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:
Ways to pick number = 1 (fourth digit - is 0) * 9 (third digit - all but 0) * 8 (second digit - all but 0 and third digit) * 7 (first digit - all but 0 and second & third digits)
Number with "0" in tens or hundreds place
Ways to pick number = 4 (fourth digit - {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick non-zero of the hundreds/tens digit) * 7 (ways to pick the thousands digit)
Number with no "0"
Ways to pick number = 4 (units digit - no 0 allowed) * 8 (tens digit - one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit - three numbers picked and 0 not allowed)
Total
9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296
I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted _________________
How many 4 digit even numbers do not use any digit more than once
1720
2160
2240
2460
2520
I think OA for this one is wrong.
Even 4-digit codes (starting with 0 is allowed) with 4 distinct digits {\(abcd\)} = 10*9*8*7/2=2520 (divided by 2 as 4-digit codes are half even half odd);
Even 4-digit codes starting with zero with 4 distinct digits (as we used zero for the first digit so there are total of 10-1=9 digits left to use) {\(0bcd\)}: {4 choices for \(d\): 2, 4, 6 or 8, zero is out as we used it for the first digit} * {9-1=8 choices for \(b\)} * {8-1=7 choices for \(c\)} = 4*8*7 = 224;
{Even 4-digit codes with 4 distinct digits} - {Even 4-digit codes with 4 distinct digits starting with zero} = {Even 4-digit numbers with 4 distinct digits} = 2520 - 224 = 2296. _________________
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number" _________________
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number"
The problem with your solution is that when we have 0 for D then the choices for A, B, and C are 9, 8 and 7 respectively and not 8, 8, and 7. So if we do the way you are doing we would have:
ABCD: If D is 0, so 1 choice for D, then choices for other letters would be: A-9, B-8, and C-7 --> 1*9*8*7=504; If D is 2, 4, 6, or 8, so 4 choices for D, then choices for other letters would be: A-8, B-8, and C-7 --> 4*8*8*7=1792;
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number"
Since your solution does not include the numbers that start with 0 (0BCD), you should be getting the same answer as Bunuel and Shrouded.
I have spend about 15 mins on this, was trying to figure out
(1) what were you doing wrong
(2) why can't we simply start from the thousandth digit and move to the right.
I believe the way Bunuel wrote is a little simpler & quicker, as long as you understand that you need to calculate all 4 digit numbers and then subtract the one's that start with zero (or an alternative that I will provide below).
(1) what I think is wrong in your approach (you are getting slightly fewer ways ) -
A B C D ( thousand , hundreds, tens, units)
D can be 0 2 4 6 8 ( any of the 5 digits )
A can be anything except (D or 0) so 8 possibilities - while doing this you have removed extra possibilities. What happens when D itself is 0 ? so you see, some of the A=0 options have already been removed when you said that A can not be same as D.
C can be anything except A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
To compensate for the above, you can do this (this is a correction to your approach or can be an alternative to Bunuel's method) -
D = 4 ways (2,4,6,8 only, 0 has been removed) A = 8 ways ( can't be equal to D and 0) B = 8 (can't be equal to A & D) C = 7 (can't be equal to A, B & D)
Gives = 1792 (4 digits numbers without repetition, but without the ones that end with 0)
D = 1 (can only be equal to 0) A = 9 (anything but D, which also covers cases where A=0) B = 8 (anything but A &D) C = 7 (anything but A, B & D)
Gives = 504 (4 digits numbers without repetition that end with 0)
Total = 1792 + 504 = 2296
Brunel - (2) why can't we simply start from the thousandth digit and move to the right - is this because we need to take the possibilities for the restrictive digits first, such as the ones digit that can only be even ?
Edit - realized after posting that Bunuel had posted a reply. _________________
Brunel - (2) why can't we simply start from the thousandth digit and move to the right - is this because we need to take the possibilities for the restrictive digits first, such as the ones digit that can only be even ?
Yes, D determines the # of choices for the rest of the digits. _________________
A can be anything except (D or 0) so 8 possibilities
C can be anything execpt A and B so 8 possibilities
B can be anything execpt ( A D C ) so 7 possibilities
total ways are 8 * 7 * 8 * 5 = 2240
The only thing different from this logic and bunuel's way is that this has to be a number that means it cannot start with a 0 as the Q is asking 4 digit even "number"
The problem with your solution is that when we have 0 for D then the choices for A, B, and C are 9, 8 and 7 respectively and not 8, 8, and 7. So if we do the way you are doing we would have:
ABCD: If D is 0, so 1 choice for D, then choices for other letters would be: A-9, B-8, and C-7 --> 1*9*8*7=504; If D is 2, 4, 6, or 8, so 4 choices for D, then choices for other letters would be: A-8, B-8, and C-7 --> 4*8*8*7=1792;
Total: 504+1792=2296.
It's better to divide D in 2 types (0 and others) i had wrong answer cause I did not separate it then I got 5*9*8*7 for D, C ,B, A.
Somebody please explain me what is wrong woth the below method.
ABCD - 4digit #
Choices for A = 9 (0 is excluded) Choices for B = 9 (one digit used for A and 0 is now included) Choices for C = 8 Choices for D = 7
# of 4 digit #s with distinct digits = 9*9*8*7 : note that it would contain equal # of even and odd numbers hence # of EVEN #s with distinct digits = 9*9*8*7 / 2 = 2268
Somebody please explain me what is wrong woth the below method.
