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 600, how many have [#permalink]
11 Feb 2012, 16:04
11
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
85% (hard)
Question Stats:
47% (02:36) correct
53% (01:40) wrong based on 201 sessions
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions?
Re: How many integers? [#permalink]
11 Feb 2012, 16:10
10
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
enigma123 wrote:
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions?
# of three-digit integers greater than 600 is 399 (999-600);
Numbers with all distinct digits: 4*9*8=288. First digit can take 4 values: 6, 7, 8 or 9, second digit can take 9 values (10 minus the one we used for the first digit) and third digit can tale 8 values (10 minus 2 digits we've already used for the first two digits);
Numbers greater than 600 which have all alike digits: 4 (666, 777, 888, 999);
{Total}-{all distinct}-{all alike}={two alike, one different} --> 399-288-4=107.
Re: Of the three-digit integers greater than 600, how many have [#permalink]
04 Oct 2012, 13:05
If the problem asks for numbers greater than 600, don't we have to start counting from 601 forward? I think we don't, but just double-checking to make sure.... thanks!
Re: Of the three-digit integers greater than 600, how many have [#permalink]
04 Oct 2012, 14:40
enigma123 wrote:
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions?
The question is about counting/combinations.
The numbers can be of the form \(AAB, \,ABA,\) or \(ABB\) where \(A\) and \(B\) are different digits and \(A\geq6\). There will be \(4\cdot9\cdot3=108\) possibilities, but we have to subtract \(1\) for the number \(600\) (which is obtained for \(A = 6\) and \(B = 0\)). Therefore, total number of possibilities \(108 - 1= 107.\)
Answer D. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: Of the three-digit integers greater than 600, how many have [#permalink]
29 Oct 2012, 15:04
Bunuel, thanks for the explanation. Mine looks more complicated, but could you please check if my solution is correct.
1) yxx - first digit can have 4 values, second - 9, third - 1 = 36 2) yyx - first digit - 4, second - 1, third - 9 = 36 3) yxy - first digit - 4, second 9, third - 1 = 36
1)+2)+3) = 108 - 1*= 107 1* - we should exclude 600
Re: Of the three-digit integers greater than 600, how many have [#permalink]
28 Dec 2013, 08:23
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Of the three-digit integers greater than 600, how many have [#permalink]
30 Dec 2014, 10:42
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Of the three-digit integers greater than 600, how many have [#permalink]
30 Dec 2014, 20:54
1
This post received KUDOS
Expert's post
Hi All,
In these types of questions, the real issue is thoroughness - make sure that you're not "missing" any of the possibilities and make sure that you're not "counting" a possibility that should NOT be counted (or accidentally counting a possibility more than once). Your ability to pattern-match will help speed you up.
With the limitations posed by this question, we COULD break the numbers down into smaller groups and then total up all of these smaller numbers (it's a slightly longer way to do things, but if you don't immediately see the more complex calculations, you can still get to the correct answer with a bit of "hand math").
Let's start with making the first 2 digits the same... 66_ 77_ 88_ 99_
Since the third digit has to be DIFFERENT from the matching pair, we have 9 options for each of the 4 groups above (you CAN'T count 666, 777, 888 or 999 - the numbers don't fit the restrictions).
Total of this group = 36
Next, let's make the first and third digits the same... 6_6 7_7 8_8 9_9
Here, we have a similar situation to the one we had above; we have 9 options for each of the 4 groups (you CAN'T count 666, 777, 888 or 999).
Total of this group = 36
Finally, let's make the second and third digits the same (I'll refer to those digits with the variable X)... 6XX 7XX 8XX 9XX
In this grouping, we have 1 "catch" - X can be any digit, BUT the number 600 is NOT permissible, since the prompt tells us for numbers GREATER THAN 600.
So, 6XX has 8 possibilities (you CAN'T count 600 or 666) 7XX, 8XX and 9XX have 9 possibilities each (you CAN'T count 777, 888 or 999)
Re: Of the three-digit integers greater than 600, how many have [#permalink]
10 Jan 2015, 13:09
Many very good approaches here! Good job everyone!
I did it in the amature's way:
So, I started calculating for 600, like this: 606 611 616 622 - 633 - 644 - 655 - 677 - 688 - 699 626 - 636 - 646 - 656 - 676 - 686 - 696
So, from here I only did 17*4= 68. By 4 because we are interested in 600, 700, 800 and 900.
Then I added the numbers after 660, that I had left out so that I wouldn't get confused: 660, 661, 662, 663, 664, 665, 667, 668, 669. This is 9*4= 36.
Adding 68+36= 104.
I saw that I was missing 3 from answer D. So, I realised that I didn't add the 600 (not allowed), 700, 800, 900.
So, 104+3 = 107.
It took a little more that 2 minutes, but again less than 2 1/2... You can't learn everything at the same time (unfortunately), so this solution kept me satisfied...
Re: Of the three-digit integers greater than 600, how many have [#permalink]
26 Sep 2015, 08:54
Bunuel wrote:
enigma123 wrote:
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72
I always struggle to solve these. What is the concept behind solving these questions?
# of three-digit integers greater than 600 is 399 (999-600);
Numbers with all distinct digits: 4*9*8=288. First digit can take 4 values: 6, 7, 8 or 9, second digit can take 9 values (10 minus the one we used for the first digit) and third digit can tale 8 values (10 minus 2 digits we've already used for the first two digits);
Numbers greater than 600 which have all alike digits: 4 (666, 777, 888, 999);
{Total}-{all distinct}-{all alike}={two alike, one different} --> 399-288-4=107.
Answer: D.
Hi Bunuel,
Can you please tell me what is wrong with my approach? Here is the way that i did:
The hundreds digit can be 6,7,8 or 9 => 4C1 Pick one more value for either tens or unit digit: 9C1 Choose 1 of 2 values above to be the repeated digits: 2C1 The permutation of the 3-digit integer: 3!/2!
Of the three-digit integers greater than 600, how many have [#permalink]
26 Sep 2015, 13:56
Expert's post
Hi vihavivi,
The big error in your calculation is in how you deal with the 'permutation'...
You described it in this way:
"The permutation of the 3-digit integer: 3!/2!"
However, this calculation includes options that are NOT allowed. For example, if you have the digits 4, 4 and 6, you could have 3 possible values: 446, 464 and 644. However, two of those values (446 and 464) are NOT greater than 600, so they should not be included.
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...