digit-positive-integers-can-be-formed-by-using-133069.html
how-many-positive-integers-of-four-different-digits-each-100898.html
how-many-even-numbers-greater-than-300-can-be-formed-with-100578.html
how-many-even-4-digit-numbers-can-be-formed-so-that-the-95371.html
m04-70602.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
of-the-three-digit-integers-greater-than-660-how-many-have-128853.html
of-the-three-digit-integers-greater-than-600-how-many-have-127390.html
how-many-3-digit-integers-can-be-chosen-such-that-none-of-126616.html
how-many-odd-three-digit-integers-greater-than-800-are-there-94655.html
how-many-positive-integers-of-four-different-digits-each-100898.html
how-many-different-7-digit-members-are-their-sum-of-whose-98698.html
a-password-to-a-certain-database-consists-of-digits-that-can-95496.html
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
a-security-company-can-use-number-1-7-to-create-a-5-digit-90731.html
of-the-integers-between-100-and-799-inclusive-how-many-do-89239.html

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135259 [2], given: 12679

Expert Post
3 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5336

Kudos [?]: 6089 [3], given: 121

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 26 Apr 2016, 02:10
3
This post received
KUDOS
Expert's post
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


Hi,
since you want to know where you have gone wrong in your method..
hundreds digit - 3-- correct
units digit - 4 -- correct but ONLY in two cases where hundreds digit is 7 and 9-- MEANS you will have to add later for hundreds digit as 8, since 8 will have 5 ODDs as units digit..
I assume you mean tens digit as 8..NO, we are looking for Non-zero, so 10 - 2( ONE of hundreds and ONE of units) - 1( zero) = 7..

Now our answer is -- 3*4*7 = 84..
But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

Total = 84+7 = 91
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6089 [3], given: 121

Current Student
avatar
Joined: 09 Mar 2014
Posts: 19

Kudos [?]: 32 [0], given: 50

Reviews Badge
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 26 Apr 2016, 11:36
Thanks Bunuel.
Thanks Chetan2u for explaining my mistake.

Kudos [?]: 32 [0], given: 50

Manager
Manager
User avatar
Joined: 12 Jan 2015
Posts: 222

Kudos [?]: 82 [0], given: 79

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 05 May 2016, 04:50
Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-
In your earlier response

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Can you please assist.

Thanks
_________________

Thanks and Regards,
Prakhar

Kudos [?]: 82 [0], given: 79

Expert Post
1 KUDOS received
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5336

Kudos [?]: 6089 [1], given: 121

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 05 May 2016, 04:56
1
This post received
KUDOS
Expert's post
PrakharGMAT wrote:
Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-
In your earlier response

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Can you please assist.

Thanks



Hi,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options -Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

you can HALF ONLY when the numbers of ODD and EVEN are same,..
here it is not the same..
UNITS digit can be any of 10, 0 to 9, but we are excluding 0.. so there are 5 ODD- 1,3,5,7,9- and 4 EVEN - 2,4,6,8..
Again when 7 and 9 are first digit , the numbers would be different and when it is 8, it will be different
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6089 [1], given: 121

Expert Post
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5336

Kudos [?]: 6089 [0], given: 121

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 05 May 2016, 05:04
Expert's post
1
This post was
BOOKMARKED
PrakharGMAT wrote:
Hi chetan2u,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Can you please assist.

Thanks


Hi PrakharGMAT,
If you want to follow approach I..
first find for 7 and 9..
hundreds digit - 7 or 9- 2 ways
units digit - 1,2,3,4,5,6,8,7/9 - so 8 --- 4 are even and 4 are ODD..
tens digit - remaining 7..
total ways = 2*8*7 = 112..
half of these will be ODD so 112/2 = 56

Now if hundreds digit is 8..
units digit = 9 but ONLY 5 will be ODD
tens digit remaining 7..
so total = 1*7*5=35

Overall = 56+35 = 91.. :) :)
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6089 [0], given: 121

Intern
Intern
avatar
Joined: 22 Mar 2016
Posts: 1

Kudos [?]: [0], given: 12

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 24 May 2016, 20:02
Hi,
Please explain how are the possible number of options "7" for tens place.

Kudos [?]: [0], given: 12

1 KUDOS received
Manager
Manager
User avatar
Joined: 18 May 2016
Posts: 67

Kudos [?]: 64 [1], given: 105

Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 01 Jun 2016, 04:27
1
This post received
KUDOS
1
This post was
BOOKMARKED
Namita06 wrote:
Hi,
Please explain how are the possible number of options "7" for tens place.


I actually had the same question, and I think the reason is as follows:

For 7AB, A & B cannot be 7, but also, AB cannot be the same, right? What it means is:
A cannot be equal B, 0 or 7, thus 10-3 = 7
B cannot be even & equal 7 or A, but we've already accounted for A & B difference above, thus B has 4 options.