ABCD - 4digit #
Choices for A = 9 (0 is excluded) Choices for B = 9 (one digit used for A and 0 is now included) Choices for C = 8 Choices for D = 7
# of 4 digit #s with distinct digits = 9*9*8*7 : note that it would contain equal # of even and odd numbers hence # of EVEN #s with distinct digits = 9*9*8*7 / 2 = 2268
I think that the red part is not correct.
Consider another example: there are 90 2-digit numbers, out of which 81 have distinct digits (minus 11, 22, 33, 44, 55, 66, 77, 88, 99 total of 9 numbers). Now, if we take your approach then even 2-digit numbers with distinct digits should be half of all 2-digit numbers with distinct digits - 81/2=not an integer.
Actual # is 90/2=45 (even 2-digit numbers) minus 4 (even 2-digt number with same digits: 22, 44, 66, 88) = 41.
You can't have zero as the first digit of a 4-digit number as in this case it'll become 3-digit number, so you don't have 10 choice for the first digit. Please see the solutions of this problem in the previous posts. _________________
Re: Counting numbers [#permalink]
24 Jun 2012, 02:23
I don't know about others but the very first solution to this problem confused me
Why to start with a zero when it says four digit number , a zero at the beginning will make it a 3 digit no.
here is a more logical approach
4 digit even numbers so first digit cannot be a zero
from 0 to 9 we have 10 digits for each of the four places, please note digits cannot be repeated so each digit has to be distinct.
even numbers will end in a 0 ,2,4,6,8 right?
so lets see if the last digit is zero then
for first digit we have 1 to 9 options{ cannot have 0 here} , second digit we have 8 options , 3 digit we have 7 options and of course units digit i .e has only one option ,0
so 9*8*7*1
same way when last digit is 2
then 8*8*7*1 { 8 options for the thousands place because it cannot be 2 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 2 }
same way when last digit is 4
then 8*8*7*1 { 8 options for the thousands place because it cannot be 4 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 4 }
when last digit is 6
then 8*8*7*1 { 8 options for the thousands place because it cannot be 6 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 6 }
when last digit is 8 then then 8*8*7*1 { 8 options for the thousands place because it cannot be 8 and 0 } { 8 options for the hundredth place because 2 digits have already been reserved } { 7 options for the tens place because 3 digits have already been reserved} { 1 options for the units place because this case is for when unit digit is 8 }
so (8*8*7*1)*4 + 9*8*7*1 = 1792 + 504 = 2296
of course explanation is long but in actual exam all I would do is,
Re: Counting numbers [#permalink]
24 Jun 2012, 03:13
shrouded1 wrote:
Consider the following cases
Number ending in "0"
Ways to pick number = 1 (fourth digit - is 0) * 9 (third digit - all but 0) * 8 (second digit - all but 0 and third digit) * 7 (first digit - all but 0 and second & third digits)
Number with "0" in tens or hundreds place
Ways to pick number = 4 (fourth digit - {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick non-zero of the hundreds/tens digit) * 7 (ways to pick the thousands digit)
Number with no "0"
Ways to pick number = 4 (units digit - no 0 allowed) * 8 (tens digit - one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit - three numbers picked and 0 not allowed)
Total
9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296
I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted
I also used the same calculation method...cant figure out my mistake!!!
Re: Counting numbers [#permalink]
24 Jun 2012, 03:16
Expert's post
aks220488 wrote:
shrouded1 wrote:
Consider the following cases
Number ending in "0"
Ways to pick number = 1 (fourth digit - is 0) * 9 (third digit - all but 0) * 8 (second digit - all but 0 and third digit) * 7 (first digit - all but 0 and second & third digits)
Number with "0" in tens or hundreds place
Ways to pick number = 4 (fourth digit - {2,4,6,8}) * 2 (either hundreds of tens digit is 0) * 8 (number of ways to pick non-zero of the hundreds/tens digit) * 7 (ways to pick the thousands digit)
Number with no "0"
Ways to pick number = 4 (units digit - no 0 allowed) * 8 (tens digit - one picked and 0 not allowed) * 7 (two picked and 0 not allowed) * 6 (thousands digit - three numbers picked and 0 not allowed)
Total
9*8*7 + 4*2*8*7 + 4*8*7*6 = 56*(9+8+24) = 2296
I suspect this answer is different from the ones you have up there because you are considering numbers starting with 0 as well, which I have not counted
I also used the same calculation method...cant figure out my mistake!!!
2296 is a correct answer, so you've made no mistake. _________________
Re: How many 4 digit even numbers do not use any digit more than [#permalink]
01 Nov 2013, 07:35
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. _________________
How many 4 digit even numbers do not use any digit more than [#permalink]
21 Sep 2015, 01:16
We must caclulate here 5 cases for the last digit: 0,2,4,6 and 8 because only in this case is the 4 digit number even. Important points here: 1st digit cannot be 0, and we are looking for 4 ditinct digits !
Case 1 last digit is equal 0: 9 x 8 x 7 x 1 = 504 Case 2 last digit is equal 2: 8 x 8 x 7 x 1 = 448 Case 3 last digit is equal 4: 8 x 8 x 7 x 1 = 448 Case 4 last digit is equal 6: 8 x 8 x 7 x 1 = 448 Case 5 last digit is equal 8: 8 x 8 x 7 x 1 = 448 -------------------------------------------------------------------------- ∑ = 2296
Actually we have a collision of restrictions only between last and first digit for value “0”, so we must test only 2 Cases: Case 1 last digit is equal 0: 9 x 8 x 7 x 1 = 504 Case 2 last digit-any other even value: 8 x 8 x 7 x 4 = 4x448 ∑ = 2296 _________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
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...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...