Last visit was: 18 May 2024, 08:48 It is currently 18 May 2024, 08:48
Toolkit
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.

# How many even 3 digit integers greater than 700 with distinct non zero

SORT BY:
Tags:
Show Tags
Hide Tags
Senior Manager
Joined: 31 Aug 2004
Posts: 289
Own Kudos [?]: 346 [304]
Given Kudos: 0
Math Expert
Joined: 02 Sep 2009
Posts: 93334
Own Kudos [?]: 624711 [194]
Given Kudos: 81899
Senior Manager
Joined: 20 Jul 2004
Posts: 318
Own Kudos [?]: 364 [62]
Given Kudos: 0
Senior Manager
Joined: 29 Mar 2012
Posts: 267
Own Kudos [?]: 1508 [45]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
30
Kudos
15
Bookmarks
Hi,

My soltuion is as follows:
You have to find the even (3 digit) numbers greater than 700, with non-zero digits as well as distict digits. So, available digits would be (1 to 9)
Even numbers starting with 7 = 1*7*4 = 28 (Hundredth digit is 7 - so, only 1 choice, unit digit can be (2, 4, 6, 8). Now tens digit will not be 7 & a digit chosen at units place - 7 possibilities)
Even numbers starting with 8 = 1*7*3 = 21
Even numbers starting with 9 = 1*7*4 = 28
Total numbers = 77

Regards,
Director
Joined: 16 Jun 2004
Posts: 510
Own Kudos [?]: 159 [23]
Given Kudos: 0
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
17
Kudos
6
Bookmarks
1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

Manager
Joined: 02 Apr 2004
Posts: 106
Own Kudos [?]: 35 [21]
Given Kudos: 0
Location: Utrecht
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
18
Kudos
2
Bookmarks
I have C as well.

the only hundred digit that are possible are 7, 8 and 9 (3)
the only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 (9)
the only unit digit that are possible are 2, 4, 6, 8 (4)

what I did is: 3 * 9 * 4 = 108

With a lot of practise these questions are better to understand.

Regards,

Alex
General Discussion
Intern
Joined: 31 Aug 2004
Posts: 9
Own Kudos [?]: 2 [2]
Given Kudos: 0
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
2
Bookmarks
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?
A 729
B 243
C 108
D 88
E 77

I agree with ur explanation but using another method i m getting D...plz tell me wat is going wrong

the no.s are even, distinct n each place has a non-zero no.....

If the 1st hundreds digit is 7.......units can be (2,4,6,8)...i.e 4 possibilities....n tens can be any of the remaining 8 digits....(7 is in hundredes n one even no. occupies tens place)....so total options 8*4=32

for 8 in hundreds........8*3 options.....coz 8 in hundreds is already an even no.....so units can be any of the 3 remaining even....so that gives 24

for 9 in hundreds...again 8*4=32

so 32+24+32=88.........

although it seems long and may not seem to be a GMAT question which takes less time........it took no more than 30 sec to solve....

why this difference in answers....plz clarify!!!!
Senior Manager
Joined: 31 Aug 2004
Posts: 289
Own Kudos [?]: 346 [16]
Given Kudos: 0
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
13
Kudos
3
Bookmarks
OA is E

Please note that these 3 digits have to be distinct so 771 for instance is not valid

It took me 10 minutes after 2 mistakes :

even integers in the range : 149
0 as last digit (including 00) = 29
0 as tens digit = 12 (702 704 706 708, ...)
Distinct digits :
7>9 : hundreds digit
1>9 : tens digit
2- 4-6-8 :units digit

=31

Result = 149-29-12-31=77
Math Expert
Joined: 02 Sep 2009
Posts: 93334
Own Kudos [?]: 624711 [5]
Given Kudos: 81899
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
2
Kudos
3
Bookmarks
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html
Intern
Joined: 03 Dec 2012
Status:Yes. It was I who let the dogs out.
Posts: 36
Own Kudos [?]: 309 [0]
Given Kudos: 27
H: B
GMAT Date: 08-31-2013
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
venksune wrote:
1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

I am clear on the explanation provided by Bunuel for this question. However I am trying to understand the original answer provided using Arithmetic Progression. Where does the ( n-1)*2 come from ? Please explain and clarify.

Posted from GMAT ToolKit
Tutor
Joined: 16 Oct 2010
Posts: 14891
Own Kudos [?]: 65346 [3]
Given Kudos: 431
Location: Pune, India
How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
2
Kudos
1
Bookmarks
hb wrote:
venksune wrote:
1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

I am clear on the explanation provided by Bunuel for this question. However I am trying to understand the original answer provided using Arithmetic Progression. Where does the ( n-1)*2 come from ? Please explain and clarify.

Posted from GMAT ToolKit

First note that the explanation you are referring to has incorrect answer. Also, this method of solving is extremely treacherous - there are far too many limitations on the acceptable numbers to use this method. You must follow Bunuel's method for this question.

As for (n - 1)*2, we are given this particular sequence: 702, 704, 706, ..... 996, 998
We need to find how many numbers there are here. One way of doing that is using arithmetic progression concept.
nth term = First term + (n - 1)*Common difference
(last term) 998 = 702 + (n - 1)*2
Here, n will be the total number of terms.