I hope it helps!
_________________

Please kindly +Kudos if my posts or questions help you!

My debrief: Self-study: How to improve from 620(Q39,V36) to 720(Q49,V39) in 25 days!

Kudos [?]: 64 [1], given: 105

4 KUDOS received
Intern
Intern
avatar
Joined: 01 May 2015
Posts: 45

Kudos [?]: 24 [4], given: 1

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 01 Jun 2016, 21:34
4
This post received
KUDOS
1
This post was
BOOKMARKED
To find odd integers greater than 700:

In 700s:
Units can be any of 4 (1,3,5,9)
Hundreds is obviously just 1 way (7)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

In 800s:
Units can be any of 5 (1,3,5,7,9)
Hundreds is obviously just 1 way (8)
So, tens can be any of the remaining 7.
So, total = 5*1*7 = 35

In 900s:
Units can be any of 4 (1,3,5,7)
Hundreds is obviously just 1 way (9)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

So, grand total = 28 + 35 + 28 = 91

Kudos [?]: 24 [4], given: 1

Intern
Intern
avatar
Joined: 28 Feb 2016
Posts: 20

Kudos [?]: 21 [0], given: 26

Schools: Wharton '19 (A)
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 05 Jun 2016, 12:44
Engr2012 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Can you please explain how you are getting to the "Number of combinations for type" 1-3?

Kudos [?]: 21 [0], given: 26

Manager
Manager
avatar
Joined: 17 Sep 2015
Posts: 95

Kudos [?]: 85 [0], given: 155

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 12 Jul 2016, 02:47
1
This post was
BOOKMARKED
Bunuel wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


Similar questions to practice:
of-the-three-digit-integers-greater-than-700-how-many-have-135188.html
how-many-integers-between-0-and-1570-have-a-prime-tens-digit-10443.html
how-many-integers-between-324-700-and-458-600-have-tens-110744.html
how-many-numbers-between-0-and-1670-have-a-prime-tens-digit-127319.html
of-the-three-digit-positive-integers-that-have-no-digits-41663.html
how-many-positive-integers-less-than-10-000-are-there-in-85291.html
a-credit-card-number-has-6-digits-between-1-to-9-the-firs-65745.html
how-many-natural-numbers-that-are-less-than-10-000-can-be-134571.html
how-many-four-digit-positive-integers-can-be-formed-by-using-133069.html
how-many-positive-integers-of-four-different-digits-each-100898.html
how-many-even-numbers-greater-than-300-can-be-formed-with-100578.html
how-many-even-4-digit-numbers-can-be-formed-so-that-the-95371.html
m04-70602.html
a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html
a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html
of-the-three-digit-integers-greater-than-660-how-many-have-128853.html
of-the-three-digit-integers-greater-than-600-how-many-have-127390.html
how-many-3-digit-integers-can-be-chosen-such-that-none-of-126616.html
how-many-odd-three-digit-integers-greater-than-800-are-there-94655.html
how-many-positive-integers-of-four-different-digits-each-100898.html
how-many-different-7-digit-members-are-their-sum-of-whose-98698.html
a-password-to-a-certain-database-consists-of-digits-that-can-95496.html
all-of-the-stocks-on-the-over-the-counter-market-are-126630.html
a-security-company-can-use-number-1-7-to-create-a-5-digit-90731.html
of-the-integers-between-100-and-799-inclusive-how-many-do-89239.html

Hope it helps.



It took me hours to solve and understand the concepts behind all these sums. It more than helped me. Atleast now when i encounter such sums i can try solving it under 2 mins.
Thanks a tonne
_________________

You have to dig deep and find out what it takes to reshuffle the cards life dealt you

Kudos [?]: 85 [0], given: 155

1 KUDOS received
Intern
Intern
avatar
B
Joined: 10 Jun 2014
Posts: 19

Kudos [?]: 19 [1], given: 119

GMAT ToolKit User
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 26 Aug 2016, 03:05
1
This post received
KUDOS
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help




i tried to solve this way:


700-1000 there 300 numbers
300/2 odds so 150 numbers
710-720-730 .... 10 *3 =30 numbers so 150-30=120
than 701-702-703 ..... 10*3=30 numbers so 120-30=90
90 is close to 91.

Kudos [?]: 19 [1], given: 119

Expert Post
Target Test Prep Representative
User avatar
S
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1789

Kudos [?]: 978 [0], given: 5

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 26 Aug 2016, 14:31
We know the integers must be greater than 700 and they must be odd; we can divide them into 6 groups: 7EO, 7OO, 8EO, 8OO, 9EO and 9OO where E denotes an even digit and O an odd digit. Let’s determine the number of integers in each group.

7EO:
The hundreds digit is 7, so it’s only 1 choice. Since all the digits are nonzero, E can be 4 numbers: 2, 4, 6 or 8. Since the digits are all different, O can be 4 numbers: 1, 3, 5 or 9. Therefore, there are 1 x 4 x 4 = 16 such integers.

