GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Jun 2019, 00:10 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

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

Author Message
TAGS:

Hide Tags

Math Expert V
Joined: 02 Aug 2009
Posts: 7755
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

pra1785 wrote:
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

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 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?

Hi..
The last digit is odd and can be any of the five digits- 1,3,4,7,9
In type 2, the Hundreds digit is 8, so units digit can be any of 5 odd digits.
But in other type, the Hundreds digit is an odd integer-7 or 9, so units digit can be any of the remaining 4 as all digits are different
_________________
Manager  S
Joined: 15 Aug 2016
Posts: 208
Location: India
Concentration: Technology, Operations
GMAT 1: 690 Q49 V35 GPA: 3.84
WE: Operations (Consulting)
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.
Math Expert V
Joined: 02 Aug 2009
Posts: 7755
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

1
Utkarsh KOhli wrote:
I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.

hi..

be careful while reading the statements. many times we all tend to overlook finer points..
same case here..
it says three digits are different and NON ZERO..
so coloured portion above will be 7 possiblities
_________________
Manager  B
Joined: 24 Jun 2017
Posts: 120
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

So we have 3 digits numbers starting with 7 , 8 and 9. No repetitive digits, no zeros, numbers should be odd so ending with odd digit
7AB, 8AB, 9AB where A even number (excluding 0) and B is odd number (excluding repetition)
even numbers are 2, 4, 6, 8
odd numbers are 1, 3, 5, ,7 ,9

So for 7AB 1*4*4 (as we have to exclude 7 from B) = 16
So for 8AB 1*3*5 (as we have to exclude 8 from A) = 15
So for 9AB 1*4*4 (as we have to exclude 9 from B) = 16

Now we can have 7BB type combinations well, where BB are both odd numbers
So for 7BB 1*4*3 (as we have to exclude 7 from B) = 12
So for 8BB 1*5*4 = 20
So for 9BB 1*4*3 (as we have to exclude 9 from B) = 12

Total 16+15+16+12+20+12=91
Manager  S
Joined: 15 Aug 2016
Posts: 208
Location: India
Concentration: Technology, Operations
GMAT 1: 690 Q49 V35 GPA: 3.84
WE: Operations (Consulting)
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

chetan2u wrote:
Utkarsh KOhli wrote:
I am confused.
7AB
8BC
9CD.

Considering 7AB.
7 is fixed and is odd. Thus B will have 4 possibilities. But I think A has 8 possibilities after we fill one value with A.
Let say we put 1 for B.
A can have 0,2,3,4,5,6,8,9 i.e 8 possibilities.
So for 1st case its 1*8*4=32
2nd case is 1*8*5=40
3rd case is same as 1st i.e 1*8*4=32
=>32+40+32=104.
Please tell me where I am going wrong.

hi..

be careful while reading the statements. many times we all tend to overlook finer points..
same case here..
it says three digits are different and NON ZERO..
so coloured portion above will be 7 possiblities

Hi Chetan.
Thank you for pointing it out. Kudos to you.
EMPOWERgmat Instructor V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14351
Location: United States (CA)
GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

Hi All,

You'll likely find it easiest to handle this calculation in 'pieces'

This question lays out the following restrictions:
1) 3-digit numbers greater than 700
2) All digits are NON-0 and DIFFERENT
3) The number must be ODD.

1st digit is 7 = 1 option
3rd digit must be ODD, but NOT 7 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 700s

Now, the 800s...

1st digit is 8 = 1 option
3rd digit must be ODD = 5 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(5)(7) = 35 numbers in the 800s

Finally, the 900s; the math here works the same as the 700s...

1st digit is 9 = 1 option
3rd digit must be ODD, but NOT 9 = 4 options
2nd digit must differ from the other two digits and not be 0 = 7 options
(1)(4)(7) = 28 numbers in the 900s

28+35+28 = 91

GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Rich Cohen

Co-Founder & GMAT Assassin Follow
Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
Intern  B
Joined: 24 May 2018
Posts: 8
Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

Show Tags

Type-I: 7AB
Type-II: 8CD
Type-III- 9EF

Type-I = 1* A * B( Odd=1,3,5,7,9. Minus 7= 1,3,5,9= 4)
So, 1 * A * 7 (A is not 7 or B= 1,2,3,4,5,6,7,8,9= 7 and B= 7)
Now Type-I is 1*7*4

Type-II = 1 * C * D (D= odd= 1,3,5,7,9 = 5/ C= 1,2,3,4,5,6,7,8,9 minus 8 and C= 7)
So, 1* 7* 5

Type-III is the same as Type-I

Type-I: 1*7*4= 28
Type-II: 1*7*5=35
Type-III- 1*7*4=28 Re: Of the three-digit positive integers whose three digits are all differ   [#permalink] 18 Oct 2018, 21:54

Go to page   Previous    1   2   [ 27 posts ]

Display posts from previous: Sort by

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