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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Of the three digit integers greater than 700, how many have [#permalink]

Show Tags

27 Sep 2010, 11:40

20

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

65% (hard)

Question Stats:

57% (01:08) correct
43% (01:18) wrong based on 338 sessions

HideShow timer Statistics

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

Hey Bunnel, your answer is super but beyond my grasping power to do by self and less than 2 mins.

So I had a fairly different approach: For a 3-digit number we have a digit in the tens, hundreds and thousands place. So counting the number of digits when tens and hundreds are same but not thousands: i.e: 711. 722...766,788,799,800 = 8+9+9 = 26 So counting the number of digits when tens and thousands are same but not hundreds: i.e: 707,717,...767,787,... = 9+9+9 = 27 So counting the number of digits when hundreds and thousands are same but not tens: i.e: 770,771,772...776,778... = 9+9+9 = 27

Total it up to 80. Answer C
_________________

Please give me kudos, if you like the above post. Thanks.

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Guys, i need help, it is 81 not 80.

abc is the 3 digit integer greater than 700:

Then a can be 7, 8, 9: we have 3 choices for a - if b=a, c#a=b: 1 choice for b and 10-1=9 choices for c: 3*1*9=27 - if b=c#a: 9 choices for b, and 1 choice for c: 3*9*1=27 - if a=c: 1 choice for c, 9 choises for b (b is different fr a): 3*1*9=27

Then we have 27+27+27=81, i cant not reason why? plz help.

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Guys, i need help, it is 81 not 80.

abc is the 3 digit integer greater than 700:

Then a can be 7, 8, 9: we have 3 choices for a - if b=a, c#a=b: 1 choice for b and 10-1=9 choices for c: 3*1*9=27 - if b=c#a: 9 choices for b, and 1 choice for c: 3*9*1=27 - if a=c: 1 choice for c, 9 choises for b (b is different fr a): 3*1*9=27

Then we have 27+27+27=81, i cant not reason why? plz help.

The red part could give you 700 but we are told that numbers must be more than 700, so subtract 1 --> 81-1=80.

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

Bunuel - Your method of pattern recognition is by far the best and quickest approach to problems such as these. Thanks for the help.

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

Re: Of the three digit integers greater than 700, how many have [#permalink]

Show Tags

09 Dec 2011, 22:19

ABC: for 700 series, PQR: for 800 and XYZ:900 A, P, and X are reserved for 7, 8 and 9 respectively. for each serise, there are three cases: 1. tens digit is same as hundreds digits, but unit digit is different. 2. Unit digit is same as hundreds digits, but tens digit is different. 3 .Both unit and tens digit are same but not = hundreds digit

Therefore, we have: for 700 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*8C1*1C1)=9+9+8=26 for 800 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*9C1*1C1)=9+9+9=27 for 900 series: (1C1*1C1*9C1)+ (1C1*9C1*1C1) +(1C1*9C1*1C1)=9+9+9=27

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

Could you elaborate on part A just a bit more please.

Of the three digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?

A) 90 B) 82 C) 80 D) 45 E) 36

Three digit number can have only following 3 patterns: A. all digits are distinct; B. two are alike and third is different; C. all three digits are alike.

We need to calculate B. B=Total - A - C

Total numbers from 700 to 999 = 299 (3-digit numbers greater than 700); A. all digits are distinct = 3*9*8=216 (first digit can have only three values 7, 8, or 9); C. all three are alike = 3 (777, 888, 999).

So, 299-216-3=80.

Answer: C.

Could you elaborate on part A just a bit more please.

Thanks

Sure.

A. all digits are distinct, are numbers like 701, 702, ...987, so all numbers from 701 to 999, inclusive which have all three different digits.

Now, the first digit can take 3 values 7, 8, or 9. The second digit can take 9 values: all 10 digits minus the one we used for the first digit. Similarly the third digit can take 8 values: all 10 digits minus the one we used for the first digit and the one we used for the second digit.

So, for A. there are 3*9*8=216 numbers.
_________________

Re: Of the three digit integers greater than 700, how many have [#permalink]

Show Tags

02 Aug 2016, 08:57

2

This post received KUDOS

Here is another way of solving this question as compared to above method.

For first digit we have 3 letters to play with (7,8,9) Next two digits we can have any letter (0,1,2,3,4,5,6,7,8,9)

CASE I (ABB) 3*9*1 = 27

CASE II (BAB) 3*9*1 = 27

CASE III (BBA) 3*1*9 =27

Total of all 3 cases above: 27+27+27 = 81

subtract 1 because we want the number to be greater than 700, the above combination include 700

Final no of ways= 81-1=80
_________________

The only time you can lose is when you give up. Try hard and you will suceed. Thanks = Kudos. Kudos are appreciated

http://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html When you post a question Pls. Provide its source & TAG your questions Avoid posting from unreliable sources.

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...