7OO:
Since all the digits are different, O can only be 4 numbers: 1, 3, 5 or 9. However, the second O can only be 3 numbers since it cannot be the same as the first O. Therefore, there are 1 x 4 x 3 = 12 such integers.

8EO:
The hundreds digit is 8, so there is only 1 option. Since all the digits are different and nonzero, E can only be 3 numbers: 2, 4, or 6. Since the last digit is O, an odd digit, O can any of the 5 odd digits: 1, 3, 5, 7 or 9. Therefore, there are 1 x 3 x 5 = 15 such integers.

8OO:
The first O can be any of the 5 odd digits: 1, 3, 5, 7 or 9. However, the second O can only be 4 numbers since it cannot be the same as the first O. Therefore, there are 1 x 5 x 4 = 20 such integers.

9EO:
This is analogous to 7EO, so there should be 16 such integers.

9OO:
This is analogous to 9EO, so there should be 12 such integers.

Therefore, there are 16 + 12 + 15 + 20 + 16 + 12 = 91 odd integers that are greater than 700 where all digits are different and nonzero.

Answer: B
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 978 [0], given: 5

Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 367

Kudos [?]: 408 [0], given: 40

GPA: 3.98
Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 30 Oct 2016, 07:40
1
This post was
BOOKMARKED
According to the constraints we have 3 options for the thousands digit (7, 8, 9), 9 options for the hundreds digit (except 0), and 5 options for the units digit (it needs to be an odd number) in total with additional constraint that all digits should be different.
Lets look at two possible cases.

First (thousands digit is odd)
2 x 7 x 4 = 56
___ ____ ____
(7,9) (9-2) (5-1)

Thousands digit is even
1 x 7 x 5 = 35
___ ____ ____
(8) (9-2) (5)

Total 56+35=91

Kudos [?]: 408 [0], given: 40

Manager
Manager
avatar
B
Joined: 13 Dec 2013
Posts: 171

Kudos [?]: 29 [0], given: 122

Location: United States (NY)
Concentration: Nonprofit, International Business
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)
Reviews Badge
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 09 Apr 2017, 11:50
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



For the 700s:
_ _ _

I determined the possible options for each slot. The number of options for the first slot is 1 (7). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 4 (1, 3, 5, 9). This means that the number of options available for the third slot is 3. 1x4x3=12.
Now, if the second slot is even, there are 4 options (0 cannot be counted). The number of options for the third slot is 4 (only one odd number is used up - by the first slot). 1x4x4=16.

12+16=28. This number of options is the same for the 900s. So, now we just need to find the number of options for the 800s:

The number of options for the first slot is 1 (8). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 5 (1, 3, 5, 7, 9). This means that the number of options available for the third slot is 4. 1x5x4=20.
Now, if the second slot is even, there are 3 options (0 cannot be counted). The number of options for the third slot is 5 (no other odd numbers used up). 1x3x5=15.

20+15=35.

Total = 28+28+35=91.

Kudos if you agree! Please comment if you disagree with the method and/or have improvements.

Kudos [?]: 29 [0], given: 122

Expert Post
Top Contributor
1 KUDOS received
Director
Director
User avatar
B
Joined: 17 Dec 2012
Posts: 623

Kudos [?]: 547 [1], given: 16

Location: India
Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 21 May 2017, 18:44
1
This post received
KUDOS
Expert's post
Top Contributor
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help

Start with the digit under most constraints
1. Units digit can have 4 values i.e., 1,3,5,9 and 1,3,5,7 when the hundreds digit is 7 and 9 resp and can have 5 values i.e, 1,3,5,7,9 when the hundreds digit is 8
2. Tens digit cannot have the value of unit and hundreds digit and cannot be 0. So can have 7 values
3. So when hundreds digit is 7 or 9, number of possibilities is 4*7 + 4*7 = 56 and when hundreds digit is 8, number of possibilities is 5*7=35
4. Total number of possibilities is 91
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Premium Material
Standardized Approaches

Kudos [?]: 547 [1], given: 16

Manager
Manager
avatar
B
Joined: 20 Jan 2016
Posts: 219

Kudos [?]: 15 [0], given: 64

Re: Of the three-digit positive integers whose three digits are all differ [#permalink]

Show Tags

New post 04 Nov 2017, 04:51
ENGRTOMBA2018 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

[Reveal] Spoiler:
What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Hi,

I am a little confused with the explanation. For number of combinations for type 2: why is it 1*7*5
Don't we have to make sure that all numbers are different, non zero and odd in type 2 as well?

What am I missing?

Kudos [?]: 15 [0], given: 64

Re: Of the three-digit positive integers whose three digits are all differ   [#permalink] 04 Nov 2017, 04:51

Go to page    1   2    Next  [ 25 posts ] 

Display posts from previous: Sort by

Of the three-digit positive integers whose three digits are all differ

  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®.