Originally posted by KarishmaB on 24 Jul 2013, 22:46.
Last edited by KarishmaB on 11 Oct 2022, 00:28, edited 1 time in total.
Intern
Joined: 21 Aug 2013
Posts: 4
Own Kudos [?]: 16 [7]
Given Kudos: 6
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
6
Kudos
1
Bookmarks
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

Attachments

File comment: The correct answer is E (see the attached file)

ABC.png [ 45.46 KiB | Viewed 94552 times ]

Intern
Joined: 17 Feb 2013
Posts: 7
Own Kudos [?]: 30 [0]
Given Kudos: 2
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
How I was able to guess the right answer:

Find no. of even 3-digit #'s greater than 700:

(1000-702)/2+1 = 150

Subtract numbers with 0 in tens place (Excluding multiples of 100):

I.e. any even number ending in 02-08. 4*3 = 12

So 150-12 = 138

Subtract numbers with 0 in units place (again, excluding multiples of 100)

I.E. 710, 720, 730, etc... 9*3 = 27

So 138-27 = 111

Subtract multiples of 100

I.e. 800,900,100

So 111-3= 108

Subtract even numbers with idential digits in the tens and units place

I.e. 722, 744, 766, 788. 4*3 = 12

So 108-12 = 96

Subtract even numbers with identical digits in hundreds and tens place

I.e. 772, 774, 882, 884, etc...4*3 = 12

96-12=88

Subtract numbers with 3 identical digits

The only one in the set is 888.

88-1= 87. Once I got to this point I stopped, because the only remaining answer choice that could possibly be right is E (77). To finish the calculation, you subtract the no. of even numbers with identical digits in the hundreds and ones place (i.e. 808, 818, 828, etc...) and you get 77.

Maybe not the ideal way to solve the problem but it's how I worked through it while never having tried a problem like this before. Took me about 4 minutes.
Manager
Joined: 03 May 2013
Posts: 94
Own Kudos [?]: 50 [1]
Given Kudos: 114
Location: India
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
1
Kudos
Alt solution

lets consider only 2,4,6 as unit digit as of now
--- (ways to fill unit 3, tens 7, hundred 3) +
now consider only 8 as unit digit
--- (ways to fill unit 1, tens 7, hundred 2) = 63+14 = 77
Retired Moderator
Joined: 10 May 2010
Posts: 796
Own Kudos [?]: 628 [4]
Given Kudos: 192
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
3
Kudos
1
Bookmarks
XYZ

No of ways to enter hundreds digit = 3 ( 7, 8, 9 ; Because the number has to be more than 700)

No of ways to enter units digit = 2, 4, 6, 8 (We cannot enter 0 as all digits are non zero, and the numbers are to be even so 1, 3, 5, 7, 9 are out)

No of ways to enter tenth digit = 10 - 3 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

Hence total number of digits = $$3 * 7 * 4$$ = 84.

Wait if we have 8 on the units and hundreds place then we are repeating digits. Hence we must remove them as per our initial conditions that all digits are distinct.

No of such numbers

No of ways to enter hundreds digit = 1 (As were considering 8 in units and hundreds place)

No of ways to enter tenth digit = 7 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

No of ways to enter tenth digit = 1 (As were considering 8 in units and hundreds place)

No of such numbers = $$1 * 7 * 1 = 7$$

Answer = $$84 - 7 = 77$$
Manager
Joined: 28 Sep 2013
Posts: 60
Own Kudos [?]: 45 [0]
Given Kudos: 45
Location: United States (NC)
Concentration: Operations, Technology
WE:Information Technology (Consulting)
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
Bunuel wrote:
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Hope it's clear.

Hi Bunuel,

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used).

Here what about 2 as second digit. For eg: 722. In this case second digit can't take 2. How are these numbers (such as 744,766,788,922,944) removed from 2*4*7 = 56 numbers?

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used).

Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?
Math Expert
Joined: 02 Sep 2009
Posts: 93334
Own Kudos [?]: 624711 [0]
Given Kudos: 81899
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
thoufique wrote:
Bunuel wrote:
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Hope it's clear.

Hi Bunuel,

Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?

We are told that X, Y and Z must be distinct non zero digits, thus XYZ cannot be 822 or 844.

As for your other question: 9 minus two digits we already used means that for Y we cannot use a digit we used for X and a digit we used for Z, so we can use only 9-2=7 values of it.

Hope it's clear.
Intern
Joined: 27 Sep 2012
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
hence "Non-distinct zero digits" shall not be eliminated. (801,800 are all valid.)???
Tutor
Joined: 16 Oct 2010
Posts: 14891
Own Kudos [?]: 65346 [0]
Given Kudos: 431
Location: Pune, India
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
guruseen wrote:
hence "Non-distinct zero digits" shall not be eliminated. (801,800 are all valid.)???

The intent of the question is that the digits are distinct and they are non-zero.
Intern
Joined: 18 Mar 2015
Posts: 20
Own Kudos [?]: 24 [2]
Given Kudos: 1
Location: India
Concentration: Technology, Entrepreneurship
GMAT 1: 730 Q50 V38
GPA: 4
WE:Engineering (Computer Software)
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
1
Kudos
1
Bookmarks
Hi,

How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

Now, we have to find 3 digit[Distinct non-zero] even integers (Last digit 2,4,6 or 8) greater than 700 (i.e., first digit has to be 7,8 or 9).

So,
1.) No. of integers greater than 700 & less than 799, satisfying above mentioned conditions are:
==> 7 _ _
==> 7 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

2.) No. of integers greater than 800 & less than 899, satisfying above mentioned conditions are:
==> 8 _ _
==> 8 _ (2,4,6)
==> 1 * 3 * 7 = 21 (3 digits--2,4 or 6 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

3.) No. of integers greater than 900 & less than 999, satisfying above mentioned conditions are:
==> 9 _ _
==> 9 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

So, adding them all we get total integers as : 28 + 21 + 21 = 77.
So, (E).

I hope that the explanation is clear.
Re: How many even 3 digit integers greater than 700 with distinct non zero [#permalink]
1   2
Moderators:
Math Expert
93334 posts
Senior Moderator - Masters Forum
3137 posts