Author 
Message 
Intern
Joined: 08 Mar 2009
Posts: 1

10
This post received KUDOS
9
This post was BOOKMARKED
How many odd threedigit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests  hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)5= 455= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4)  4 = 36  4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!!



Intern
Joined: 10 Sep 2008
Posts: 36

Hello, I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd threedigit integers greater than 800 are there such that all their digits are different? 72= Answer I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you.



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: M25 #20 [#permalink]
Show Tags
18 Mar 2009, 22:00
dczuchta wrote: Hello, I know this answer has been recently discussed but that explanation confuses me even more. ____________________________________________________________________________________ How many odd threedigit integers greater than 800 are there such that all their digits are different? 72= Answer I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you. First place = 8 or 9 = 2 Second place = 0  7 and either 8 or 9 = 8 Third place = 1, 3, 5, 7 and 9 but there could be odd integer in first or second place. so; 1. if 8 in first place and 4 even integers in second place = 1x4x5 = 20 2. if 8 in first place and one of 5 odds in second place and one of remaining 4 odds in third place = 1x5x4 = 20 3. if 9 in first place and one of 5 evens in second place and one of 4 odds in third place = 1x5x4 = 20 4. if 9 in first place and one of 4 odds in second place and one of remaining 3 odds in third place = 1x4x3 = 12 total = 72.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Intern
Joined: 28 Mar 2009
Posts: 23

Re: M25#20 [#permalink]
Show Tags
23 May 2009, 18:34
14
This post received KUDOS
I used the following method during my test:
If 8 is first digit, there are 5 ways to fill third digit (1,3,5,7 or 9) and 8 ways to fill 2nd digit: 5x8 = 40
If 9 is first digit, there are 4 ways to fill 3rd digit (1,3,5 or 7) and 8 ways to fill 2nd digit: 4x8 = 32
40+32 = 72



Intern
Joined: 20 Jun 2009
Posts: 8

Re: M25#20 [#permalink]
Show Tags
28 Jun 2009, 14:19
Hello, Can you please help me understand as how there are 8 digits in the middle  If first can be 8 then middle is 0,1,2,3,4,5,6,7,9  which totals to 9. Can you please help



Intern
Joined: 16 Feb 2006
Posts: 30
Location: ZURICH

Re: M25#20 [#permalink]
Show Tags
29 Jun 2010, 07:27
If first is 8 and last can be either of 3,5,7,9 (odd numbers only) so in middle it will have only 8 numbers left to be filled out of 09 From your choices above (then middle is 0,1,2,3,4,5,6,7,9  which totals to 9), you have taken all the 4 odd numbers (3,5,7,9) at same time. Which cannot be possible.
_________________
TRY N TRY UNTIL U SUCCEED



Manager
Affiliations: NCC,SAE,YHIA
Joined: 04 May 2010
Posts: 51
Location: Mumbai , India
WE 1: 3 years international sales & mktgprojects

Re: M25#20 [#permalink]
Show Tags
29 Jun 2010, 09:02
Constraints : 1. Three digit number 2. Starting with 8 or 9 3. No repetition 4. Odd When 8 is first digit: 1. Second number even (0,2,4,6), third number (1,3,5,7,9) total cases (4*5 =20) 2. Second number odd (1,3,5,7,9), third number (odd numbers 1=4)  total cases (5*4=20) When 9 is first digit: 3. Second number even (0,2,4,6,8), third number odd (1,3,5,7) total cases (5*4 = 20) 4. Second number is odd (1,3,5,7), third number odd (odd numbers 1 =3)  total cases (4*3 = 12) Total number of possible arrangements = 20+20+20+12 = 72 This is lengthy procedure and is risky because you may miss one case & answer will be completely different , IMO direct application of P&C theory will be better approach under time constraints .
_________________
Sun TzuVictorious warriors win first and then go to war, while defeated warriors go to war first and then seek to win.



Intern
Joined: 26 May 2010
Posts: 1

Re: M25#20 [#permalink]
Show Tags
29 Jun 2010, 09:55
3
This post received KUDOS
8 _ _ You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,9}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72
9 _ _ You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,8}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72
But you only need odd numbers so (72 + 72)/2 = 72. Answer is C.



Director
Joined: 21 Dec 2009
Posts: 583
Concentration: Entrepreneurship, Finance

Re: M25#20 [#permalink]
Show Tags
30 Jun 2010, 07:20
4
This post received KUDOS
I hope my approach is correct anyway; Let N = total no of digits with different individual digits N= 2x9x8 = 144 no of even = no of odds = 72. (D)
_________________
KUDOS me if you feel my contribution has helped you.



Intern
Joined: 26 Apr 2010
Posts: 5

Re: M25#20 [#permalink]
Show Tags
03 Jul 2010, 16:52
1
This post received KUDOS
I just used slot method: 2 choices for the hundreds digit (8 or 9), 9 choices for the tens digit (0,1,2,3,4,5,6,7, 8 or 9), and 4 choices for the tens digit (1 or 9, 3 or 9, 5 or 9 , 7 or 9) = 2 x 9 x 4 = 72.



Manager
Joined: 20 Dec 2010
Posts: 247
Schools: UNC Duke Kellogg

Re: M25#20 [#permalink]
Show Tags
02 Jul 2011, 07:40
Between 800 and 899  (1)*(8)*(5) = 40 (units place  1,3,5,7,9) Between 900 and 999  (1)*(8)*(4) = 32 (units place  1,3,5,7)
72



Manager
Joined: 11 Jul 2009
Posts: 167
WE: Design (Computer Software)

Re: M25#20 [#permalink]
Show Tags
03 Jul 2011, 06:11
first digit can be selected from (8,9) in 2 ways second distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 9 ways since one digit is already selected Third distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 8 ways since two digits are already selected . So total number of ways digits can be slected is 2*8*9 But we need to consider odd number of digits so (2*8*9)/2=72 SO...ANSWER is C
_________________
Kaustubh



Math Expert
Joined: 02 Sep 2009
Posts: 39656

Re: M25#20 [#permalink]
Show Tags
26 Mar 2012, 08:20
5
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
k8llykh wrote: How many odd threedigit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests  hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)5= 455= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4)  4 = 36  4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!! In the range 800  900:1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900  999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*4*8 = 32. Total: 40+32 = 72. Answer: C.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 15 Sep 2009
Posts: 265

Re: M25#20 [#permalink]
Show Tags
05 Jul 2012, 05:21
1
This post received KUDOS
Slot method. If hundreds digit is 8, there are 5 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*5=40 If hundreds digit is 9, there are 4 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*4=32 Total= 72 Answer is C. Cheers, Der alte Fritz.
_________________
+1 Kudos me  I'm half Irish, half Prussian.



Intern
Joined: 03 Feb 2012
Posts: 7

Re: M25#20 [#permalink]
Show Tags
05 Jul 2012, 10:52
1
This post received KUDOS
I just culculated (2*9*8)/2 = 72
(2*9*8) to get all nummers with all their digits different from 800 to 999 and than divided it by 2 to get only the odd numbers



Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 229

Re: M25#20 [#permalink]
Show Tags
08 Jul 2012, 09:58
k8llykh wrote: How many odd threedigit integers greater than 800 are there such that all their digits are different? (A) 40 (B) 56 (C) 72 (D) 81 (E) 104 Source: GMAT Club Tests  hardest GMAT questions Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker. For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part: For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)5= 455= 40 Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4)  4 = 36  4 = 32 possibilites. Add the 2 together, and you get 40+32=72. The answer! Hope that helps!! good problem...every time i get such problems as these, i try to solve in conservative(old manual) method.....i should try to find the right method which is also the faster method....ANSWER is 72 and no doubt abt it.
_________________
Regards, Harsha
Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat
Satyameva Jayate  Truth alone triumphs



Intern
Joined: 08 Jun 2012
Posts: 2

Re: M25#20 [#permalink]
Show Tags
23 Jul 2012, 12:05
Bunuel wrote: k8llykh wrote: How many odd threedigit integers greater than 800 are there such that all their digits are different?
(A) 40 (B) 56 (C) 72 (D) 81 (E) 104
Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.
For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:
For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)5= 455= 40
Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4)  4 = 36  4 = 32 possibilites.
Add the 2 together, and you get 40+32=72. The answer!
Hope that helps!! In the range 800  900:1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900  999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*4*8 = 32. Total: 40+32 = 72. Answer: C. Hi Geek, I wonder how u r solving all questions with proper explanations... HAts off for tat... How u r solving by these strategy? Does any math strategy is there for this types of problem and could you pls help me to read those things.... In the range 800  900:1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*5*8 = 40. In the range 900  999:1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: 1*4*8 = 32. Cheers, Sharbu



Manager
Joined: 14 Jun 2012
Posts: 65

Re: M25#20 [#permalink]
Show Tags
31 Aug 2012, 20:49
I am kicking myself for getting this problem wrong on account of a really bad silly mistake. I missed reading the "odd" in the question stem and ended up with getting an answer not in the given choices. The good thing is that I knew how to approach it using the combinatrics method but on account of poor reading of the question stem I ended up choosing a wrong answer. Bunuel's way of selecting the 3rd digit after selecting the first (8/9) was particularly helpful and quick. Missed this question only because of not reading with full concentration. Real bad mistake.
_________________
My attempt to capture my BSchool Journey in a Blog : tranquilnomadgmat.blogspot.com
There are no shortcuts to any place worth going.



Intern
Joined: 04 Jul 2013
Posts: 11
GMAT Date: 10082013

Re: M25#20 [#permalink]
Show Tags
06 Jul 2013, 00:30
2
This post received KUDOS
Hi everyone, most fastest way to solve this question is
8_ _, different odd no. will be 8 _ _. first digit is 8, for odd number third digit will be (1,3,5,7,9) i.e. 5 possibility 2nd digit we can fill from (09), since we are looking for different digit so choice for 2nd digit will 10(09)  1st digit2nd digit with 8 possibility 8 _(8) _(5) = 8*5 = 40
similarly for 9 _ _
for 3rd digit choices are {1,3,5,7} for 2nd digit choices are 10  1st digit3rd digit = 8
so 8*4=32
so total = 72 different odd digits
Another simple wat to solve is b/w 800900 we have 50 odd no. now we will substract the odd no. like {11,33,55,77,99}= 5 and 881,83,85,87,89 =5 i.e. 5055 =40
b/w 900999 we have 50 odd no substract {11,33,55,77,99}=5 and 991,93,95,97,99 = 5 and like 919,929 .....989=8 50558 = 32 so different odd number will be 40+32 = 72










